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Chapter 18: Inference about One Population Mean STAT 1450 18.0 Inference about One Population Mean Connecting Chapter 18 to our Current Knowledge of Statistics βΈ We know the basics of confidence interval estimation (Chapter 14) and tests of significance (Chapter 15). Nuances that we should be aware of were also presented (Chapter 16). 18.0 Inference about One Population Mean Parameters and their Point Estimates Measure Sample Statistic and Point Estimate Population Parameter Mean π₯ ΞΌ s Ο π π Standard Deviation Proportion of Successes βΈ In the coming chapters, we will either find confidence intervals for the population parameters, or, conduct tests of significance regarding their hypothesized values . βΈ In either case, the point estimates will help us in our endeavors. 18.0 Inference about One Population Mean Inference when Ο is unknown βΈ It is unlikely that the population standard deviation Ο will be known and the population mean ΞΌ will not be known. βΈ Chapter 14 taught us that π₯ is the best point estimate of µ. Similarly, s can estimate Ο. 18.0 Inference about One Population Mean Inference when Ο is unknown βΈ In Chapter 11 when Ο was known we used π₯βπ π§=π π ο§ This statistic follows a Standard Normal Z-distribution βΈ When Ο is not known we can use s instead: π‘= π₯βπ π π ο§ This statistic is not quite Normal. It follows a t-distribution. 18.1 Conditions for Inference about a Population Mean Conditions for Inference about a Mean βΈ The conditions for inference about a mean are listed on page 437 of the text. Random sample: ο§ Do we have a random sample? ο§ If not, is the sample representative of the population? ο§ If not a representative sample, was it a randomized experiment? 18.1 Conditions for Inference about a Population Mean Conditions for Inference about a Mean Large enough population : sample ratio: ο§ Is the population of interest β₯ 20 times βnβ? The population is from a Normal Distribution. ο§ If the population is not from a Normal Distribution, then the sample size must be βlarge enoughβ with a shape similar to the Normal Distribution; then we apply the Central Limit Theorem. 18.1 Conditions for Inference about a Population Mean Standard Error βΈ When the standard deviation of a statistic is estimated from data, the result is called the standard error of the statistic. βΈ The standard error of the sample mean is π π . βΈ Now the sample standard deviation will replace Ο. 18.1 Conditions for Inference about a Population Mean Example: Standard Error βΈ A random sample of 49 students reported receiving an average of 7.2 hours of sleep nightly with a standard deviation of 1.74. What is the standard deviation of the mean? 18.1 Conditions for Inference about a Population Mean Example: Standard Error βΈ A random sample of 49 students reported receiving an average of 7.2 hours of sleep nightly with a standard deviation of 1.74. What is the standard deviation of the mean? ο§ (sample) standard deviation of 1.74 ο§ ππ‘ππππππ πππππ (π . π. ) = 1.74 49 = 0.2486 18.2 The t Distributions The t-distribution βΈ Draw an SRS of size n from a large population that has the Normal distribution with mean ΞΌ and standard deviation Ο. The one-sample t statistic π₯βπ π‘=π π has the t distribution with n β 1 degrees of freedom. 18.1 Conditions for Inference about a Population Mean Standard Error βΈ Now the sample standard deviation will replace s; allowing us to use the one-sample t statistic for confidence intervals and tests of significance. βΈ As mentioned earlier, the t-distribution is βnot quite Normal.β 18.2 The t Distributions The t Distributions βΈ Here is a plot of two t distributions (dashed) and the standard Normal distribution (solid): 18.2 The t Distributions T-distribution Compared to the Z-Distribution Similarities Differences Symmetric about 0 T has thicker tails Single-peaked & Bellshaped Varies based upon βdegrees of freedomβ 18.2 The t Distributions T-distribution Compared to the Z-Distribution βΈ Poll: The t2 curve has thicker dashes. The t9 curve has smaller dashes. Z is the solid curve. What would you anticipate happening to tdf as the degrees of freedom (df) increase? a) tdf will not be affected b) tdf will approach Z c) tdf will become further from Z. 18.2 The t Distributions T-distribution Compared to the Z-Distribution βΈ Poll: The t2 curve has thicker dashes. The t9 curve has smaller dashes. Z is the solid curve. What would you anticipate happening to tdf as the degrees of freedom (df) increase? a) tdf will not be affected b) tdf will approach Z c) tdf will become further from Z. 18.2 The t Distributions Using Table C βΈ βWhat is happening to these values as we increase the βdfβ?β 2/3. If the bottom row contains z*, and t-critical values increase as we βmove up the chart,β should we expect intervals based upon t to be larger or smaller than those based upon z? Suppose df=37, then use row for df=30. Walk through this example. 18.2 The t Distributions Using Table C Notes: 1. The t-distribution critical values decrease as the degrees of freedom (df) increase. 2. The final row includes 1.645, 1.96, & 2.576. These are βcommon confidence levelsβ & z*. 3. Confidence intervals based upon βtβ will be slightly wider than those based upon βz.β 4. Be conservative. When the exact df is not listed, βround downβ and use the closest df that does not exceed the df that is desired. 18.3 The One-sample t Confidence Interval The One-sample t Confidence Interval βΈ Draw an SRS of size n from a large population having unknown mean ΞΌ. βΈ A level C confidence interval for ΞΌ is π₯± π‘β π π βΈ where t* is the critical value for the t(n β 1) density curve with area C between β t* and t*. This interval is exact when the population distribution is Normal and is approximately correct for large n in other cases. 18.3 One-Sample t Confidence Intervals The One-sample t Confidence Interval βΈ The one-sample t confidence interval is used to estimate means. βΈ Its form is similar to previous forms of confidence intervals: estimate ± margin of error π₯ ±(1.96 or another z-score) π₯ ± π‘ β π π Introduction of Confidence Intervals π π General Form (when Ο is known) Now (s is unknown). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels βΈ Hemoglobin (Hb) levels are normally distributed, and should neither be too large, nor too small. A random sample of 11 boys from an underserved country had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Compute a 90% confidence interval for the average hemoglobin level for boys from this particular country. 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 1. Components οΌDo we have an SRS? Yes. Stated as a random sample. Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 1. Components οΌDo we have an SRS? Yes. Stated as a random sample. Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. οΌLarge enough population: sample ratio? Yes. . Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 1. Components οΌDo we have an SRS? Yes. Stated as a random sample. Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. οΌLarge enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 1. Components οΌDo we have an SRS? Yes. Stated as a random sample. Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. οΌLarge enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. οΌt-distribution? Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 1. Components οΌDo we have an SRS? Yes. Stated as a random sample. Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. οΌLarge enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. οΌt-distribution? Yes. n = 11 < 40 β¦ Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 1. Components οΌDo we have an SRS? Yes. Stated as a random sample. Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. οΌLarge enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. οΌt-distribution? Yes. n = 11 < 40 β¦ but data is approximately Normal, so we can use t-distribution. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 t*(90%, 10) = 1.812 3. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. 90% df=11-1=10 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 t*(90%, 10) = 1.812 3. 4. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. Construct the confidence interval. 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 t*(90%, 10) = 1.812 4. Interval. 11.3 ± 1.812 1.5 11 = 3. 4. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. Construct the confidence interval. 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 t*(90%, 10) = 1.812 4. Interval. 11.3 ± 1.812 1.5 11 = 11.3 ± .82 3. 4. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. Construct the confidence interval. 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 4. Interval. 11.3 ± 1.812 t*(90%, 10) = 1.812 1.5 11 3. 4. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. Construct the confidence interval. = 11.3 ± .82 = (10.48,12.12) 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 4. Interval. 11.3 ± 1.812 t*(90%, 10) = 1.812 1.5 11 3. 4. 5. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. Construct the confidence interval. *Interpret* the interval. = 11.3 ± .82 = (10.48,12.12) 18.3 One-Sample t Confidence Intervals Example: 90% CI for Hb levels 2. Components. π = 11.3, s = 1.5, n = 11 Steps for SuccessConstructing Confidence Intervals for m (s unknown). 1. 2. 3. Select t*. df =(n-1)=10 4. Interval. 11.3 ± 1.812 t*(90%, 10) = 1.812 1.5 11 3. 4. 5. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). Identify the 3 key components of the confidence interval (mean, s.d., n). Select t*. Construct the confidence interval. *Interpret* the interval. = 11.3 ± .82 = (10.48,12.12) 5. Interpret. We are 90% confident that the mean hemoglobin level for boys from this country is between 10.48 g/dL and 12.12 g/dL 18.3 One-Sample t Confidence Intervals Technology Tips β Computing Confidence Intervals (s unknown) Technology Tips β Computing Confidence Intervals (s unknown) βΈ TI-83/84: STAT ο TESTS ο TInterval ο Enter ο§ Select Stats. Enter s, π₯, n, and the confidence level. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) βΈ JMP: Enter the data. Analyze ο Distribution. ο§ βClick-and-Dragβ (the appropriate variable) into the βY, Columnsβ box. Click on OK. ο§ Click on the red upside-down triangle next to the title of the variable from the βY,Columnsβ box. Proceed to βConfidence Intervalβ -> ο§ Select the appropriate confidence level. 18.3 One-Sample t Confidence Intervals Technology Tips β Computing 90% Confidence Intervals (π unknown) βΈ TI-83/84 ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. (for this example) ο§ Inpt >> STATS ο§ π : 11.3 >> s: 1.5 ο§ C-Level : 90 ο§ Calculate (ENTER) >> n : 11 18.3 One-Sample t Confidence Intervals Technology Tips β Computing 90% Confidence Intervals (π unknown) βΈ TI-83/84 ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. (for this example) ο§ Inpt >> STATS ο§ π : 11.3 >> s: 1.5 ο§ >> n : 11 C-Level : 90 ο§ Calculate (ENTER) (10.48, 12.12) 18.3 One-Sample t Confidence Intervals Technology Tips β Computing 90% Confidence Intervals (π unknown) βΈ TI-83/84 ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. (for this example) ο§ Inpt >> STATS ο§ π : 11.3 >> s: 1.5 ο§ >> n : 20 C-Level : 90 ο§ Calculate (ENTER) (10.728, 11.88) 18.3 One-Sample t Confidence Intervals Technology Tips β Computing 90% Confidence Intervals (π unknown) βΈ TI-83/84 ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. (for this example) ο§ Inpt >> STATS ο§ π : 11.3 >> s: 1.5 ο§ C-Level : 90 ο§ Calculate (ENTER) >> n : 33 Larger n, narrower interval. (10.858, 11.742) 18.4 One-Sample t Test The One-sample t Test of Significance βΈ Draw an SRS of size n from a large population that has the Normal distribution with mean ΞΌ and standard deviation Ο. The one-sample t statistic π₯βπ π‘=π π has the t distribution with n β 1 degrees of freedom. 18.4 One-Sample t Test The One-sample t Test of Significance βΈ To test the hypothesis π»π : π = ππ , compute the one-sample t statistic π₯ β ππ π‘= π π 18.4 One-Sample t Test The One-sample t Test of Significance βΈ The p-value for a test of H0 against ο§ π»π : π < π0 is π π β€ π‘ . ο§ π»π : π > π0 is π(π β₯ π‘). ο§ π»π : π = π0 is 2π(π β₯ π‘ ). ο§ These P-values are exact if the population distribution is Normal and are approximately correct for large n in other cases. 18.4 One-Sample t Test Example: Hemoglobin levels βΈ Recall from earlier that Hb levels are normally distributed. Our original example featured a random sample of 11 boys from an underserved country that had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Is there significant evidence, at the .05 level of significance that the average Hb level for boys from this country is below 12, which results in ___________? 18.4 One-Sample t Test Example: Hemoglobin levels βΈ Recall from earlier that Hb levels are normally distributed. Our original example featured a random sample of 11 boys from an underserved country that had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Is there significant evidence, at the .05 level of significance that the average Hb level for boys from this country is below 12, which results in ___anemia________? 15.4 Tests for a Population Mean Steps for Success β Conducting Tests of Significance Steps for SuccessConducting Tests of Significance 1. Set up your Hypotheses. 2. Check your Conditions. 3. Compute the Test Statistic. 4. Compute the P-Value. 5. Make a Decision. 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. b) List all given information from the data collected. n=11, π₯ = 11.3, sd=1.5 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. b) List all given information from the data collected. n=11, π₯ = 11.3, sd=1.5 c) State the null (H0) and alternative (HA) hypotheses. 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. b) List all given information from the data collected. n=11, π₯ = 11.3, sd=1.5 c) State the null (H0) and alternative (HA) hypotheses. H0: µ = 12 Ha :µ < 12 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. b) List all given information from the data collected. n=11, π₯ = 11.3, sd=1.5 c) State the null (H0) and alternative (HA) hypotheses. d) Specify the level of significance. Ξ± =.05 H0: µ = 12 Ha :µ < 12 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. b) List all given information from the data collected. n=11, π₯ = 11.3, sd=1.5 c) State the null (H0) and alternative (HA) hypotheses. H0: µ = 12 Ha :µ < 12 d) Specify the level of significance. Ξ± =.05 e) Determine the type of test. Left-tailed Right-tailed Two-Tailed 18.4 One-Sample t Test Example: Hemoglobin levels State: Are boys from this underserved country anemic (i.e., Hb m < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country. b) List all given information from the data collected. n=11, π₯ = 11.3, sd=1.5 c) State the null (H0) and alternative (HA) hypotheses. H0: µ = 12 Ha :µ < 12 d) Specify the level of significance. Ξ± =.05 e) Determine the type of test. Left-tailed Right-tailed Two-Tailed 18.4 One-Sample t Test Example: Hemoglobin levels Plan: f) Sketch the region(s) of βextremely unlikelyβ test statistics. 18.4 One-Sample t Test Example: Hemoglobin levels Solve: 18.4 One-Sample t Test Example: Hemoglobin levels Solve: a) Check the conditions for the test you plan to use. ο§ Random Sample? Yes. Stated as a random sample. 18.4 One-Sample t Test Example: Hemoglobin levels Solve: a) Check the conditions for the test you plan to use. ο§ Random Sample? Yes. Stated as a random sample. ο§ Large enough population: sample ratio? Yes. The number of boys is arbitrarily large; therefore, N > 20*11 = 220. 18.4 One-Sample t Test Example: Hemoglobin levels Solve: a) Check the conditions for the test you plan to use. ο§ Random Sample? Yes. Stated as a random sample. ο§ Large enough population: sample ratio? Yes. The number of boys is arbitrarily large; therefore, N > 20*11 = 220. ο§ Large enough sample; Normal or t-distribution? Yes. n = 11 < 40. But data is approximately Normal, so we can use t-distribution. 18.4 One-Sample t Test Example: Hemoglobin levels Solve: b) Calculate the test statistic π₯ β π0 π‘= = π π 18.4 One-Sample t Test Example: Hemoglobin levels Solve: b) Calculate the test statistic π₯ β π0 11.3 β 12 π‘= = = β1.548 π π 1.5 11 18.4 One-Sample t Test Example: Hemoglobin levels Solve: b) Calculate the test statistic c) Determine (or approximate) the P-Value. 1.548 π₯ β π0 11.3 β 12 π‘= = = β1.548 π π 1.5 11 ο§ -1.372 > -1.548 > -1.812 P-value 18.4 One-Sample t Test Example: Hemoglobin levels Solve: b) Calculate the test statistic c) Determine (or approximate) the P-Value. 1.548 π₯ β π0 11.3 β 12 π‘= = = β1.548 π π 1.5 11 ο§ -1.372 > -1.548 > -1.812 ο§ .10 > P-value > .05 P-value 18.4 One-Sample t Test Example: Hemoglobin levels Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). 18.4 One-Sample t Test Example: Hemoglobin levels Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). 18.4 One-Sample t Test Example: Hemoglobin levels Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). Because the approximate P-value is larger than 0.05, we fail to reject the null hypothesis. 18.4 One-Sample t Test Example: Hemoglobin levels Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). Because the approximate P-value is larger than 0.05, we fail to reject the null hypothesis. b) Interpret the decision in the context of the original claim. 18.4 One-Sample t Test Example: Hemoglobin levels Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). Because the approximate P-value is larger than 0.05, we fail to reject the null hypothesis. b) Interpret the decision in the context of the original claim. There is not enough evidence (at a=.05) that, boys from this country are typically anemic (i.e., mean Hb <12 g/dl). 18.4 One-Sample t Test Technology Tips β Conducting Tests of Significance (s unknown) TI-83/84: STAT ο TESTS ο TTest ο Enter. ο§ Select Stats. Enter m0 s, π₯, n, and the confidence level. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) JMP: Enter the data. Analyze ο Distribution.βClick-and-Dragβ (the appropriate variable) into the βY, Columnsβ box. Click on OK. ο§ Click on the red upside-down triangle next to the title of the variable from the βY,Columnsβ box. ο§ Proceed to βConfidence Intervalβ -> Select the appropriate confidence level. 18.4 One-Sample t Test Example: Hemoglobin levels (n= 11, 20, and 33) βΈ Letβs now use technology to conduct the test of significance at a = .05 for the three different sample sizes (n =11, 20, and 33). We will particularly focus on the test statistic, p-value, and decision at a = .05. 18.4 One-Sample t Test Technology Tips β Conducting Tests of Significance (Ο unknown) βΈ TI-83/84. STAT ο TESTS ο TTest ο Enter. ο§ Select Stats. Enter π0 , π , π₯ and n. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) (for this example) ο§ Inpt >> STATS ο§ m0: 12 >> π : 11.3 >> s: 1.5 >> n : 11 >> m : < ο§ Calculate (ENTER) t= - 1.548 p= .076 18.4 One-Sample t Test Technology Tips β Conducting Tests of Significance (Ο unknown) βΈ TI-83/84. STAT ο TESTS ο TTest ο Enter. ο§ Select Stats. Enter π0 , π , π₯ and n. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) (for this example) ο§ Inpt >> STATS ο§ m0: 12 >> π : 11.3 >> s: 1.5 ο§ Calculate (ENTER) >> n : 11 >> m : < Fail to reject H0, .076 > .05 => p-value > a . There is not enough evidence (at a = .05) to conclude that boys from this country are typically anemic. t= - 1.548 p= .076 18.4 One-Sample t Test Technology Tips β Conducting Tests of Significance (Ο unknown) βΈ TI-83/84. STAT ο TESTS ο TTest ο Enter. ο§ Select Stats. Enter π0 , π , π₯ and n. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) (for this example) ο§ Inpt >> STATS ο§ m0: 12 >> π : 11.3 >> s: 1.5 >> n : 20 ο§ Calculate (ENTER) t= - 2.087 p= .025 >> m : < 18.4 One-Sample t Test Technology Tips β Conducting Tests of Significance (Ο unknown) βΈ TI-83/84. STAT ο TESTS ο TTest ο Enter. ο§ Select Stats. Enter π0 , π , π₯ and n. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) (for this example) ο§ Inpt >> STATS ο§ m0: 12 >> π : 11.3 >> s: 1.5 ο§ Calculate (ENTER) >> n : 20 >> m : < Reject H0, .025 < .05 => p-value < a . There is enough evidence (at a = .05) to conclude boys from this country are typically anemic. t= - 2.087 p= .025 18.4 One-Sample t Test Technology Tips β Conducting Tests of Significance (Ο unknown) βΈ TI-83/84. STAT ο TESTS ο TTest ο Enter. ο§ Select Stats. Enter π0 , π , π₯ and n. Select Calculate. ο§ (Note: Select Data when π₯ and n are not provided. Then enter the list where the data are stored.) (for this example) ο§ Inpt >> STATS ο§ m0: 12 >> π : 11.3 >> s: 1.5 ο§ Calculate (ENTER) >> n : 33 >> m : < Reject H0, .006 < .05 => p-value < a . There is enough evidence (at a = .05) to conclude boys from this country are typically anemic. t= - 2.68 p= .006 18.6 Matched Pairs t Procedures Matched Pairs t Procedures βΈ One way to demonstrate that a treatment causes an observed effect is to use a matched pairs experiment. βΈ In a matched pairs design ο§ subjects are matched in pairs and each treatment is given to one subject in each pair or ο§ observations are taken on the same subject before-and-after some treatment. 18.6 Matched Pairs t Procedures Matched Pairs t Procedures βΈ To compare the responses to the two treatments in a matched pairs design, find the difference between the responses within each pair. Then apply the one-sample t procedures to these differences. 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Some researchers claim that music relaxes students and reduces stress while studying. 12 students were selected at random. Their initial resting pulse rate (beats/minute) was obtained, and each person participated in a month-long music-listening, relaxation therapy program. A final resting pulse rate was taken at the end of the experiment. The data are given below. βΈ Is there any evidence that music reduced the mean pulse rate, and consequently, reduced stress? Assume the underlying distributions are normal and use a 0.025 level of significance. Subject Initial Pulse Rate Final Pulse Rate Amy Bob Cal Dee Eve Fay Gus Hal Ike Joe Kel Moe 67 71 67 83 70 75 71 68 72 88 78 70 61 72 70 76 58 61 74 59 61 64 71 77 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Music βreducesβ pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate β Initial Pulse Rate 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Music βreducesβ pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate β Initial Pulse Rate All three of these expressions are equivalent.: 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Music βreducesβ pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate β Initial Pulse Rate All three of these expressions are equivalent.: ο§ D= βDiffβ < 0 (implies that the pulse rate has βreduced.β) 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Music βreducesβ pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate β Initial Pulse Rate All three of these expressions are equivalent.: ο§ D= βDiffβ < 0 (implies that the pulse rate has βreduced.β) ο§ Final Pulse Rate β Initial Pulse Rate < 0 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Music βreducesβ pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate β Initial Pulse Rate All three of these expressions are equivalent.: ο§ D= βDiffβ < 0 (implies that the pulse rate has βreduced.β) ο§ Final Pulse Rate β Initial Pulse Rate < 0 ο§ Final Pulse Rate < Initial Pulse Rate 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Music βreducesβ pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate β Initial Pulse Rate All three of these expressions are equivalent.: ο§ D= βDiffβ < 0 (implies that the pulse rate has βreduced.β) ο§ Final Pulse Rate β Initial Pulse Rate < 0 ο§ Final Pulse Rate < Initial Pulse Rate ο§ DiffFinal-Initial= DiffFinal<Initial this is preferred since it directly aligns with the hypothesis presented. 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Compute the sample statistics for the DiffFinal-Initial check the distribution for our assumptions. Subject Amy Bob Cal Dee Eve Fay Gus Hal Ike Joe Kel Moe Initial Pulse Rate Final Pulse Rate 67 71 67 83 70 75 71 68 72 88 78 70 61 72 70 76 58 61 74 59 61 64 71 77 1 3 D (final-initial) -6 -7 -12 -14 3 -9 -11 -24 -7 7 18.6 Matched Pairs t Procedures Example: Music relaxation therapy Diff(Final-Initial) Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N -6.3333 8.7316908 2.520622 -0.785482 -11.88118 12 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ State: What is the practical question that requires a statistical test? Does music reduce stress (as measured by pulse rates)? µFinal - Initial < 0 or µD < 0 where D = Final Pulse Rate β Initial Pulse Rate 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Plan: a. Identify the parameter. m = mean difference between final and initial pulse rates 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Plan: a. Identify the parameter. m = mean difference between final and initial pulse rates b. List all given information from the data collected. n=12, π₯ = β6.333, s = 8.732 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Plan: a. Identify the parameter. m = mean difference between final and initial pulse rates b. List all given information from the data collected. n=12, π₯ = β6.333, s = 8.732 c. State the null (H0) and alternative (Ha) hypotheses. H0: mDiff = 0 d. Specify the level of significance. a = .025 HA: mDiff < 0 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Plan: a. Identify the parameter. m = mean difference between final and initial pulse rates b. List all given information from the data collected. n=12, π₯ = β6.333, s = 8.732 c. State the null (H0) and alternative (Ha) hypotheses. H0: mDiff = 0 d. Specify the level of significance. HA: mDiff < 0 a = .025 e. Determine the type of test. Left-tailed Right-tailed f. Sketch the region(s) of βextremely unlikelyβ test statistics. Two-Tailed 18.4 One-Sample t Test Example: Music relaxation therapy Solve: a) Check the conditions for the test you plan to use. ο§ Random Sample? Yes. Stated as a random sample. 18.4 One-Sample t Test Example: Music relaxation therapy Solve: a) Check the conditions for the test you plan to use. ο§ Random Sample? Yes. Stated as a random sample. ο§ Large enough population: sample ratio? Yes. The number of students is arbitrarily large; therefore, N > 20*12 = 240. 18.4 One-Sample t Test Example: Music relaxation therapy Solve: a) Check the conditions for the test you plan to use. ο§ Random Sample? Yes. Stated as a random sample. ο§ Large enough population: sample ratio? Yes. The number of students is arbitrarily large; therefore, N > 20*12 = 240. ο§ Large enough sample; Normal or t-distribution? Yes. n = 12 < 40. But the data seem to be approximately Normal, so we will attempt to use the t-distribution. 18.4 One-Sample t Test Example: Hemoglobin levels Solve: b) Calculate the test statistic π₯ β π0 β6.33 β 0 π‘= = π π 8.732 12 18.4 One-Sample t Test Example: Hemoglobin levels Solve: b) Calculate the test statistic π₯ β π0 β6.33 β 0 π‘= = = β2.511 π π 8.732 12 18.4 One-Sample t Test Example: Hemoglobin levels Solve: π₯ β π0 β6.33 β 0 π‘= = = β2.511 π π 8.732 12 b) Calculate the test statistic c) Determine (or approximate) the P-Value. 18.4 One-Sample t Test Example: Hemoglobin levels Solve: π₯ β π0 β6.33 β 0 π‘= = = β2.511 π π 8.732 12 b) Calculate the test statistic c) Determine (or approximate) the P-Value. ο§ -2.328 > -2.511 >-2.718 2.511 18.4 One-Sample t Test Example: Hemoglobin levels Solve: π₯ β π0 β6.33 β 0 π‘= = = β2.511 π π 8.732 12 b) Calculate the test statistic c) Determine (or approximate) the P-Value. 2.511 ο§ -2.328 > -2.511 >-2.718 ο§ .02 > P-Value >.01 P-value 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). Because the approximate P-value is less than 0.025, we reject the null hypothesis. 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). Because the approximate P-value is less than 0.025, we reject the null hypothesis. b) Interpret the decision in the context of the original claim. 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Conclude: a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0). Because the approximate P-value is less than 0.025, we reject the null hypothesis. b) Interpret the decision in the context of the original claim. There is enough evidence at the a=.025 level to conclude that music reduces stress (and lowers pulse rates). 18.6 Matched Pairs t Procedures Question βΈ Based upon our data, are matched pairs based upon dependent samples? a) Yes. (The data from 2nd variable are related to the data from the 1st variable). b) No. (The data from 2nd variable are not related to the data from the 1st variable). c) Not sure 18.6 Matched Pairs t Procedures Question βΈ Based upon our data, are matched pairs based upon dependent samples? a) Yes. (The data from 2nd variable are related to the data from the 1st variable). b) No. (The data from 2nd variable are not related to the data from the 1st variable). c) Not sure 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Construct a 95% confidence interval for the average difference in initial and final pulse rates. 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11, .95)=2.201 by looking across the df = 11 row and down the 95% confidence level column. 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11, .95)=2.201 by looking across the df = 11 row and down the 95% confidence level column. x ο± tοͺ s 8.732 = ο6.333 ο± 2.201 n 12 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11, .95)=2.201 by looking across the df = 11 row and down the 95% confidence level column. x ο± tοͺ s 8.732 = ο6.333 ο± 2.201 = ο6.333 ο± 5.548 = ο0.785 to -11.881 n 12 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11, .95)=2.201 by looking across the df = 11 row and down the 95% confidence level column. x ο± tοͺ s 8.732 = ο6.333 ο± 2.201 = ο6.333 ο± 5.548 = ο0.785 to -11.881 n 12 We are 95% confident that the mean difference in final and initial pulse rates (with Final minus Initial) is between -0.785 and -11.881 bpm. Or, we are 95% confident that participating in the relaxation experiment reduce oneβs pulse rate by an average of 0.7852 to 11.881 bpm. 18.6 Matched Pairs t Procedures Constructing a 95% confidence interval for the average difference in initial and final pulse rates. βΈ Weβll use the TI-83/84 first this time. Then verify via the formula. βΈ Enter the differences { -6, 1, 3, -7, -12, -14, 3, -9, -11, -24, -7, 7} in L1. ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. ο§ Inpt >> DATA ο§ List: L1 >> Freq: 1 ο§ C-Level : 95 ο§ Calculate (ENTER) 18.6 Matched Pairs t Procedures Constructing a 95% confidence interval for the average difference in initial and final pulse rates. βΈ TI-83/84 βΈ Enter the differences { -6, 1, 3, -7, -12, -14, 3, -9, -11, -24, -7, 7} in L1. ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. ο§ Inpt >> DATA ο§ List: L1 >> Freq: 1 ο§ C-Level : 95 ο§ Calculate (ENTER) ( -11.88, -0.7855) 18.6 Matched Pairs t Procedures Constructing a 95% confidence interval for the average difference in initial and final pulse rates. βΈ TI-83/84 ο§ STAT >> TESTS >> TInterval >> Enter ο§ Note: Select Data when π₯ and π are not provided. Then enter the list where the data are stored. (for this example) ο§ Inpt >> STATS ο§ π : -6.33 >> s: 8.732 >> π : -6.33 >> n : 12 automatically) ο§ C-Level : 95 ο§ Calculate (ENTER) ( -11.88, -0.7855) (these should populate 18.6 Matched Pairs t Procedures Example: Music relaxation therapy βΈ Technology output for the hypothesis test and confidence interval: Technology output for the hypothesis test and confidence interval: Test Mean Hypothesized Value 0 Actual Estimate -6.3333 DF 11 Std Dev 8.73169 t Test Test Statistic -2.5126 Prob > |t| 0.0289* Prob > t 0.9856 Prob < t 0.0144* Confidence Intervals Parameter Estimate Mean -6.33333 Std Dev 8.731691 Lower CI -11.8812 6.185487 Upper CI -0.78548 14.82535 1-Alpha 0.950 0.950 18.7 Robustness of t Procedures Robustness of t Procedures βΈ A confidence interval or significance test is called robust if the confidence interval or P-value does not change very much when the conditions for use of the procedure are violated. βΈ Except in the case of small samples, the condition that the data are an SRS from the population of interest is more important than the condition that the population distribution is Normal. 18.7 Robustness of t Procedures Robustness of t Procedures βΈ The t-procedures guard against non-Normality except when there is strong skewness or outliers present. βΈ When the data are not from a Normal distribution we also need to consider the sample size: Sample size less than 15 The t procedures can be used if the data close to Normal (roughly symmetric, single peak, no outliers)? If there is clear skewness or outliers then, do not use t. Sample size between 15 and 40 The t procedures can be used except in the presences of outliers or strong skewness. Sample size is at least 40 The t procedures can be used even for clearly skewed distributions. 18.7 Robustness of t Procedures Robustness of t Procedures βΈ Note that we have changed the βlarge enough sampleβ condition to be adaptable to the situations that we encounter. This is because t procedures are robust against violations of Normality. 18.7 Robustness of t Procedures Robustness of t Procedures βΈ Example: The number of text messages sent daily for 25 college students are below. Can we safely use t-procedures? 10 10 12 15 17 18 22 23 24 25 25 25 26 27 28 30 42 51 53 75 103 118 120 130 135 18.7 Robustness of t Procedures Robustness of t Procedures βΈ Example: The number of text messages sent daily for 25 college students are below. Can we safely use t-procedures? 10 10 12 15 17 18 22 23 24 25 25 25 26 27 28 30 42 51 53 75 103 118 120 130 135 βΈ No. The sample size is 25, but there are clear outliers and strong skewness. Five-Minute Summary βΈ List at least 3 concepts that had the most impact on your knowledge of inference about a population mean. _____________ _________________ _______________