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İŞL 276 Fundamentals of Hypothesis Testing: One-Sample Tests What is a Hypothesis? A hypothesis is a claim (assumption) about a population parameter: population mean Example: The mean monthly cell phone bill of this city is μ = $42 population proportion Example: The proportion of adults in this city with cell phones is π = 0.68 The Null Hypothesis, H0 States the claim or assertion to be tested Example: Average number of mobile phone in a Turkish family is equal to 3 ( H0 : μ 3 ) Is always about a population parameter, not about a sample statistic H0 : μ 3 H0 : X 3 The Null Hypothesis, H0 (continued) Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until proven guilty Refers to the status quo Always contains “=” , “≤” or “” sign May or may not be rejected The Alternative Hypothesis, H1 Is the opposite of the null hypothesis e.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: μ ≠ 3 ) Challenges the status quo Never contains the “=” , “≤” or “” sign May or may not be proven Is generally the hypothesis that the researcher is trying to prove Hypothesis Testing Process Claim: the population mean age is 50. (Null Hypothesis: H0: μ = 50 ) Population Is X 20 likely if μ = 50? If not likely, REJECT Null Hypothesis Suppose the sample mean age is 20: X = 20 Now select a random sample Sample Reason for Rejecting H0 Sampling Distribution of X 20 μ = 50 If H0 is true If it is unlikely that we would get a sample mean of this value ... ... if in fact this were the population mean… X ... then we reject the null hypothesis that μ = 50. Level of Significance, Defines the unlikely values of the sample statistic if the null hypothesis is true Defines rejection region of the sampling distribution Is designated by , (level of significance) Typical values are 0.01, 0.05, or 0.10 Is selected by the researcher at the beginning Provides the critical value(s) of the test Level of Significance and the Rejection Region Level of significance = H0: μ = 3 H1: μ ≠ 3 /2 Two-tail test /2 0 Upper-tail test H0: μ ≥ 3 H1: μ < 3 Lower-tail test Rejection region is shaded 0 H0: μ ≤ 3 H1: μ>3 0 Represents critical value Hypothesis Tests for the Mean Hypothesis Tests for Known (Z test) Unknown (t test) Z Test of Hypothesis for the Mean (σ Known) Convert sample statistic ( X ) to a Z test statistic Hypothesis Tests for Known σKnown (Z test) The test statistic is: X μ Z σ n Unknown σUnknown (t test) Critical Value Approach to Testing For a two-tail test for the mean, σ known: Convert sample statistic ( statistic ) Determine the critical Z values for a specified level of significance from a table or computer Decision Rule: If the test statistic falls in the rejection region, reject H0 ; otherwise do not reject H0 X ) to test statistic (Z Two-Tail Tests There are two cutoff values (critical values), defining the regions of rejection H0: μ = 3 H1 : μ 3 /2 /2 X 3 Reject H0 -Z Lower critical value Do not reject H0 0 Reject H0 +Z Upper critical value Z 6 Steps in Hypothesis Testing 1. 2. State the null hypothesis, H0 and the alternative hypothesis, H1 Choose the level of significance, , and the sample size, n 3. Determine the appropriate test statistic and sampling distribution 4. Determine the critical values that divide the rejection and nonrejection regions 6 Steps in Hypothesis Testing (continued) 5. Collect data and compute the value of the test statistic 6. Make the statistical decision and state the managerial conclusion. If the test statistic falls into the nonrejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem Hypothesis Testing Example Test the claim that the true mean of mobile phone in a Turkish family is equal to 3. Assume σ = 0.8 and sample results are n = 100, X = 2.84 1. State the appropriate null and alternative hypotheses H0: μ = 3 H1: μ ≠ 3 (This is a two-tail test) 2. Specify the desired level of significance and the sample size Suppose that = 0.05 and n = 100 are chosen for this test Hypothesis Testing Example (continued) 3. Determine the appropriate technique σ is known so this is a Z test. 4. Determine the critical values For = 0.05 the critical Z values are ±1.96 5. Collect the data and compute the test statistic Suppose the sample results are n = 100,X = 2.84 (σ = 0.8 is assumed known) So the test statistic is: Z X μ 2.84 3 .16 2.0 σ 0.8 .08 n 100 Hypothesis Testing Example (continued) 6. Is the test statistic in the rejection region? = 0.05/2 Reject H0 if Z < -1.96 or Z > 1.96; otherwise do not reject H0 Reject H0 -Z= -1.96 = 0.05/2 Do not reject H0 0 Reject H0 +Z= +1.96 Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region Hypothesis Testing Example (continued) 6(continued). Reach a decision and interpret the result = 0.05/2 Reject H0 -Z= -1.96 = 0.05/2 Do not reject H0 0 Reject H0 +Z= +1.96 -2.0 Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of TVs in US homes is not equal to 3 p-Value Approach to Testing p-value: Probability of obtaining a test statistic more extreme ( ≤ or ) than the observed sample value given H0 is true Also called observed level of significance Smallest value of for which H0 can be rejected p-Value Approach to Testing (continued) Convert Sample Statistic (e.g., (e.g., Z statistic ) )X to Test Statistic Obtain the p-value from a table or computer Compare the p-value with If p-value < , reject H0 If p-value , do not reject H0 p-Value Example Example: How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is = 3.0? X = 2.84 is translated to a Z score of Z = -2.0 P(Z 2.0) 0.0228 P(Z 2.0) 0.0228 /2 = 0.025 /2 = 0.025 0.0228 0.0228 p-value = 0.0228 + 0.0228 = 0.0456 -1.96 -2.0 0 1.96 2.0 Z p-Value Example (continued) Compare the p-value with If p-value < , reject H0 If p-value , do not reject H0 Here: p-value = 0.0456 = 0.05 Since 0.0456 < 0.05, we reject the null hypothesis /2 = 0.025 /2 = 0.025 0.0228 0.0228 -1.96 -2.0 0 1.96 2.0 Z Connection to Confidence Intervals • For sample mean is 2.84 and σ = 0.8 and n = 100, the 95% confidence interval is: 0.8 2.84 - (1.96) to 100 0.8 2.84 (1.96) 100 2.6832 ≤ μ ≤ 2.9968 Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at = 0.05 One-Tail Tests In many cases, the alternative hypothesis focuses on a particular direction H0: μ ≥ 3 H1: μ < 3 H0: μ ≤ 3 H1 : μ > 3 This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3 This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3 Lower-Tail Tests H0: μ ≥ 3 There is only one critical value, since the rejection area is in only one tail H1 : μ < 3 Reject H0 -Z Do not reject H0 0 μ Critical value Z X Upper-Tail Tests There is only one critical value, since the rejection area is in only one tail Z _ X H0: μ ≤ 3 H1: μ > 3 Do not reject H0 0 Zα Reject H0 μ Critical value Ex:Upper-Tail Z Test for Mean ( Known) A phone industry manager thinks that customer monthly cell phone bills have increased, and now average over $52 per month. The company wishes to test this claim. (Assume = 10 is known) Form hypothesis test: H0: μ ≤ 52 the average is not over $52 per month H1: μ > 52 the average is greater than $52 per month (i.e., sufficient evidence exists to support the manager’s claim) Example: Find Rejection Region (continued) Suppose that = 0.10 is chosen for this test Find the rejection region: Reject H0 = 0.10 Do not reject H0 0 1.28 Reject H0 Reject H0 if Z > 1.28 Review:One-Tail Critical Value What is Z given = 0.10? 0.90 Standardized Normal Distribution Table (Portion) 0.10 = 0.10 Z .07 .08 .09 1.1 .8790 .8810 .8830 0.90 1.2 .8980 .8997 .9015 z 0 1.28 Critical Value = 1.28 1.3 .9147 .9162 .9177 Example: Test Statistic (continued) Obtain sample and compute the test statistic Suppose a sample is taken with the following results: n = 64, X = 53.1 (=10 was assumed known) Then the test statistic is: Xμ 53.1 52 Z 0.88 σ 10 n 64 Example: Decision (continued) Reach a decision and interpret the result: Reject H0 = 0.10 Do not reject H0 1.28 0 Z = 0.88 Reject H0 Do not reject H0 since Z = 0.88 ≤ 1.28 i.e.: there is not sufficient evidence that the mean bill is over $52 p -Value Solution Calculate the p-value and compare to (continued) (assuming that μ = 52.0) p-value = 0.1894 Reject H0 = 0.10 0 Do not reject H0 1.28 Z = 0.88 Reject H0 P( X 53.1) 53.1 52.0 P Z 10/ 64 P(Z 0.88) 1 0.8106 0.1894 Do not reject H0 since p-value = 0.1894 > = 0.10 t Test of Hypothesis for Mean (σ Unknown) Convert sample statistic ( X) to a t test statistic Hypothesis Tests for Known σKnown (Z test) Unknown σUnknown (t test) The test statistic is: t n-1 X μ S n Example: Two-Tail Test( Unknown) The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in X = $172.50 and S = $15.40. Test at the = 0.05 level. (Assume the population distribution is normal) H0: μ = 168 H1: μ 168 Example Solution: Two-Tail Test H0: μ = 168 H1: μ 168 = 0.05 n = 25 is unknown, so use a t statistic Critical Value: t24 = ± 2.0639 /2=.025 /2=.025 Reject H0 -t n-1,α/2 -2.0639 t n1 Do not reject H0 Reject H0 0 1.46 t n-1,α/2 2.0639 X μ 172.50 168 1.46 S 15.40 n 25 Do not reject H0: not sufficient evidence that true mean cost is different than $168 Hypothesis Tests for Proportions Involves categorical variables Two possible outcomes “Success” (possesses a certain characteristic) “Failure” (does not possesses that characteristic) Fraction or proportion of the population in the “success” category is denoted by π Proportions (continued) Sample proportion in the success category is denoted by p X number of successes in sample p n sample size When both nπ and n(1-π) are at least 5, p can be approximated by a normal distribution with mean and standard deviation (1 ) μp σp n Hypothesis Tests for Proportions The sampling distribution of p is approximately normal, so the test statistic is a Z value: Z pπ π(1 π ) n Hypothesis Tests for p nπ 5 and n(1-π) 5 nπ < 5 or n(1-π) < 5 Not discussed in this chapter Z Test for Proportion An equivalent form to the last slide, but in terms of the number of successes, X: X n Z n (1 ) Hypothesis Tests for X X5 and n-X 5 X<5 or n-X < 5 Not discussed in this chapter Example: Z Test for Proportion A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the = 0.05 significance level. Check: n π = (500)(.08) = 40 n(1-π) = (500)(.92) = 460 Z Test for Proportion: Solution Test Statistic: H0: π = 0.08 H1: π 0.08 Z = 0.05 n = 500, p = 0.05 Decision: Critical Values: ± 1.96 Reject p .05 .08 2.47 (1 ) .08(1 .08) n 500 Reject Reject H0 at = 0.05 Conclusion: .025 .025 -1.96 -2.47 0 1.96 z There is sufficient evidence to reject the company’s claim of 8% response rate. p-Value Solution (continued) Calculate the p-value and compare to (For a two-tail test the p-value is always two-tail) Do not reject H0 Reject H0 Reject H0 /2 = .025 p-value = 0.0136: /2 = .025 P(Z 2.47) P(Z 2.47) 0.0068 0.0068 -1.96 Z = -2.47 0 2(0.0068) 0.0136 1.96 Z = 2.47 Reject H0 since p-value = 0.0136 < = 0.05 Example 1 Price/earnings ratios for stocks. Theory: stable rate of P/E in market = 13. If P/E (market) < 13, you should invest in the stock market. If P/E (market) > 13, you should take your money out. We have a sample of 50 = 12.1. Historical σ = 3.0456 44 Estimate Steps Can we estimate if the population P/E is 13 or not? Common steps: Set hypothesis: H0: μ = 13 HA: μ ≠ 13 Select = .05. 45 Calculating Test Statistic 3.0456 Standard Error x 0.4307 n 50 z value 46 x - 0 x 12.1 13 2.0896 0.4307 p-value Approach 47 Calculate the p-value. We will calculate for the lower tail → then make an adjustment for the upper tail. p-value, Two-Tailed Test p(z > 2.09) = ?? p(z < –2.09) = ?? Z=–2.09 0 Z=2.09 We can just calculate one value, and double it. 48 z Calculating the p-value cont’d Find 2.090 on the Standard Normal Distribution tables: z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 … 1.9 2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817 2.1 … 49 p-value, Two-Tailed Test 0.4817 p(z > 2.09) = 0.5 -.4817 = 0.0183 p(z < –2.09) = ?? Z=–2.09 0 Z=2.09 Doubling the value, we find the p-value = 0.0366 50 z Should We Reject the Null Hypothesis? Yes! p-value = 0.0366 < = 0.05. There is only a 3.66% chance that the measured price/earnings ratio sample mean of 12.1 is not equal to the stable rate of 13 by random chance. 51 Example #2: You want to test the hypothesis that a treatment conducted at the ward you are working on in a hospital is more beneficial than the average treatment used in other wards of the hospital.Given that the mean wellness score of the people on your ward is 87, and that the mean for the entire hospital (N = 20) is 76 and the standard deviation of population is 15, is your ward significantly better? State Ho and H1 State if you’re using a one- or two-tailed test and why. State the p-value that supports your claim. Hypothesis Testing w/ One Sample 87 76 11 z 3.27 15 3.36 20 Smaller Portion = p < .0006 Example A family therapist states that parents talk to their teenagers an average of 27 minutes per week. Surprised by that claim, a psychologist decided to collect some data on the amount of time parents spend in conversation with their teenage children. For the n = 12 parents, the study revealed that following times (in minutes) devoted to conversation in a week: Example Do the psychologists findings differ significantly from the therapist’s claim? If so, is the family expert’s claim an overestimate or underestimate of actual time spent talking to children? Use the 0.05 level of significance with two tails. Mean = 24.58, s2 = 12.24, (s/√n) = 1.00 29 22 19 25 27 28 21 22 24 26 30 22 Hypothesis Testing w/ One Sample Example #2: H0 = μ = 27 Critical t (df = 11) = ±2.201 Reject H0, and conclude that the data are significantly different from the therapist’s claim J) Hypothesis Tests, 2 Means: ’s Unknown Two datasets –> is the mean value of one larger than the other? Is it larger by a specific amount? μ1 vs. μ2 –> μ1 – μ2 vs. D0. Often set D0 = 0 –> is μ1 = μ2? 57 Example: Female vs. Male Salaries Saskatchewan 2001 Census data: - only Bachelor’s degrees - aged 21-64 - work full-time - not in school Men: M = $46,452.48, sM = 36,260.1, nM = 557. Women: W = $35,121.94, sW = 20,571.3, nW = 534. M – W = $11,330.44 } our point estimate. Is this an artifact of the sample, or do men make significantly more than women? 58 Hypothesis, Significance Level, Test Statistic 1. 2. 3. We will now ONLY use the p-value approach, and NOT the critical value approach. Research hypothesis: men get paid more: H0: μM – μW < 0 H1: μM – μW > 0 Select = 0.05 Compute test t-statistic: t ( xM xW ) D0 sW2 sM2 nM nW (46452.48 35121.94) 0 6.375 1314794928 426178351 557 534 59 4. a. Compute the Degrees of Freedom Can compute by hand, or get from Excel: 2 s s n1 n2 = 888 df 2 2 2 2 1 s1 1 s2 n1 1 n1 n2 1 n2 2 1 2 2 60 4. b. Computing the p-value Degrees of Freedom 0.20 0.10 0.02 0.025 0.01 0.005 .845 .842 1.290 1.282 1.660 1.645 1.984 1.960 2.364 2.326 2.626 2.576 … 100 6.375 up here somewhere The p-value <<< 0.005. 61 5. Check the Hypothesis Since the p-value is <<< 0.05, we reject H0. We conclude that we can accept the alternative hypothesis that men get paid more than women at a very high level of confidence (greater than 99%). 62 Excel t-Test: Two-Sample Assuming Unequal Variances Male Female Mean 46452.47935 35121.94195 Variance Observations 1314794928 557 423178351.3 534 Hypothesized Mean Diff. df t Stat P(T<=t) one-tail 0 888 6.381034789 1.41441E-10 .00000000014 t Critical one-tail P(T<=t) two-tail 1.646571945 2.82883E-10 t Critical two-tail 1.962639544 63