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Notation 8-4 Testing a Claim About a Mean This section presents methods for testing a claim about a population mean. n = sample size x = sample mean μ x = population mean When σ is not known, we use a “t test” that incorporates the Student t distribution. Test Statistic Requirements 1) The sample is a simple random sample. t= 2) Either or both of these conditions is satisfied: The population is normally distributed or n > 30. 11 Running the Test P-values: Use technology or use the Student t distribution in Table A-3 with degrees of freedom df = n – 1. Critical values: Use the Student t distribution with degrees of freedom df = n – 1. x − μx s n Important Properties of the Student t Distribution 1. The Student t distribution is different for different sample sizes (see Figure 7-5 in Section 7-3). 2. The Student t distribution has the same general bell shape as the normal distribution; its wider shape reflects the greater variability that is expected when s is used to estimate σ. 3. The Student t distribution has a mean of t = 0. 4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1. 5. As the sample size n gets larger, the Student t distribution gets closer to the standard normal distribution. Example Example - Continued Listed below are the measured radiation emissions (in W/kg) corresponding to a sample of cell phones. Requirement Check: 1. We assume the sample is a simple random sample. Use a 0.05 level of significance to test the claim that cell phones have a mean radiation level that is less than 1.00 W/kg. 0.38 0.55 1.54 1.55 0.50 0.60 0.92 0.96 1.00 0.86 2. The sample size is n = 11, which is not greater than 30, so we must check a normal quantile plot for normality. 1.46 The summary statistics are: x = 0.938 and s = 0.423 . Example - Continued Example - Continued The points are reasonably close to a straight line and there is no other pattern, so we conclude the data appear to be from a normally distributed population. Step 1: The claim that cell phones have a mean radiation level less than 1.00 W/kg is expressed as μ < 1.00 W/kg. 22 Step 2: The alternative to the original claim is μ ≥ 1.00 W/kg. Step 3: The hypotheses are written as: H 0 : μ = 1.00 W/kg H1 : μ < 1.00 W/kg Example - Continued Step 4: The stated level of significance is α = 0.05. Step 5: Because the claim is about a population mean μ, the statistic most relevant to this test is the sample mean: x . Example - Continued Step 6: Calculate the test statistic and then find the P-value or the critical value from Table A-3: t= x − μ x 0.938 − 1.00 = = −0.486 s 0.423 n 11 Example - Continued Example - Continued Step 7: Critical Value Method: Because the test statistic of t = –0.486 does not fall in the critical region bounded by the critical value of t = –1.812, fail to reject the null hypothesis. Step 7: P-value method: Technology, such as a TI-83/84 Plus calculator can output the P-value of 0.3191. Since the P-value exceeds α = 0.05, we fail to reject the null hypothesis. Example Finding P-Values Assuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the Pvalue corresponding to the given results. Step 8: Because we fail to reject the null hypothesis, we conclude that there is not sufficient evidence to support the claim that cell phones have a mean radiation level that is less than 1.00 W/kg. 33 Example – Confidence Interval Method We can use a confidence interval for testing a claim about μ. For a two-tailed test with a 0.05 significance level, we construct a 95% confidence interval. For a one-tailed test with a 0.05 significance level, we construct a 90% confidence interval. a) In a left-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = –2.007. b) In a right-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = 1.222. c) In a two-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = –3.456. Example – Confidence Interval Method Using the cell phone example, construct a confidence interval that can be used to test the claim that μ < 1.00 W/kg, assuming a 0.05 significance level. Note that a left-tailed hypothesis test with α = 0.05 corresponds to a 90% confidence interval. Using methods described in Section 7.3, we find: 0.707 W/kg < μ < 1.169 W/kg Example – Confidence Interval Method Because the value of μ = 1.00 W/kg is contained in the interval, we cannot reject the null hypothesis that μ = 1.00 W/kg . Based on the sample of 11 values, we do not have sufficient evidence to support the claim that the mean radiation level is less than 1.00 W/kg. 44