5 Running the TestP-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.
6 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.
7 ExampleListed below are the measured radiation emissions (in W/kg) corresponding to a sample of cell phones.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.The summary statistics are:0.380.551.541.550.500.600.920.961.000.861.46
8 Example - Continued Requirement Check: We assume the sample is a simple random sample.The sample size is n = 11, which is not greater than 30, so we must check a normal quantile plot for normality.
9 Example - ContinuedThe 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.
10 Example - ContinuedStep 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.Step 2: The alternative to the original claim is μ ≥ 1.00 W/kg.Step 3: The hypotheses are written as:
11 Example - ContinuedStep 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: .
12 Example - ContinuedStep 6: Calculate the test statistic and then find the P-value or the critical value from Table A-3:
13 Example - ContinuedStep 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.
14 Example - ContinuedStep 7: P-value method: Technology, such as a TI-83/84 Plus calculator can output the P-value of Since the P-value exceeds α = 0.05, we fail to reject the null hypothesis.
15 ExampleStep 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.
16 Finding P-ValuesAssuming that neither software nor a TI-83 Plus calculator is available, use Table A-3 to find a range of values for the P-value corresponding to the given results.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 =c) In a two-tailed hypothesis test, the sample size is n = 12, and the test statistic is t = –3.456.
17 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.
18 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 < W/kg
19 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.