Hypothesis Testing Involving One Population Chapter 11

11.3 Hypothesis Testing Involving a Population Mean: σ Known We will skip over section 11.2 (hypothesis testing involving a proportion) and start hypothesis testing using sample means. In problems in this section, we will be provided with a population mean, μ, and a population standard deviation, σ. H 0 and H a will be formulated with a population mean using the words “population mean.” The standard error formula: The z-score of the test statistic:

11.3 Hypothesis Testing Involving a Population Mean: σ Known There are 2 types of hypothesis tests involving a population mean. 1. The population standard deviation will be provided in the problem In this case, use the z-score of the test statistic. 2. The population standard deviation will not be given, but the sample standard deviation will be given. Then, we’ll use the t-score of the test statistic.

Example 11.8 A tire manufacturer advertises that its brand of radial tires has a mean life of 40,000 miles with a population standard deviation of 1,500 miles. A consumer’s research team decides to investigate this claim after receiving several complaints from people who believe this advertisement is false (i.e. the mean life of 40,000 is too high). If the research team tests 100 of these radial tires and obtains a sample mean tire life of 39,750, is the advertisement legitimate? Use α =5%.

Example 11.8 Step 1: Formulate the hypotheses. H 0 : The population mean tire life is 40,000 miles H 0 : μ =40,000 miles H a : The population mean tire life is less than 40,000 miles. H a : μ <40,000 miles

Example 11.8 Step 2: Determine the model to test the null hypothesis, H 0. The model will be the sampling distribution of the mean. Since n>30, the sampling distribution will be approximately normal with the following values:

Example 11.8 Step 3: Formulate the decision rule. Directional or non-directional? We have a directional test in the left tail since H a is trying to show μ <40,000 miles. For a left tailed test with a significance level of 5%, z C is from page 543.

Example 11.8 Step 3: Formulate the decision rule. Decision rule: Reject H 0 if the z-score of the sample mean is less than

Example 11.8 Step 4: Analyze the sample data The problem gives us We compute the test statistic- the z-score of the sample mean.

Example 11.8 Step 5: State the conclusion. Since the z-score of the sample mean, z=-1.67, is less than the critical z-score z LC =-1.65, we reject H 0 and accept H a at α =5%.

Example 11.8 Answer the question. Is the advertisement claim that the mean life of a brand of radial tires is 40,000 legitimate? No, the research team can conclude that the advertisement was not legitimate.

Example 11.8 Could the null hypothesis, H 0, be rejected at a smaller significance level, α ? Maybe, let’s suppose that α was given as 4.8%, then z c = invnorm(.048) = In this case, we can still reject the H 0. But if α was given as 4%, then z c = invnorm(.04) = and H 0 would not be rejected.

Example 11.8 What is the smallest significance level, α, for which H 0 can be rejected? The area to the left of the test statistic, will give the smallest α for which H 0 can be rejected. Normalcdf( , -1.67) = = 4.75%

P-value approach to Hypothesis Testing , or 4.75%, is called the p-value for this problem. It is also called the observed significance level of a hypothesis test. If we compare the observed significance level, p-value, to the given significance level, α, of the problem, we can come to a conclusion for this hypothesis test without using critical z- scores. In general, if the p-value is less than α, then we can reject the null hypothesis.

P-value approach to Hypothesis Testing When using the p-value approach, the steps to a hypothesis test change slightly: Step 1. Formulate the hypotheses. Step 2. Determine the model to test the null hypothesis. Step 3. Formulate a decision rule. “Reject the null hypothesis if the p-value of the [sample mean or sample proportion] is less than α ” Step 4. Analyze the sample data. (calculate the p- value of the sample mean or proportion) Step 5. State the conclusion. Either reject H o and accept H a, or fail to reject H o.

P-value approach to Hypothesis Testing 1TT on the left 1TT on the right

Example 11.8 using p-values Step 1: Formulate the hypotheses. H 0 : The population mean tire life is 40,000 miles. μ =40,000 miles H a : The population mean tire life is less than 40,000 miles. μ <40,000 miles Step 2: The model will be the sampling distribution of the mean.

Example 11.8 using p-values Step 3: Decision rule: Reject H 0 if the p-value of the sample mean is less than 5%

Example 11.8 using p-values Step 4. Analyze the sample data. (find the p-value of the sample mean) STAT  TESTS  Z-Test Step 5. State the conclusion. Reject the null hypothesis since the p-value of the sample mean, , is less than α = 5%. The research team can conclude that the advertisement was not legitimate

Example 11.8 using p-values Figure comparing the size of α and the p-value. Since the p-value can “fit inside” α, the p-value is smaller and the test statistic is in the reject H 0 region. P-value = = 4.78% α =5% Fail to reject H 0 Reject H 0

Example 11.7 An automatic opening device for parachutes has stated a mean release time of 10 seconds with a population standard deviation of 3 seconds. A local parachute club decides to test this claim against the alternative hypothesis that the release time is not 10 seconds. The club purchases 36 of these devices and finds that the mean release time is 8.5 seconds. Does this sample result indicate that the opening device does not have a mean release time of 10 seconds? Use α =1%.

Example 11.7 Step 1: Formulate the hypotheses. H 0 : The population mean release time is 10 seconds. H 0 : μ =10 seconds H a : The population mean release time is not 10 seconds. H a : μ ≠10 seconds

Example 11.7 Step 2: Determine the model to test the null hypothesis, H 0. The model will be the sampling distribution of the mean. Since n>30, the sampling distribution will be approximately normal with the following values:

Example 11.7 Step 3: Formulate the decision rule. Directional or non-directional? We have a non-directional test since H a is trying to show μ ≠10 seconds There are two tails with two z-scores in a non-directional test, z LC and z RC. For a significance level of 1%, z LC and z RC are and 2.58.

Example 11.7 Step 3: Formulate the decision rule. Decision rule: Reject H 0 if the z-score of the sample mean is either less than or greater than 2.58.

Example 11.7 Step 4: Analyze the sample data The problem gives us We compute the test statistic- the z-score of the sample mean.

Example 11.7 Step 5: State the conclusion. Since the z-score of the sample mean, z=-3, is less than the critical z-score z LC =-2.58, we reject H 0 and accept H a at α =1%.

Example 11.7 Answer the question. Does the sample result indicate that the opening device does not have a mean release time of 10 seconds? Yes, the sample mean of 8.5 seconds indicates that the population mean release time for the opening device is not 10 seconds.

Example 11.7 using p-values Step 1: Formulate the hypotheses. H 0 : The population mean release time is 10 seconds. μ =10 seconds H a : The population mean release time is not 10 seconds. μ ≠10 seconds Step 2: The model will be the sampling distribution of the mean. Since n > 30, the sampling distribution will be approximately normal.

Example 11.7 using p-values Step 3: Decision rule: Reject H 0 if the p- value of the sample mean is less than 1%. Note: there is no need to label critical z- scores when using the p-value approach.

Example 11.7 using p-values Step 4. Analyze the sample data. (find the p-value of the sample mean) STAT  TESTS  Z-Test Step 5. State the conclusion. Reject the null hypothesis since the p-value of the sample mean, , is less than α = 1%. The sample mean of 8.5 seconds indicates that the population mean release time for the opening device is not 10 seconds