Section 7-2 Hypothesis Testing for the Mean (n  30) Objective: SWBAT Find a p value and use them to test a mean ц. How to use P values for a z test. How to find critical values and rejection regions in a normal distribution. How to use rejection regions for a z test.

Using p values to make decisions Decision Rule Based on P-value To use a P value to make a conclusion in a hypothesis test, compare the p value with α. 1. If P < α then reject Ho 2. If P > α then fail to reject Ho

Example 1 Interpreting a P- value The P-value for a hypothesis test is P = 0.0237. What is your decision if the Level of significance is (1) α =0.05 and (2) α = 0.01? Solution Because 0.0237 < 0.05 you should reject the null hypothesis. 2. Because 0.0237 > 0.01 you should fail to reject the null hypothesis. Try it yourself The P – value for a hypothesis test is P = 0.0347. What is your decision if the Level of significance is (1) α = 0.01 and (2) α = 0.05 Compare the P-value with the level of significance. b. Make your decision.

Finding the P – value for a Hypothesis Test After determining the hypothesis test’s standardized test statistic and the test’s statistic’s area, Do one of the following to find the P-value. For the left tailed test, P = (Area in left tail) For the right tailed test, P = (Area in right tail) c. For a two tailed test, P = 2(Area in tail of test statistic)

Example 2 Finding a P - Value for the left tailed test Find the P – value for a left tail hypothesis test with a test statistic of z = -2.23. Decide whether to reject Ho if the level of significance is α = 0.01 Solution: From table 4 in the front of your book the area corresponding to Z= -2.23 is 0.0129 which is the area in the left tail. So the P-value for a left tailed hypothesis test with a test statistic of z=-2.23 is P=0.0129. Because 0.0129 > 0.01. You should fail to reject Ho . Try it Yourself: Find the P-value for a left tailed hypothesis test with a test statistic of z = -1.62. Decide whether to reject Ho if the level of significance is α = 0.05

Finding a P - Value for a Two tailed test Example 3 Finding a P - Value for a Two tailed test Find the P – value for a two tailed hypothesis test with a test statistic of z = 2.14. Decide whether to reject Ho if the level of significance is α = 0.05 Solution: From table 4 in the front of your book the area corresponding to Z= 2.14 is 0.9838 which is the area in the right tail. The area in the right tail is 1- .9838 = 0.0162. So the P-value for a two tailed hypothesis test with a test statistic of z=2.14 is P= 2(0.0162)= 0.0324 . Because 0.0324 < 0.05. You should reject Ho . Try it Yourself: Find the P – value for a two tailed hypothesis test with a test statistic of z = 2.31. Decide whether to reject Ho if the level of significance is α = 0.01 P VALUE FOR A Z = 2.31 IS .9896 So the value of P = 1- .9896 = .0104 and P = 2(0.104)= 0.0208 And since 0.0208 > .01 Then you fail to reject Ho

The z-Test for a Mean The z-test is a statistical test for a population mean. The z-test can be used: (1) if the population is normal and s is known or (2) when the sample size, n, is at least 30. The test statistic is the sample mean and the standardized test statistic is z. The formula for z should be familiar from the Central Limit theorem. When n  30, use s in place of .

The z-Test for a Mean (P-value) A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. At = 0.05, do you have enough evidence to reject the company’s claim? 1. Write the null and alternative hypothesis. The fact that the sampling distribution is normal comes from the Central Limit Theorem studied in Chapter 5. 2. State the level of significance. = 0.05 3. Determine the sampling distribution. Since the sample size is at least 30, the sampling distribution is normal.

σ= 10 4. Find the test statistic and standardize it. n = 52 s = 10 Test statistic 5. Calculate the P-value for the test statistic. Since n is at least 30, use s in place of the population standard deviation. Since this is a right-tail test, the P-value is the area found to the right of z = 1.44 in the normal distribution. From the table P = 1 – 0.9251 Area in right tail P = 0.0749. z = 1.44

Since 0.0749 > 0.05, fail to reject H0. 6. Make your decision. Compare the P-value to . Since 0.0749 > 0.05, fail to reject H0. In other words we ACCEPT the Ho (Null hypothesis) 7. Interpret your decision. There is not enough evidence to reject the claim that the mean sodium content of one serving of its cereal is no more than 230 mg. The claim was the null hypothesis. When the claim is the null hypothesis there will either be enough evidence to reject the claim or not enough evidence to reject the claim. You can never prove that the null hypothesis is true.

Example 4: Ha: : ц < 30 min. Hypothesis Testing Using P- Values In an advertisement a pizza shop claims that its mean delivery time is less than 30 minutes. A random selection of 36 delivery times has a sample mean of 28.5 minutes and a standard deviation of 3.5 minutes. I s there enough evidence to support the claim at α = 0.01? Use a P value. Ho : ц > 30 minutes Ha: : ц < 30 min. the level of significance. α = 0.01

Rejection Area -2.57 From Table 4 the P - value = .0051 Since the Alternative hypothesis indicates this Is a left tailed test. So P = .0051 and .0051 < .01 You should decide to reject the Ho (The null hypothesis). Interpretation: At the 1% level of significance you have sufficient evidence to conclude that the mean delivery time is less than 30 min.

Rejection Regions Sampling distribution for z z0 Critical Value z0 The rejection region is the range of values for which the null hypothesis is not probable. It is always in the direction of the alternative hypothesis. Its area is equal to . Using rejection regions is another way to make a decision about a hypothesis test. Show that the larger the value of alpha, the larger the rejection area will be. In a two tail test, the rejection region is divided between the left and right tails. A critical value separates the rejection region from the non-rejection region.

Critical Values The critical value z0 separates the rejection region from the non-rejection region. The area of the rejection region is . Rejection region Rejection region z0 z0 Find z0 for a right-tail test with = .05. Find z0 for a left-tail test with = .01. Students will need the normal distribution table here. For those using TI-83, this would be a good time to describe the inverse normal distribution. Students will be introduced to other sampling distributions in later sections. For the left tail test look up the cumulative area of .01. The closest is.0099 with a z-score of –2.33 For the right tail test, look up the cumulative area of .95. The closest are .9495 and .9505. Choose the z-score in the middle. For the two tail test, look up the cumulative area of .005 for the left tail value. The closest areas are .0049 and .0051 with s-scores of –2.57 and z=-2.58. Choose the midpoint. Rejection region Rejection region z0 = 1.645 z0 = –2.33 –z0 = –2.575 and z0 = 2.575 z0 z0 Find –z0 and z0 for a two-tail test with = .01.

Using the Critical Value to Make Test Decisions 1. Write the null and alternative hypothesis. Write H0 and Ha as mathematical statements. Remember H0 always contains the = symbol. 2. State the level of significance. This is the maximum probability of rejecting the null hypothesis when it is actually true. (Making a type I error.) These steps remain the same for all hypothesis tests. Discuss the normal sampling distribution that is the one for the sample mean with a large sample. 3. Identify the sampling distribution. The sampling distribution is the distribution for the test statistic assuming that the equality condition in H0 is true and that the experiment is repeated an infinite number of times.

4. Find the critical value. 5. Find the rejection region. The critical value separates the rejection region of the sampling distribution from the non-rejection region. The area of the critical region is equal to the level of significance of the test. Rejection Region z0 A right tail test is shown here. 6. Find the test statistic. Perform the calculations to standardize your sample statistic.

If the claim is the alternative hypothesis, you will 7. Make your decision. If the test statistic falls in the critical region, reject H0. Otherwise, fail to reject H0. 8. Interpret your decision. If the claim is the null hypothesis, you will either reject the claim or determine there is not enough evidence to reject the claim. If the claim is the alternative hypothesis, you will either support the claim or determine there is not enough evidence to support the claim.

The z-Test for a Mean A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. At = 0.05, do you have enough evidence to reject the company’s claim? 1. Write the null and alternative hypothesis. The fact that the sampling distribution is normal comes from the Central Limit Theorem studied in Chapter 5. 2. State the level of significance. = 0.05 3. Determine the sampling distribution. Since the sample size is at least 30, the sampling distribution is normal.

5. Find the rejection region. Since Ha contains the > symbol, this is a right-tail test. Rejection region 4. Find the critical value. 5. Find the rejection region. z0 1.645 6. Find the test statistic and standardize it. n = 52 = 232 s = 10 The critical value is 1.645. The cumulative area is .9500. Regardless of the decision rule that is used, the decision will be the same. 7. Make your decision. z = 1.44 does not fall in the rejection region, so fail to reject H0 8. Interpret your decision. There is not enough evidence to reject the company’s claim that there is at most 230 mg of sodium in one serving of its cereal.

Using the P-value of a Test to Compare Areas Area to the left of z 0.1093 = 0.05 z0 = –1.645 Rejection area 0.05 z = –1.23 P = 0.1093 z0 z For a P-value decision, compare areas. If reject H0. If fail to reject H0. The decision will be the same regardless of whether the test statistic is compared to the critical value or if the P-value of the test statistic is compared to the level of significance. When using critical value compare z-scores. When using P-values, compare areas. For a critical value decision, decide if z is in the rejection region If z is in the rejection region, reject H0. If z is not in the rejection region, fail to reject H0.

Homework 1-20 pgs. 357-358 Day2: 21-33 all pgs. 358-359 35-45 all pg