7.2 Hypothesis Testing for the Mean (Large Samples. Statistics Mrs. Spitz Spring 2009. Objectives/Assignment. How to find critical values in a normal distribution How to use the z-test to test a mean How to find P-values and use them to test a mean Assignment: pp. 324-327 #1-36.
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Reminder: If you cannot find the exact area in Table 4, use the area that is closest. For instance, in Ex. 1, the area closest to 0.01 is 0.0099
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Ho: ≥ $45,000
Ha: < $45,000 (Claim)
Because the test is a left-tailed test and the level of significance is = 0.05, the critical value is zo = -1.645 and the rejection region is z < -1.645. Because the sample size is at least 30, the standardized test statistic for the z-test is:
Ho: = $8390
Ha: $8390 (Claim)
Because the test is a two-tailed test and the level of significance is = 0.05, the critical values are
-zo = -1.96 and are zo = 1.96 the rejection region is
z < -1.96 and z > 1.96. Because n ≥ 30, the standardized test statistic for the z-test is:
Ho: ≥ 30 minutes
Ha: < 30 minutes (Claim)
The standardized test statistic for the z-test is:
Ho: = $143,260
Ha: $143,260 (Claim)
The level of significance is = 0.05. Using the z-test, the standardized test statistic is:
Using Table 4, the area corresponding to
z = -1.51 is 0.0655. Because the test is a two-tailed test, the P-value is equal to twice the area to the left of z = -1.51. So,
Because the P-value is greater than , you should fail to reject the null hypothesis. So there is not enough evidence at the 5% level of significance to conclude that the mean franchise investment is not $143,260.