Calculate the SEMean ii. Construct a 98% confidence interval for the population scores of Mrs. Paul’s students. iii. State the null and alternate hypothesis for this test.
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i. Calculate the SEMean
ii. Construct a 98% confidence interval for the population scores of Mrs. Paul’s students.
iii. State the null and alternate hypothesis for this test.
iv. Calculate the missing value ** and use tables to approximate the value of ***.
v. Using a 1% level of significance, state the decision criteria for this test
vi. State the conclusion for the test and why
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- A dentist wanted to determine if a relationship existed between childhood fluoride exposure and cavities. She took a sample of adults in her practice and counted how many cavities each person had in his or her permanent teeth. She also determined how many years of childhood each person was exposed to tap water with fluoride. The minimum value on this variable was 0 and the maximum was 18. Using the data below, an alpha of .05 and a two-tailed test, was there a significant relationship? Years of fluoride Number of cavities 0 10 18 1 2 7 12 3 3 4 10 5 M = 7.5s =6.98 M = 5.00s = 3.16Mrs. Paul wants to test that claim that, on average, her students average mark is 85. She pulls therecords of 100 of per past students over a three-year period. The grades inputted into MINITABfor analysis. Exhibit 1 below shows the results of an analysis carried out on the data obtained. Exhibit 1 One-Sample Z: Level I/IITest of mu = 85 vs mu not = 85The assumed sigma = 15.0 Variable N Mean StDev SEMean Z PStudents Score 100 75.4 12.99 * ** *** i. Calculate the SEMean ii. Construct a 98% confidence interval for the population scores of Mrs. Paul’s students.iii. State the null and alternate hypothesis for this test. iv. Calculate the missing value ** and use tables to approximate the value of ***.v. Using a 1% level of significance, state the decision criteria for this test…Mrs. Paul wants to test that claim that, on average, her students average mark is 85. She pulls therecords of 100 of per past students over a three-year period. The grades inputted into MINITABfor analysis. Exhibit 1 below shows the results of an analysis carried out on the data obtained.Exhibit 1One-Sample Z: Level I/IITest of mu = 85 vs mu not = 85The assumed sigma = 15.0 Variable N Mean StDev SE Mean Z PStudents Score 100 75.4 12.99 * ** *iv. Calculate the missing value ** and use tables to approximate the value of ***.v. Using a 1% level of significance, state the decision criteria for this test vi. State the conclusion for the test and why
- Mrs. Paul wants to test that claim that, on average, her students average mark is 85. She pulls therecords of 100 of per past students over a three-year period. The grades inputted into MINITABfor analysis. Exhibit 1 below shows the results of an analysis carried out on the data obtained.Exhibit 1One-Sample Z: Level I/IITest of mu = 85 vs mu not = 85The assumed sigma = 15.0 Variable N Mean StDev SE Mean Z P Students Score 100 75.4 12.99 * ** ***i. Calculate the SEMeanii. Construct a 98% confidence interval for the population scores of Mrs. Paul’s students.iii. State the null and alternate hypothesis for this test.iv. Calculate the missing value ** and use tables to approximate the value of ***v. Using a 1% level of significance, state the decision criteria for this test vi. State the conclusion for the test and whyA random sample of 8080 eighth grade students' scores on a national mathematics assessment test has a mean score of 265265. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 260260. Assume that the population standard deviation is 3030. At alphaαequals=0.140.14, is there enough evidence to support the administrator's claim? Complete parts (a) through (e).