Question

Question 11 1 pts In 2012, the mean weight of women was reported at 2012 = 162 lbs. Suppose that an investigator hypothesizes that weights are even higher in 2015. She collected a sample of 29 women in the year of 2015, from which she measured a mean weight of 158 lbs with a standard deviation of 10 lbs. An two-tailed alpha level = .05 is applied. Select ALL correct statement(s) below: The investigator aims to compare a sample mean (2015) to a population mean (2012). D A one-sample t test should be used to test this hypothesis. The degrees of freedom for this test is 28. The critical value for this test is 2.048 Based on the obtained t statistic and the critical t statistic, the investigator fails to reject the null hypothesis. The conclusion of this study is: in 2015, women's mean weight is significantly different from that in 2012.

Answer 1

Answer)

As the population s.d is unknown here we will use t distribution to estimate the answers

Null hypothesis Ho : u = 162

Alternate hypothesis Ha : u > 162

Alpha = 0.05

Degrees of freedom is = n-1 = 28

For df 28 and alpha 0.05

T crtical value from t table is = 1.701

Rejection region is if test statistics is > 1.701

Reject Ho

Test statistics is = (sample mean - claimed mean)/(s.d/√n)

Test statistics = (158 - 162)/(10/√29) = -2.15

As the test statistics is not greater than 1.701

We fail to reject ho

So we do not have enough evidence to conclude that mean weight is.greater than 162

So

Correct answers are

The investigator aims to compare sample mean (2015) with population mean

A one sample t test should.be used

The degrees of freedom are 28

Based on the test the investigator fails to reject the null hypothesis Ho