Someone claims that firstyear students in New Jerse public u

Someone claims that first-year students in New Jerse public universities are as tall on average as first-year students in Pennsylvania public universities. To test that claim, you collect 100 students from each of the two states and obtain the following results: The mean height for New Jersey students is 65 inches and the mean height for Pennsylvania students is 64 inches. assume that population standard deviations are known and are 8 inches for both New Jersey and Pennsylvania. Using a significance level of 0.05, what is/are the critical values?

A. There are two critical values. They are -2.58 and +2.58

b. There is only one critical value. It is +2.58.

c. There is only one critical value. It is +1.96

d. There are two critical values. They are -1.96 and +1.96.

Continuing with the example about first-year universty students form the previous uestion, what is the sample value of the test statistic?

14.3019 (or -14.3019 dependent on how you set up the test statistic)

0.8839 (or -0.8839 dependent on how you set up the test statistic)

1.2357 (or -1.2357 dependent on how you set up the test statistic)

1.64 (or -1.64 dependent on how you set up the test statistic)

Using the critical values and the sample value tha toy uobtained in the previous two questions, what is your decision regaridng the null hypothesis?

Do not reject the null hypothesis.

Reject the null hypothesis. Accept the alternative hypothesis.

Stil continuing with the first-year university student example, what is the p-value given the information above?

0.1884

0.3768

0.4105

0.9500

Using the p-value you obtained in question 5 and the information given in question 2, what is your decision regarding the null hypothesis? (Please keep in mind that we always get the same decision with the p-value as with the 6-step procedure that you carried out previously.)

Reject the null hypothesis and accept the alternate hypothesis.

Do not reject the null hypothesis.

Solution

a)

dF =99 , significance = 0.05

Answer : d. There are two critical values. They are -1.96 and +1.96.

b)

z= (65-64) / sqrt( 64/100 + 64/100) = 0.8839

c)

Failed to reject null hypothesis .

d)

P-value = 2(1-0.8116) = 0.3768

e)

Failed to reject null hypothesis since p ( 0.3768) > 0.05.