Question

The two data sets in the table below are dependent random samples. The population of ( x − y ) (x-y) differences is approximately normally distributed. A claim is made that the mean difference ( x − y ) (x-y) is less than -31.4.

X 25 32 48 37 39 34 37

Y 73 64 66 80 78 67 84

For each part below, enter only a numeric value in the answer box. For example, do not type "z =" or "t =" before your answers. Round each of your answers to 3 places after the decimal point.

(a) Calculate the value of the test statistic used in this test. Test statistic's value =

(b) Use your calculator to find the P-value of this test. P-value =

(c) Use your calculator to find the critical value(s) used to test this claim at the 0.1 significance level. If there are two critical values, then list them both with a comma between them. Critical value(s) =

(d) What is the correct conclusion of this hypothesis test at the 0.1 significance level?

-There is not sufficient evidence to warrant rejection the claim that the mean difference is less than -31.4

-There is not sufficient evidence to support the claim that the mean difference is less than -31.4

-There is sufficient evidence to warrant rejection the claim that the mean difference is less than -31.4

-There is sufficient evidence to support the claim that the mean difference is less than -31.4

Answer 1

As the claim is that the mean difference is less than -31.4, we can formulate the null and alternative hypotheses as:

H0: >= -31.4

H1: < -31.4

Now, as the samples are dependent, we carry out 2-sample t-test.

Mean of sample 1, = 36, Mean of sample 2, = 73.14

Std deviation of sample 1 s1 = 7.024, Std deviation of sample 2, s2 = 7.755

a) Test statistic, t =  [ () ] / sqrt[(s1^2/n1) + (s2^2/n2)]

= -9.39 (plugging in values calculated above)

b) P-value ~ 0 (using t-score calculator and calculating p-value using t-score = -9.39 and d.o.f. = 7+7-2 = 12)

c) Critical value of t-score at 0.1 significance level for this left tailed test (there is a single such critical value since this is a left tailed test)

= -1.356

d) Since the absolute value of t-statistic, | t | >> 1.356 (abs. critical t-score), we reject the null hypothesis at the 0.1 significance level. Hence, there is not sufficient evidence to warrant rejection the claim that the mean difference is less than -31.4.