Question

Jessika, Jenna and Lauren conducted a "tea test" comparing expensive loose leaf tea to some cheap dollar store tea. Ten people scored the loose leaf tea: 6, 1, 6, 4, 1, 6, 8, 5, 3, 7. Those same 10 people scored the cheap tea: 8, 8, 4, 6, 4, 8, 4, 3, 7, 3. Conduct a t-test to determine whether there is a statistically significant difference between the two teas. Use the .05 significance level. What is your decision regarding the null hypothesis?

Fail to reject the null hypothesis because p = .05. There is a small statistically significant difference between the two teas.

Reject the null hypothesis because p < .05. There is a statistically significant difference between the two teas as expected.

Fail to reject the null hypothesis because p > .05. There is a statistically significant difference between the two teas.

Fail to reject the null hypothesis because p < .05. There is no statistically significant difference between the two teas.

Reject the null hypothesis because p > .05. There is no statistically significant difference between the two teas.

Fail to reject the null hypothesis because p > .05. There is no statistically significant difference between the two teas.

None of the above

Jessika, Jenna and Lauren conducted a "tea test" comparing expensive loose leaf tea to some cheap dollar store tea. Ten people scored the loose leaf tea: 6, 1, 6, 4, 1, 6, 8, 5, 3, 7. Those same 10 people scored the cheap tea: 8, 8, 4, 6, 4, 8, 4, 3, 7, 3. Conduct a paired t-test to determine whether there is a statistically significant difference between the two teas. Use the .05 significance level. What is the p-value for this two-tailed test? (Answer with only three decimal places, like this: 5.555) ____

On a paired t-test, which hypothesis states there is no difference between the mean difference weights of different brands of cookies and zero?

null hypothesis

alternate hypothesis

research hypothesis

decision rule

critical hypothesis

Answer 1

The hypotheses for the test are:

H₀: = 0

H_a: != 0

(where is the mean population difference in tea scores)

Using a paired value T-test, the T-statistic for the given samples is,

T = -0.694

The p-value for the above T score for d.o.f= 10+10-2 = 18 is,

p = 0.505 (can be calculated using the T.TEST() function on the above samples in Excel for 2-tailed distribution and paired samples test option)

Hence, we fail to reject the null hypothesis because p > 0.05. There is no statistical difference between the the two teas.

As shown above, in a paired t-test, the null hypothesis states there is no difference between the mean difference weights of different brands of cookies and zero.