Question

Jessika, Jenna and Lauren conducted a "tea test" comparing expensive loose leaf tea to some cheap...

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

Homework Answers

Answer #1

The hypotheses for the test are:

H0: = 0

Ha: != 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.

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