Question

II. USE THE INFORMATION BELOW TO ANSWER PROBLEMS 9-11 When I was a kid, I remember my grandparents buying copper bracelets to reduce arthritis pain. Researchers of the time, wondered if there was a difference between the use of copper and a sham treatment invented for the experiment. Pain was measured using a visual analog scale. Below is the TI- 84 output for the data. Assume equal population variances and assume the populations are independent and normally distributed. Sample Size Mean Copper 20 0.49 units Sham 20 0.44 units 90% CI for Mu Copper - Mu Sham: (-0.59, 0.69) Choose the best interpretation of the confidence interval in the TI-84 output. a) We are 90% confident that the pain reduction for the Copper group is greater than for b) We are 90% confident that the pain reduction for the Copper group is less than for the 9. the Sham group. d) Sham group We are 90% confident that there is no statistically significance difference in pain reduction for the Copper group and the Sham group. This Juestion can't be answered based on this information. None of the above conclusions are correct. Pick the formula that was used to compute the interval given: ** 10. a) c) (*. - *) tt .-Pt2 0:00 None of the above Leave Pick the appropriate tor z value used in the above interval. 1 PAGE 1 OF 1 BB WORDS

Answer 1

ANSWERS:

The hypotheses for this test.

Ho = Mu copper - Mu sham = 0

Ha = Mu copper - Mu sham  $\neq$ 0

9. We are 90% confident that there is no statisitcally significant difference in pain reduction for the copper group and the sham group (OPTION C)

Reason:  If a confidence interval includes the null value, then there is no statistically meaningful or statistically significant difference between the groups. If the confidence interval does not include the null value, then we conclude that there is a statistically significant difference between the groups .Here the null value is 0 which is included in the interval (-0.59,0.69)

10. OPTION C

$\overline{X1} - \overline{X2} \: \pm t\sqrt{S_{p}^{2}(1/n1+1/n2)}$

Reason : Since the sample mean and standard deviation are given and the population variance are equal and populations are independent and normallly distributed hence we use two sample t test for equal variances.

11. tc​ = 1.686. (OPTION A)

Reason : df=n1​+n2​−2=20+20−2=38.

The critical value for α=0.1 and df=38 degrees of freedom is tc​=t1−α/2;df = ​1.686.

Hope this helps.