Construct a 95% confidence interval for the mean difference of the population given the following information. Would you reject or fail to reject the claim that there is no difference in the mean?
$d̄=-0.728$
$s_{d}=1.34$
$n=10$

• A. The 95% confidence interval = $\left(-1.152,-0.304\right)$; Fail to reject the claim that there is no difference in the mean.
• B. The 95% confidence interval = $\left(-1.686,0.230\right)$; Fail to reject the claim that there is no difference in the mean.
• C. The 95% confidence interval = $\left(-1.686,0.230\right)$; Reject the claim that there is no difference in the mean.
• D. The 95% confidence interval = $\left(-1.152,-0.304\right)$; Reject the claim that there is no difference in the mean.

Answer: (B)

Verified step by step guidance

1. Step 1: Identify the given information. The sample mean difference (d̄) is -0.728, the standard deviation of the differences (s_d) is 1.34, and the sample size (n) is 10. The confidence level is 95%.
2. Step 2: Calculate the standard error of the mean difference (SE_d). The formula for the standard error is SE_d = s_d / sqrt(n), where s_d is the standard deviation and n is the sample size.
3. Step 3: Determine the critical t-value for a 95% confidence level with degrees of freedom (df = n - 1). Use a t-distribution table or statistical software to find the t-value corresponding to a two-tailed test with df = 9.
4. Step 4: Calculate the margin of error (ME) using the formula ME = t * SE_d, where t is the critical t-value and SE_d is the standard error calculated in Step 2.
5. Step 5: Construct the confidence interval using the formula: Confidence Interval = d̄ ± ME. Substitute the values of d̄, ME, and calculate the lower and upper bounds of the interval. Finally, compare the interval to the claim (0 difference) to decide whether to reject or fail to reject the null hypothesis.