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?
$\bar{d}=-0.728$
$s_d=1.34$
$n=10$

• A. The 95% confidence interval = $(-1.152,-0.304)$; Fail to reject the claim that there is no difference in the mean.
• B. The 95% confidence interval = $(-1.686,0.230)$; 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 = $(-1.152,-0.304)$; 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 is \( \bar{d} = -0.728 \), the standard deviation of the differences is \( s_d = 1.34 \), and the sample size is \( n = 10 \). The confidence level is 95%.
2. Step 2: Calculate the standard error of the mean difference using the formula \( SE = \frac{s_d}{\sqrt{n}} \). Substitute \( s_d = 1.34 \) and \( n = 10 \) into the formula.
3. Step 3: Determine the critical value (\( t^* \)) for a 95% confidence interval with \( n-1 = 9 \) degrees of freedom. Use a t-distribution table or statistical software to find \( t^* \).
4. Step 4: Compute the margin of error (ME) using the formula \( ME = t^* \cdot SE \). Multiply the critical value by the standard error calculated in Step 2.
5. Step 5: Construct the confidence interval using the formula \( \text{CI} = \bar{d} \pm ME \). Subtract and add the margin of error to the sample mean difference \( \bar{d} \) to find the lower and upper bounds of the confidence interval. Finally, compare the interval to zero to determine whether to reject or fail to reject the claim that there is no difference in the mean.