A teacher claims her students’ average test score is 75. A researcher suspects it’s different. A sample of 25 students has a mean score of 78 with a standard deviation of 6.
$\left(A\right)$ Create a $90\%$ confidence interval for the mean test score.

• A. $\left\lbrack75.250,80.750\right\rbrack$
• B. $\left\lbrack75.948,80.052\right\rbrack$
• C. $\left\lbrack74.950,81.050\right\rbrack$
• D. $\left\lbrack74.550,81.450\right\rbrack$

Answer: (B)

Verified step by step guidance

1. Identify the sample statistics: sample mean \( \bar{x} = 78 \), sample standard deviation \( s = 6 \), and sample size \( n = 25 \).
2. Since the population standard deviation is unknown and the sample size is small, use the t-distribution to construct the confidence interval.
3. Determine the degrees of freedom \( df = n - 1 = 24 \) and find the t-critical value \( t_{\alpha/2} \) for a 90% confidence level from the t-distribution table.
4. Calculate the standard error of the mean using the formula: \[ SE = \frac{s}{\sqrt{n}} \]
5. Construct the confidence interval using the formula: \[ \left[ \bar{x} - t_{\alpha/2} \times SE, \quad \bar{x} + t_{\alpha/2} \times SE \right] \]