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. $(75.250,80.750)$
• B. $(75.948,80.052)$
• C. $(74.950,81.050)$
• D. $(74.550,81.450)$

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. Determine the confidence level, which is 90%, and find the corresponding critical value from the t-distribution with \( n - 1 = 24 \) degrees of freedom. This critical value is denoted as \( t^* \).
3. Calculate the standard error of the mean (SEM) using the formula: \[ SEM = \frac{s}{\sqrt{n}} \]
4. Compute the margin of error (ME) by multiplying the critical t-value by the standard error: \[ ME = t^* \times SEM \]
5. Construct the confidence interval for the population mean using the formula: \[ \left[ \bar{x} - ME, \ \bar{x} + ME \right] \]