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.
$\left(B\right)$ At the $0.10$ significance level, test the claim.

• A. $P-value=0.02$, fail to reject $H_0$. There is not enough evidence to suggest $\mu\ne75$
• B. $P-value=0.08$, reject $H_0$. There is enough evidence to suggest $\mu\ne75$
• C. $P-value=0.02$, reject $H_0$. There is enough evidence to suggest $\mu \ne 75$
• D. $P-value=0.08$, fail to reject $H_0$. There is not enough evidence to suggest $\mu\ne75$

Answer: (C)

Verified step by step guidance

1. Identify the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). Here, \(H_0: \mu = 75\) (the teacher's claim) and \(H_a: \mu \neq 75\) (the researcher's suspicion that the mean is different).
2. Determine the significance level \(\alpha = 0.10\) and note that this is a two-tailed test because the alternative hypothesis is \(\mu \neq 75\).
3. Calculate the test statistic using the formula for a t-test since the population standard deviation is unknown and the sample size is small (\(n=25\)): \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \(\bar{x} = 78\), \(\mu_0 = 75\), \(s = 6\), and \(n = 25\).
4. Find the degrees of freedom, which is \(df = n - 1 = 24\), and use the t-distribution to find the p-value corresponding to the calculated test statistic.
5. Compare the p-value to the significance level \(\alpha\): if \(p\)-value \(\leq \alpha\), reject \(H_0\); otherwise, fail to reject \(H_0\). This decision will tell you whether there is enough evidence to support the claim that the mean is different from 75.