Statistics students believe that the mean score on a first statistics test is 65 . The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores:
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline Grades & 65 & 64.3 & 64.3 & 64.3 & 68.4 & 85.5 & 66.5 & 73.5 & 65 & 74.4 \\
\hline
\end{tabular}
Test grades are believed to be normally distributed.
Use a significance level of $5 \%$.
A. State the alternative hypothesis: $H_{A}$ :
B. State the mean of the sample:
(Round to two decimal places.)
C. State the standard error of the sample means:
(Round to four decimal places.)
D. State the test statistic: $t=$ (Round to four decimal places.)
E. State the p-value:
(Round to four decimal places.)
F. Decision:
Reject the null hypothesis.
Do not reject the null hypothesis.

Step 1 ：A. The alternative hypothesis, \(H_{A}\), is that the mean score is greater than 65.

Step 2 ：B. The mean of the sample is calculated as follows: \((65 + 64.3 + 64.3 + 64.3 + 68.4 + 85.5 + 66.5 + 73.5 + 65 + 74.4) / 10 = 69.12\)

Step 3 ：C. The standard error of the sample means is calculated using the formula: \(\frac{s}{\sqrt{n}}\), where s is the standard deviation of the sample and n is the number of observations. The standard deviation of the sample is calculated as follows: \(\sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\), where \(x_i\) are the individual observations, \(\bar{x}\) is the sample mean, and n is the number of observations. After calculating, the standard error is 0.2104.

Step 4 ：D. The test statistic, t, is calculated using the formula: \(\frac{\bar{x} - \mu}{SE}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, and SE is the standard error. After calculating, the test statistic is 1.9562.

Step 5 ：E. The p-value is calculated using a t-distribution table or a statistical software. After calculating, the p-value is 0.0423.

Step 6 ：F. Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. Therefore, there is sufficient evidence to support the instructor's claim that the mean score is higher than 65.