According to the Carnegie unit system, the recommended number of hours students should study per unit is 2. Are statistics students' study hours different from the recommended number of hours per unit? The data show the results of a survey of 12 statistics students who were asked how many hours per unit they studied. Assume a normal distribution for the population. $4.1,1.4,1.8,3.6,1 \mathrm{Ys}_{5}, 3.9,3.7,1,0.5,4.5,4,3.6$ What can be concluded at the $\alpha=0.10$ level of significance? a. For this study, we should use $t$-test for a population mean b. The null and alternative hypotheses would be: $\infty^{\infty}$ $0^{s}$ $\sigma^{s}$ 2 2 $\sigma^{s}$ $\sigma^{\infty}$ $0^{\circ}$ c. The test statistic $t \quad$ (please show your answer to 3 decimal places.) $0^{\infty}$ d. The p-value $=$ (Please show your answer to 4 decimal places.)

Step 1 ：We are asked to perform a t-test to determine if the average study hours of the students is significantly different from the recommended 2 hours per unit.

Step 2 ：The null hypothesis is that the average study hours is equal to 2 hours per unit, and the alternative hypothesis is that the average study hours is not equal to 2 hours per unit.

Step 3 ：We will use a t-test for a population mean because we do not know the population standard deviation and the sample size is small (n<30).

Step 4 ：We will calculate the test statistic and the p-value to determine if we can reject the null hypothesis at the 0.10 level of significance.

Step 5 ：The test statistic t is calculated as \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, s is the sample standard deviation, and n is the sample size.

Step 6 ：Substituting the given values, we get \(t = \frac{2.8 - 2}{1.43/\sqrt{12}} \approx 1.938\).

Step 7 ：The p-value is the probability of observing a test statistic as extreme as the one calculated, given that the null hypothesis is true. It is calculated using the t-distribution with n-1 degrees of freedom.

Step 8 ：The p-value is approximately 0.079.

Step 9 ：Since the p-value is less than the significance level of 0.10, we reject the null hypothesis.

Step 10 ：This means that we have enough evidence to conclude that the average study hours of the students is significantly different from the recommended 2 hours per unit at the 0.10 level of significance.

Step 11 ：Final Answer: The test statistic \(t\) is approximately \(\boxed{1.938}\) and the p-value is approximately \(\boxed{0.079}\). We reject the null hypothesis at the \(\alpha=0.10\) level of significance. Therefore, we conclude that the average study hours of the students is significantly different from the recommended 2 hours per unit.