The average fruit fly will lay 388 eggs into rotting fruit. A biologist wants to see if the average will be fewer for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. \[ 391,371,389,387,398,378,364,380,392,375,362,357 \] What can be concluded at the the $\alpha=0.10$ level of significance level of significance? a. For this study, we should use $t$-test for a population mean b. The null and alternative hypotheses would be: 388 $0^{s}$ $0^{6}$ os 388 $0^{s}$ $\sigma^{\infty}$ $\sigma^{s}$ c. The test statistic $t+\sqrt{ }=$ (please show your answer to 3 decimal places.) $0^{s}$ d. The $p$-value $=$ (Please show your answer to 4 decimal places.)

Step 1 ：We are given a set of data representing the number of eggs laid by genetically modified fruit flies. The data is as follows: \[391, 371, 389, 387, 398, 378, 364, 380, 392, 375, 362, 357\]

Step 2 ：We are asked to perform a t-test to determine if the average number of eggs laid by these genetically modified fruit flies is significantly different from the average number of eggs laid by normal fruit flies, which is 388.

Step 3 ：The null hypothesis for this test is that the average number of eggs laid by the genetically modified flies is equal to 388. This can be represented as: \[H_0: \mu = 388\]

Step 4 ：The alternative hypothesis is that the average number of eggs laid by the genetically modified flies is not equal to 388. This can be represented as: \[H_1: \mu \neq 388\]

Step 5 ：We calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s) from our data. The sample mean is approximately 378.667 and the sample standard deviation is approximately 13.200.

Step 6 ：We then calculate the t-statistic using the formula: \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\] where \(\mu\) is the population mean, \(\bar{x}\) is the sample mean, s is the sample standard deviation, and n is the sample size. The calculated t-statistic is approximately -2.449.

Step 7 ：We also calculate the p-value, which is the probability of observing a result as extreme as the one we have (or more extreme) if the null hypothesis is true. The calculated p-value is approximately 0.0323.

Step 8 ：Since the p-value is less than the significance level of 0.10, we reject the null hypothesis. This means that there is significant evidence at the 0.10 level to suggest that the average number of eggs laid by the genetically modified fruit flies is different from 388.

Step 9 ：Final Answer: \[\boxed{t = -2.449, p = 0.0323}\]