Question

The average fruit fly will lay 381 eggs into rotting fruit. A biologist wants to see if the average will change 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. 393, 398, 355, 365, 354, 384, 381, 386, 351, 350, 387, 394, 390, 370 What can be concluded at the the α = 0.10 level of significance level of significance?

1. For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
2. The null and alternative hypotheses would be:

H0:H0:  ? p μ  ? = > ≠ <       

H1:H1:  ? μ p  ? ≠ = > <    

3. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? ≤ >  αα
6. Based on this, we should Select an answer accept reject fail to reject  the null hypothesis.

Answer 1

Answer a)

In this case, we will be using t-test for a population mean.

Answer b)

The null and alternative hypotheses would be:

H0: μ = 381

H1: μ ≠ 381

Answer c)

For calculating test statistic, first, we need to calculate mean and standard deviation

Step 1: Find the mean

Mean = (393+398+355+365+354+384+381+386+351+350+387+394+390+370)/14 = 375.5714

The test statistic t = -1.160

Answer d)

P-value corresponding to t = -1.60 and df = 14-1 = 13 is 0.2669 (Obtained using calculator. screenshot below)

Answer e)

In this case α = 0.10, therefore

The p-value is > α

Answer f)

Since p-value = 0.2669 > α = 0.10, we fail to reject null hypothesis