Scenario #2 Mouse Genetics A study of 25 genes involved in spermatogenesis (sper

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Scenario #2 Mouse Genetics

A study of 25 genes involved in spermatogenesis (sperm formation) found their locations in the mouse genome. The study was carried out to test a prediction of evolutionary theory that such genes should occur disproportionately often on the X chromosome (more often than expected by chance). If genes for spermatogenesis occurred “randomly” throughout the genome, then we would expect only 6.1% of them to fall on the X chromosome because the X chromosome contains 6.1% of the genes in the genome. As it turned out, 10 of the 25 spermatogenesis genes were on the X chromosome (Wang et al. 2001). Do the results support the hypothesis that spermatogenesis genes occur preferentially on the X chromosome? Use hypothesis testing to evaluate your claim.

a.State the null and alternative hypothesis

b.Calculate the appropriate test statistics. You may use R if needed.

c.Suppose you calculate the p-value to be less than 0.0001, using a binomial probability distribution. Interpret the p-value in the context of the problem (actually interpret what the p-value measures, not whether you reject or fail to reject)

d.Do you reject or fail to reject the null hypothesis? Explain why you made your choice beyond the comparison of the p-value and significance level

e.State the conclusion of the hypothesis test in the context of the problem. (Remember the five parts)

f.In order to conduct our test, we made an assumption that the data collection followed a binomial experiment. Identify each assumption of the binomial experiment and evaluate if it is satisfied in this situation (explain in context of problem).

i.Assumption 1:

ii.Assumption 2:

iii.Assumption 3:

iv.Assumption 4:

v.Assumption 5:

Scenario #3 Germination Rates

You heard that a new voracious variety of native bunch grass seed has been developed for ecologically friendly landscaping surrounding golf courses. The goal of your research, as greenhouse manager at a prestigious golf course, is to evaluate the seed germination rate claim of 75% or if it is less than the reported rate. To evaluate the germination rate, you sow 20 seeds in a greenhouse so that each seed will be exposed to identical conditions. You then recorded the total number of seedlings that emerged. You observed 13 seeds germinate out of the 20. Use hypothesis testing to evaluate your claim.

a.State the null and alternative hypothesis

b.Calculate the appropriate test statistics. You may use R if needed.

c.Suppose you calculate the p-value to be 0.2142, using a binomial probability distribution. Interpret the p-value in the context of the problem (actually interpret what the p-value measures, not whether you reject or fail to reject)

d.Do you reject or fail to reject the null hypothesis? Explain why you made your choice beyond the comparison of the p-value and significance level.

e.State the conclusion of the hypothesis test in the context of the problem. (Remember the five parts)

g.In order to conduct our test, we made an assumption that the data collection followed a binomial experiment. Identify each assumption of the binomial experiment and evaluate if it is satisfied in this situation (use context of problem in explanations).

i.Assumption 1:

ii.Assumption 2:

iii.Assumption 3:

iv.Assumption 4:

v.Assumption 5:

Explanation / Answer

Solution:-

2)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.061

Alternative hypothesis: P > 0.061

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.0479

z = (p - P) /

z = 7.08

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than 0.0001.

Thus, the P-value = almost 0.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we do not have sufficient evidence that only 6.1% of them to fall on the X chromosome.

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