Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Two-Sample Hypothesis The results of a mathematics placement exam at two differe

ID: 3321877 • Letter: T

Question

Two-Sample Hypothesis The results of a mathematics placement exam at two different campuses of Mercy College follow Campus Sample Size Mean Population Std. Deviation 330 310 2 31 Answer the following Assume you are trying to conclude that the exam scores at Campus 1 are highe than the exam scores at Campus 2 a. Determine if the appropriate hypothesis test should be an upper tail, lower tail, o 2-tail test. b. Determine the value of the test statistic. C. Determine the critical z or t value using a level of significance of 5%. d. Determine the p-value e. State your conclusion: Reject Ho or Do not Reject Ho? (Using a significance level of 5%). f. What is the probability of making a Type I Error?

Explanation / Answer

SOlution:-

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

Null hypothesis: 1< 2
Alternative hypothesis: 1 > 2

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample z-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.5933
z = [ (x1 - x2) - d ] / SE

z = 3.37

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a z statistic of 3.37.

Therefore, the P-value in this analysis is 0.0004

Interpret results. Since the P-value (0.0004) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that scores at campus 1 are higher than scores at campus 2.

The probability of type I error is 0.05.