Please label all answers with the appropriate alphabet (a, b, c, or d) Biostatis
ID: 3292163 • Letter: P
Question
Please label all answers with the appropriate alphabet (a, b, c, or d) Biostatistics 2nd Exam October 12, 2016 Name HU ID # 0-15 . Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith (Experience Teacher) and Mrs. C students, and Mrs. Jones had 25 students. At the end of the year, ( Jones (First Year Teacher. After the assignment, Mrs. Smith had 30 each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15. (Assume that student performance is approximately normal.) (a) Choose an appropriate Test Statistic to test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective (no difference in mean grades) teachers at a 5% level of Ssignificance. (b) Give the Null and Alternative Hypotheses. (c) 9a-Draw a curve showing the Critical Value(s) and make a Statistical Decision. (d) write a one-sentence ConclusionExplanation / Answer
Solution:-
1)
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 two-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.9661
DF = 53
t = [ (x1 - x2) - d ] / SE
t = - 7.25
tcritical = + 2.006
Since t value of the test lies in the rejection region, hence we have to reject the null hypothesis.
- 2.006 > t > 2.006
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size 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 the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 53 degrees of freedom is more extreme than - 7.25; that is, less than - 7.25 or greater than 7.25.
Thus, the P-value = less than 0.00001
Interpret results. Since the P-value (0.00001) is less than the significance level (0.05), we cannot accept the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that Mrs. smith and Mrs. Jones are equally effective teachers.