Question

For the three questions below…..think z scores and the area under the curve.

3. A nursing professor was curious as to whether the students in a very large class she was teaching who turned in their tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 tests was 78. Did the students turning in their tests first score significantly different from the mean? Explain. 8 points

4. Do college students from private schools have higher than average test-taking skills? A researcher administered a standard measure of test-taking skills to a randomly chosen student from a private university and found him to have a score of 68. The average on this measure for the population in general is 60 with a standard deviation of 10. What should you conclude about whether this student from a private university has higher test-taking ability? Explain. 8 points

5. Does computer-assisted instruction help community college health science students with reading difficulties to learn reading skills at a faster than normal rate? A researcher arranged for one of these students to have access to a set of computer-learning programs instead of the normal reading curriculum for one term. At the end of the term, the researcher tested her on a standardized reading ability test on which the mean for students with reading difficulties is 36 with a standard deviation of 6. The test participant scored 55. What would you conclude? Explain. 8 points

Answer 1

Solution:-

3)

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

Null hypothesis: u = 75
Alternative hypothesis: u 75

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 2.23607

DF = n - 1

D.F = 19
t = (x - u) / SE

t = 1.342

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 19 degrees of freedom is less than -1.342 or greater than 1.342.

Thus, the P-value = 0.195

Interpret results. Since the P-value (0.195) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that students turning in their tests first score significantly different from the mean.