Question

(4) 114 pts] Grades on a standardized test are known to have a mean of 1000 for students in Canada. The test is administered to 453 randomly selected students in Alberta; in this sample, the mean is 1013 and the sample standard deviation is 108 (a) Construct a 95% confidence interval for the average test score for Alberta students. (b) Is there statistically significant evidence (at the 5% level) that Alberta students per- form differently than other students in Canada? (c) Another 503 students are selected at random from Alberta. They are given a 3-hour preparation course before the test is administered. Their average test score is 1019 with a standard deviation of 95. (i) Construct a 95% confidence interval for the average test score for the change in rage test score associated with the prep course. (ii) Is there statistically significant evidence that the prep course helped? (d) The original 453 students are given the prep course and then are asked to take the test a second time. The average change in their test scores is 9 points, and the standard deviation of the change i s 60 points. (i) Construct a 95% confidence interval for the change in average test scores. (ii) Is there statistically significant evidence that students will perform better on their second attempt after taking the prep course? (iii) Students may have performed better in their second attempt because of the in their first attempt. prep course or because they gained test-taki Describe an experiment that would quantify these two effects. ing experience

Answer 1

Solution:-

a) 95% confidence interval for the average test score for Alberta students is  C.I = ( 1003.05, 1022.95).

$C.I = \bar{x}\pm z_{\alpha /2}\ast \frac{\sigma }{\sqrt{n}}$

C.I = 1013 + 1.96*5.07428

C.I = 1013 + 9.9456

C.I = ( 1003.05, 1022.95)

b)

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

Null hypothesis: u = 1000
Alternative hypothesis: u $\neq$ 1000

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

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

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

SE = s / sqrt(n)

S.E = 5.07428
z = (x - u) / SE

z = 2.56

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 z statistic less than -2.56 or greater than 2.56.

Thus, the P-value = 0.01.

Interpret results. Since the P-value (0.01) 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 prep course helped.

c)

i) 95% confidence interval for the average test score for Alberta students is  C.I = ( 1010.698, 1027.302).

$C.I = \bar{x}\pm z_{\alpha /2}\ast \frac{\sigma }{\sqrt{n}}$

C.I = 1019 + 1.96*4.23584

C.I = 1019 + 8.30225

C.I = ( 1010.698, 1027.302)

ii)

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

Null hypothesis: u = 1000
Alternative hypothesis: u $\neq$ 1000

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

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

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

SE = s / sqrt(n)

S.E = 4.23584
z = (x - u) / SE

z = 4.49

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 z statistic less than -4.49 or greater than 4.49.

Thus, the P-value = less than 0.001

Interpret results. Since the P-value (almost 0) 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 prep course helped.