Question

A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 261 students was x¯¯¯x¯ = 130 minutes. Suppose that we know that the studey time follows a Normal distribution with standard deviation σσ = 65 minutes in the population of all first-year students at this university. Regard these students as an SRS from the population of all first-year students at this university. Does the study give good evidence that students claim to study more than 2 hours per night on the average?

(a) State null and alternative hypotheses in terms of the mean study time in minutes for the population.
(b) What is the value of the test statistic z?
(c) Can you conclude that students do claim to study more than two hours per weeknight on the average?

(a) H0:  Ha:  (Type in "mu" as the substitute for μμ and "!=" for ≠≠.)
(b) z:
(c) Conclusion:  (Answer with "Yes/Y" or "No/N".)

A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. The mean score on the final exam for a course using the old edition is 75. Ten randomly selected people who used the new text take the final exam. Their scores are shown in the table below.

Person	A	B	C	D	E	F	G	H	I	J
Test Score	85	78	70	88	67	96	74	90	81	93

Use a 0.010.01 significance level to test the claim that people do better with the new edition. Assume the standard deviation is 10.3. (Note: You may wish to use statistical software.)

(a) What kind of test should be used?

A. Two-Tailed
B. One-Tailed
C. It does not matter.

(b) The test statistic is  (rounded to 2 decimals).

(c) The P-value is

(d) Is there sufficient evidence to support the claim that people do better than 75 on this exam?

A. Yes
B. No

(e) Construct a 9999% confidence interval for the mean score for students using the new text.
<μ<<μ<

Answer 1

1)

a) H₀: < 120

    H₁: > 120

b) The test statistic z = ()/()

                                 = (130 - 120)/(65/)

                                 = 2.50

P-value = P(Z > 2.50)

             = 1 - P(Z < 2.50)

             = 1 - 0.9938

            = 0.0062

At 0.05 significance level, since the P-value is less than the significance level(0.0062 < 0.05), so we should reject the null hypothesis.

c) Yes, we can conclude that students do claim to study more than two hours per weeknight on the average.

2) a) Option - B) One-Tailed

b) = (85 + 78 + 70 + 88 + 67 + 96 + 74 + 90 + 81 + 93)/10 = 82.2

The test statistic z = ()/()

                              = (82.2 - 75)/(10.3/)

                              = 2.21

c) P-value = P(Z > 2.21)

                 = 1 - P(Z < 2.21)

                  = 1 - 0.9864

                  = 0.0136

d) since the P-value is greater than the significance level(0.0136 > 0.01), so we should not reject the null hypothesis.

No, there is not sufficient evidence to support the claim that people do better than 75 on this exam.

e) At 99% confidence level, the critical value is z_0.005 = 2.58

The 99% confidence interval is

+/- z_0.005 *

= 82.2 +/- 1.96 * 10.3/

= 82.2 +/- 6.384

= 75.816, 88.584