Question

1. ) An instructor believes that students do not retain as much information from a lecture on a Friday compared to a Monday. To test this belief, the instructor teaches a small sample of college students some preselected material from a single topic on statistics on a Friday and on a Monday. All students received a test on the material. The differences in exam scores for material taught on Friday minus Monday are listed in the following table.

Difference Scores (Friday − Monday)
+3.3
+4.5
+6.3
+1.1
−1.7

(a) Find the confidence limits at a 95% CI for these related samples. (Round your answers to two decimal places.)
to

(b) Can we conclude that students retained more of the material taught in the Friday class?

Yes, because 0 lies outside of the 95% CI.No, because 0 is contained within the 95% CI.    

2;) Listening to music has long been thought to enhance intelligence, especially during infancy and childhood. To test whether this is true, a researcher records the number of hours that eight high-performing students listened to music per day for 1 week. The data are listed in the table.

Music Listening Per Day (in hours)
4.1
4.8
4.9
3.7
4.3
5.6
4.1
4.3

(a) Find the confidence limits at a 95% CI for this one-independent sample. (Round your answers to two decimal places.)
to  hours per day

(b) Suppose the null hypothesis states that students listen to 3.5 hours of music per day. What would the decision be for a two-tailed hypothesis test at a 0.05 level of significance?

Retain the null hypothesis because the value stated in the null hypothesis is within the limits for the 95% CI.Reject the null hypothesis because the value stated in the null hypothesis is outside the limits for the 95% CI.    Reject the null hypothesis because the value stated in the null hypothesis is within the limits for the 95% CI.Retain the null hypothesis because the value stated in the null hypothesis is outside the limits for the 95% CI.

Answer 1

a) = (3.3 + 4.5 + 6.3 + 1.1 + (-1.7))/5 = 2.7

s_d = sqrt(((3.3 - 2.7)^2 + (4.5 - 2.7)^2 + (6.3 - 2.7)^2 + (1.1 - 2.7)^2 + (-1.7 - 2.7)^2)/4) = 3.1016

At 95% confidence interval the critical value is t^* = 2.777

The 95% confidence interval is

+/- t^* * s_d/

= 2.7 +/- 2.777 * 3.1016/

= 2.7 +/- 3.85

= -1.15, 6.55

b) Since the interval contains 0, so we cannot conclude that students retained more of the material taught in the Friday class.

2)a) = (4.1 + 4.8 + 4.8 + 3.7 + 4.3 + 5.6 + 4.1 + 4.3)/8 = 4.4625

s = sqrt(((4.1 - 4.4625)^2 + (4.8 - 4.4625)^2 + (4.8 - 4.4625)^2 + (3.7 - 4.4625)^2 + (4.3 - 4.4625)^2 + (5.6 - 4.4625)^2 + (4.1 - 4.4625)^2 + (4.3 - 4.4625)^2)/7) = 0.5878

At 95% confidence interval the critical value is t^* = 2.365

The 95% confidence interval is

+/- t^* * s/

= 4.4625 +/- 2.365 * 0.5878/

= 4.4625 +/- 0.4915

= 3.971, 4.954

= 3.97, 4.95

b) Reject the null hypothesis because the value stated in the null hypothesis is outside the limits for the 95% CI.