Solutions HWA6

12.16:

1. H0: mu = 60

2. HA: mu > 60

3. Alpha = .05

4. x-bar_s = 66.3, t_s = (66.3 - 60)/(s/sqrt[n]) = 6.3/(8.4/sqrt[12]) = 2.598

5. P-value = Pr(t_{11} > t_s) = Pr(t > 2.598) = tcdf(2.598,99999999,11) = .0124 < alpha, so reject H0.

Using Table 4 we get .01 < P-value < .02.

6.. There is sufficient evidence (.01 < P-value < .02) to conclude that population average drying time is greater than 60 minutes.

12.22

1. H0: mu = 1.2

2. HA: mu not = 1.2

3. Alpha = .05

4. x-bar_s = 1.2275, t_s = (1.2275 – 1.2)/(s/sqrt[n]) = .0275/(.03864/sqrt[12]) = 2.4653

5. P-value = Pr(|t_{11}| > t_s) = 2*Pr(t > 2.4653) = 2*tcdf(2.4653,99999999,11) = .0314 < alpha, so reject H0.

Using Table 4 we get .02 < P-value < .04.

6.. There is sufficient evidence (.02 < P-value < .04) to conclude that population average thickness is greater than 1.2 mm.

12.24

1. H0: pi = .60

2. HA: pi > .60

3. Alpha = .05

4. K_s = 70, z_s = (69.5 - 60)/(sqrt[n(.60)(.40)]) = 1.9392

5. P-value = Pr(Z > 1.9392) = .0262 < alpha, so reject H0.

6.. There is sufficient evidence (P-value < .0262) to conclude that player has improved. He makes better than 60% of his shots.