10.3

1.

  1. t001 = 2.602

  2. -t0.05 = -1.729

  3. +/- t0.025 = +/- 2.179

2.

  1. t0.1 = 1.323

  2. -t0.01 = -1.304

  3. +/- t0.01 = +/- 2.449

3.

  1. t0 = -1.379

  2. -t0.05 = -1.714

  3. Skip this problem.

  4. Type answer here. ot enough evidence to reject the null hypothesis, as the test statistic is greater than the critical value.

4.

  1. t0 = 2.674

  2. t0.1 = 1.318

  3. Skip this problem.

(d)Test value is greater than t0.1, so there is not enough evidence to reject the null hypothesis.

5.

  1. t0 = 2.502

  2. -t0.005 = -2.819, +t0.005 = 2.189

  3. Skip this problem.

  4. There is not enough evidence to reject the null hypothesis, asthe test stat is between the two critical values.

  5. There is not enough evidence to reject the null hypothesis.

11.

Ho: u = 67

Ha: u > 67

  1. The probability of getting a sample mean of $73+ is 0.02.

  2. The P value is lower than the alpha, so we reject the null hypothesis.

13.

Ho: u = 22

Ha: u > 22

  1. Sampleing is random due to the large population.

  2. t0 = 2.176, > t0.05 = 1.660. So we reject the nll hypothesis.

  3. There is sufficient informaton to say that those who complete the curriculum score above 22 on the ACT.

15.

Test statistic = t0 = -2.553 > -t0.01 = -2.718,

There is sufficient evidence to reject the null hypotehsis.

17.

t= 0.813 < t0.05 = 1.685.

There is sufficient evidence to reject the null hypothesis.

19.

LB = 35.44 UB = 42.36. These are the bounds for the 95% confidence interval.

There is not sufficient information to reject the null hypothesis.

10.4

1.

  1. 28.869

  2. 14.041

  3. chi0.975 = 16.047 chi0.025 = 45.722

3.

  1. 20.496

  2. 13.091

  3. Skip this problem

  4. chi0 is larger than chi0.95, so the null hypothesis is not rejected.

13.

Note. The sample standard deviation of the data shown is.\(15.205043\). You will have to knit this to see the number though.

chi0 = 6.422 chi0.95 = 3.325, p-value = 0.303

There is not sufficient information to substantiate rejecting the null hypothesis.

15.

chi0 = 15.639 chi0.9 = 15.659 p value = 0.0993

There is sufficient information to substantiate rejecting the null hypothesis.