10.3

1.

  1. Critical value = 2.602

  2. Critical value = -1.729

  3. Critical value = =/- 2.179

2.

  1. Critical value = 1.321

  2. Critical value = -2.426

  3. Critical value = +/-2.738

3.

  1. Test statistic = -1.38

  2. Critical value = -1.714

  3. Skip this problem.

  4. We do not reject the null hypothesis.The test is left-tailed but the test statistic is greater than the critical value.

4.

  1. Test statistic = 2.67

  2. Critical values = 1.318

  3. Skip this problem.

  4. We reject the null hypothesis because the test statistic is located within the critical region.

5.

  1. Test statistic = 2.53

  2. Critical values = -2.819, 2.819

  3. Skip this problem.

  4. We do not reject the null hypothesis because the test statistic is located between the critical values.

  5. Lower bound = 99.39; Upper bound = 110.21; We do not reject the null hypothesis because the 99% Confidence Interval includes the mean value of 100.

11.

Ho: Mean = $67

Ha: Mean > $67

  1. We have a 0.02 probability of obtaining a sample mean of $73 or higher from a population with a mean of $67. If we obtained 100 simple random samples of size n=40 from a population with a mean of $67, then we would expect about 2 of these samples to result in sample means of $73 or higher.

  2. Since the P-value is small compared to alpha, we reject the null hypothesis. We can conclude with the evidence provided that the mean dollar amount withdrawn from a PayEase ATM is more than the mean amount from a standard ATM.

13.

Ho: Mean = 22

Ha: Mean > 22

  1. The sample is random, and the sample size is large; n = 200, so n is greater than or equal to 30. This means we can reasonably assume that the sample is small relative to the population, so the scores are independent.

  2. Classical Approach: Test statistic = 2.176; Critical value = 1.660; We reject the null hypothesis since the test statistic is located within the critical region.

  3. We can conclude with the evidence that students who complete the core curriculum are scoring above 22 on the math portion of the ACT.

15.

Test statistic = -4.55

We reject the null hypothesis because the test statistic is located within the critical region.

17.

Test statistic = 0.813

We do not reject the null hypothesis because the test statistic does not end up being located in the critical region.

19.

Confidence Interval: Lower bound = 35.44 years; Upper bound = 42.36 years.

Since the interval includes 40.7 years, we do not reject the null hypothesis because there is no significant evidence to prove that the mean age of a death-row inmate has changed since 2002.

10.4

1.

  1. Critical value = 28.869

  2. Critical value = 14.041

  3. Critical values = 16.047 and 45.722

3.

  1. Test statistic = 20.496

  2. Critical value = 13.091

  3. Skip this problem

  4. We do not reject the null hypothesis because the test statistic is greater than the critical value.

13.

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

Test statistic = 6.422

We do not reject the null hypothesis because there is not enough evidence at the alpha = 0.05 level of significance to conclude that the standard deviation of wait-time is less than 18 seconds.

15.

Test statistic = 15.64

We reject the null hypothesis. We can conclude with the evidence at the alpha = 0.10 level of significance that Derrick Rose is a more consistent player than other shooting guards in the NBA.