10.3

1.

  1. t0.01 = 2.602.

  2. -t0.05 = -1.729.

  3. ±t0.025 = ±2.179.

2.

  1. Critical value: 1.321.

  2. Critical value: -2.426.

  3. Critical value: ±2.738.

3.

  1. t0 = -1.379.

  2. -t0.05 = -1.714.

  3. Skip this problem.

  4. No enough evidence for the researcher to reject the null hypothesis because it is a left-tailed test and the test statistic is greater than the critical value (-1.379 > -1.714).

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 greater than the critical value (2.67>1.318) in a right tailed test.

5.

  1. t0 = 2.502.

  2. -t0.05 = -2.819;t0.05 = 2.819.

  3. Skip this problem.

  4. No enough evidence for the researcher to reject the null hypothesis, since the test statistic is between the critical values (-2.819 < 2.502 < 2.819).

  5. Lower bound: 99.39, Upper bound: 110.21. We do not reject the statement in the null hypothesis because the 99% confidence interval includes 100.

11.

Ho: μ = $67.

Ha: μ > $67.

  1. There is a 0.02 probability of obtaining a sample mean of $73 or higher from a population whos mean is $67. So, if we ontained 100 simple random samples of size n = 40 from a population those mean is $67, we would expect about 2 of these samples to result in sample means of $73 or higher.

  2. Because the P-value is low (P-value = 0.02 < a = 0.05) we reject the statement in the null hypothesis. There is sufficient evidence to conclude that the mean dollar amount withdrawn from a PayEase ATM is more than the mean amount from a standard ATM (more than $65).

13.

Ho: μ = 22.

Ha: μ > 22.

  1. The sample is random. The sample size is large, n = 200 > 30. We can assume that the sample is small reltive to the population, so the scores are independent.

  2. Classical approach: t0 = 2.176 > t0.05 = 1.660; reject the null hypothesis.

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

15.

Test Statistic = -4.553.

Reject the null hypothesis because there is sufficient evidence to conclude that the mean hippocampal volume in alcoholic adolescents is less than the normal mean volume of 9.02 cm^3.

17.

Test Statistic = 0.813.

Do not reject the null hypothesis because there is not sufficient evidence to conclude that the mean FICO score of high-income individuals is greater than that of the general population. Meaning, it is not unlikely to obtain a mena credit score of 714.2 or higher even though the true population mean credit score is 703.5.

19.

95% confidence interval: Lower Bound: 35.44 years, Upper Bound: 42.36 years.

Due to the interval including 40.7 years, there is not enough evidence to conclude that the mean age of 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;45.722.

3.

  1. Test Statistic = 20.496.

  2. Critical Value = 13.091.

  3. Skip this problem

  4. Do not reject the null hypothesis because test statistic > 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.

Do not reject the null hypothesis at the a = 0.05 level of significance. There is not sufficient evidence at the a = 0.05 level of significance to conclude that the standard deviation wait-time is less than 18.0 seconds.

15.

Test Statistic = 15.639.

Reject the null hypothesis at the a = 0.10 level of significance. There is sufficient evidence at the a = 0.10 level of significance to conclude that Rose is a more consistent plater than other shooting guards in the NBA.