# Hypotheses, Type I and Type II Errors, and Statistical Tests

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## ) Hypotheses, Type I and Type II Errors, and Statistical Tests

For each of the following tests, state the hypotheses, identify if it is a right-, left-, or two-tailed test and write Type I Error and Type II Error pertaining to the problem.

1. a) A professor of statistics states that the average student spends 3 hours studying for the midterm exam. Ho: μ = 3 vs. Ha: μ ≠ 3 Two-tailed test Type I Error: Concluding the average student doesn’t study 3 hours for midterm when in fact that is false. Type II Error: Not concluding the average student doesn’t study 3 hours for midterm when in fact that is true.
2. b) A spouse stated that the average amount of money spent on Christmas gifts for immediate family members is above \$1,200. Ho: μ = 1200 (or μ ≤ 1200) vs. Ha: μ > 1200 Right-tailed test Type I Error: Concluding the average amount of money spent on Christmas gifts is above \$1,200 when in fact that is false. Type II Error: Not concluding the average amount of money spent on Christmas gifts is above \$1,200 when in fact that is true.
3. c) The proportion of female students at UNF is less than 60%. Ho: p = .6 (or p ≥ .6) vs. Ha: p Left-tailed test Type I Error: Concluding the proportion of female students at UNF is less than 60% when in fact that is false. Type II Error: Not concluding the proportion of female students at UNF is less than 60%, when in fact that is true.
4. d) The average American drinks 2.5 cups of coffee per day. Ho: μ = 2.5 vs. Ha: μ ≠ 2.5 Two-tailed test Type I Error: Concluding the average American doesn’t drink 2.5 cups of coffee per day when in fact that is false. Type II Error: Not concluding the average American doesn’t drink 2.5 cups of coffee per day when in fact that is true.
5. e) The proportion of students that pass Mr. Johnson’s class is 70%. Ho: p = 0.7 vs. Ha: p ≠ 0.7 Two-tailed test Type I Error: Concluding the proportion of students passing Mr. Johnson’s class is not 70% when in fact, that is false. Type II Error: Not concluding the proportion of students passing Mr. Johnson’s class is not 70% when in fact, that is true.
6. f) The manager of the U. of Iowa bookstore claims that the average student spends less than \$400 per semester at the university’s bookstore. Ho: μ = 400 (μ ≥ 400) vs. Ha: μ Left-tailed test Type I Error: Concluding the average student spends less than \$400 per semester when in fact that is false. Type II Error: Not concluding the average student spends less than \$400 per semester when in fact that is true.
7. g) The average American adult drinks less than 3 cups of coffee per day. Ho: μ = 3(μ≥ 3) vs. Ha: μ Left-tailed test Type I Error: Concluding the average American drinks less than 3 cups of coffee per day when in fact that is false. Type II Error: Not concluding the average American drinks less than 3 cups of coffee per day when in fact that is true.
8. h) The average housewife works more than 40 hours per week in house related activities. Ho: μ = 40 (or μ ≤ 40) vs. Ha: μ > 40 Right-tailed test Type I Error: Concluding the average housewife works more than 40 hours per week, when in fact that is false. Type II Error: Not concluding the average housewife works more than 40 hours per week, when in fact that is true.
9. i) The average employee calls in sick 3 times a year. Ho: μ = 3 vs. Ha: μ ≠ 3 Two-tailed test Type I Error: Concluding the average employee doesn’t call in sick 3 times a year, when in fact that is false. Type II Error: Not concluding the average employee doesn’t call in sick 3 times a year, when in fact that is true.

10.1) The thickness of a plastic film (in mils) on a substrate material is thought to be influenced by the temperature at which the coating is applied. Eleven substrates are coated at 125 F, resulting in a sample mean coating thickness of 101.28 and a sample standard deviation of 15.08. Another 13 substrates are coated at 150 F, for which sample mean is 101.70 and standard deviation 20.15 are observed. It was originally suspected that reducing the process temperature would reduce mean coating thickness. Assume the populations are normal. (Assume population variances unknown, but equal). a) Do the data support the claim? Use 1% significance level. Hypotheses Ho: μ1= μ2 (μ1–μ2 = 0) Ha: μ122 because populations are normal and σ1, σ2 is unknown. Calculate Test Statistic = –0.057 from Excel output. P-value= P(T22 220 .20, which implies that p-value >0 .20. Decision: p-value >0.20 > α = 0.01 à Do not reject Ho à Conclusion: There is not enough statistical evidence at 1% significance level to infer that the population mean coating thickness decreases as the coating temperature decreases. b) Find a 95% CI on the difference in mean coating thickness. 95% C.I. for μ1–μ2 is (101.28 –101.70) ± t.025,22 se = –.42 ±2.074(7.3836) = –.42 ± 15.3136 = –15.73 to 14.89. We are 95% sure that the difference between the mean coating thickness at 125F and the mean coating thickness at 150F is between –15.73 and 14.89. 10.2) Eleven employees were put under the care of the company nurse because of high cholesterol readings. The nurse lectured them on the dangers of this condition and put them on a new diet. Shown are the cholesterol readings of the 11 employees both before the new diet and one month after use of the diet began. Assume differences in cholesterol readings are normally distributed in the population. First we calculate the differences and them we calculate the mean and the standard deviation of the differences to input them in the Excel FILE provided in Canvas. a)Construct a 98% confidence interval to estimate the populations mean difference of cholesterol readings for people who are involved in this program. 98% C.I. for for μ1–μ2 is 28.545 ± t.01,10 se= 28.545 ± 2.764(7.1508) =28.545 ± 19.76 = 8.78 to 48.31.We are 98% sure that average difference of cholesterol readings is between 8.78 and 48.31. b) Is there enough statistical evidence at 5% significance level to infer that the cholesterol reading has dropped by more than 15 after a month of using the new diet? Hypotheses Ho: =15 Ha: >15(right-tail) Significance level α =0.05 Test Statistic àd àT10 because differences are normally distributed. Calculate Test Statistic= 1.894 from Excel output. P-value= P(T10> 1.894). In the T-table, for 10 degrees of freedom, 1.894 is between 1.812 and 2.228 corresponding to probabilities of .05 and .025, respectively. Therefore, .025 10> 1.894)