3. (20 points) It is supposed that the average number of hours an adult sleeps from Friday night to Saturday morning is 7.5 hours. A researcher believes that in general college students sleep less that this from Friday night to Saturday morning. The researcher selects 40 students at random and finds that on average they sleep 6.5 hours with a (sample) standard deviation of 1.95 hours. (Assume that the population is approximately normally distributed.) $H_o$: (a) Set up a hypothesis to be tested at the $\alpha = 0.01$ level of significance. $H_1$: (b) What is the critical value for the test? Sketch a graph of the critical region. (c) What is the value of the test variable (the test statistic)? (d) What is the $P$-value? (e) What is the conclusion? 4. (20 points) Suppose that 66% of Americans heat their homes with natural gas. A random sample of 200 homes found that 135 used natural gas. Does the sample support the claim that the percentage has changed. Use the $\alpha = 0.10$ level of significance. (a) Set up a hypothesis to be tested. $H_o$: $H_1$: (b) What is the critical value for the test? Sketch a graph of the critical region. (c) What is the value of the test variable (the test statistic)? (d) What is the $P$-value? (e) What is the conclusion?
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01 level of significance. Ho: μ = 7.5 (null hypothesis) H1: μ < 7.5 (alternative hypothesis) For the natural gas study: a) Set up a hypothesis to be tested. Ho: p = 0.66 (null hypothesis) H1: p ≠ 0.66 (alternative hypothesis) Show more…
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A random sample of 10 students contains the following observations, in hours, for time spent studying in the week before final exams: 28, 57, 42, 35, 61, 39, 55, 46, 49, 38. Assume that the population distribution is normal. Test at the 5% significance level the null hypothesis that the population mean is 40 hours against the alternative that it is higher. (NOTE: We need to assume that class sizes are normally distributed in order to use the t distribution because the sample size n=10 < 25, and the Central Limit Theorem does not apply.) A. Since the test statistic equals 1.50, and the critical t value is 1.833, we fail to reject the null hypothesis and conclude that there is not enough evidence to determine that hours of work is truly higher than 40 hours/week. B. Since the test statistic equals 1.83, and the critical t value is 1.50, we reject the null hypothesis and conclude that there is enough evidence to determine that hours of work is higher than 40 hours/week. C. Since the test statistic equals 1.50, and the critical t value is 2.262, we fail to reject the null hypothesis and conclude that there is not enough evidence to determine that hours of work is truly higher than 40 hours/week. D. Since the test statistic equals 1.50, and the critical t value is 1.812, we fail to reject the null hypothesis and conclude that there is not enough evidence to determine that hours of work is truly higher than 40 hours/week.
Adi S.
1- A random sample of 10 students contains the following observations, in hours, for time spent studying in the week before final exams: 28, 57, 42, 35, 61, 39, 55, 46, 49, 38. Assume that the population distribution is normal. Test at the 5% significance level the null hypothesis that the population mean is 40 hours against the alternative that it is higher. (NOTE: We need to assume that class sizes are normally distributed in order to use the t distribution because the sample size n=10 < 25, and the Central Limit Theorem does not apply.) A. Since the test statistic equals 1.50, and the critical t value is 1.833, we fail to reject the null hypothesis and conclude that there is not enough evidence to determine that hours of work is truly higher than 40 hours/week. B. Since the test statistic equals 1.83, and the critical t value is 1.50, we reject the null hypothesis and conclude that there is enough evidence to determine that hours of work is higher than 40 hours/week. C. Since the test statistic equals 1.50, and the critical t value is 2.262, we fail to reject the null hypothesis and conclude that there is not enough evidence to determine that hours of work is truly higher than 40 hours/week. D. Since the test statistic equals 1.50, and the critical t value is 1.812, we fail to reject the null hypothesis and conclude that there is not enough evidence to determine that hours of work is truly higher than 40 hours/week. 2- Screws Inc. has asked for assistance in determining whether the manufacturing process is operating correctly. When the process is operating properly, it produces screws whose weights are normally distributed with a population mean of 5 grams and a population standard deviation of 0.1 grams. The manager wishes to know if the recent change to a new materials supplier has lowered the weight of the screws. In a random sample of 16 screws, the sample mean weight is 4.962 grams. Is there evidence to conclude that lower weight screws are now being produced at a significance level of 5%? The p-value is _______. What is your conclusion?
Maitreya E.
A researcher is testing a hypothesis of a single mean. The critical t value for α = .05 and a two-tailed test is +2.0796. The observed t value from sample data is -2.92. The decision made by the researcher based on this information is to _____ the null hypothesis. a. reject b. not reject c. redefine d. change alternate hypothesis e. restate A researcher wants to estimate the proportion of the population which possesses a given characteristic. A random sample of size 800 is taken resulting in 360 items which possess the characteristic. The point estimate for this population proportion is _______. a. 0.55 b. 0.45 c. 0.35 d. 0.65 e. 0.70 A researcher wants to estimate the population proportion with a 90% level of confidence. She estimates from previous studies that the population proportion is no more than .30. The researcher wants the estimate to have an error of no more than .02. The necessary sample size is at least _______. a. 29 b. 47 c. 298 d. 1421 e. 1500 Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers. Accordingly, he plans a sample of waiting times for walk-in customers. If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is ________. a. 0.9974 b. 0.4987 c. 0.9772 d. 0.4772 e. 0.5000 A researcher wants to conduct a before/after study on 11 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be -2.40 with a sample standard deviation of 1.21. Assume that the differences are normally distributed in the population. The degrees of freedom for this test are _______. a. 11 b. 2 c. 9 d. 20 e. 10
Keondre P.
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