A large manufacturing company investigated the service it received from suppliers and discovered that, in the past, 32% of all materials shipments were received late. However, the company recently installed a just-in-time system in which suppliers are linked more closely to the manufacturing process. A random sample of 110 deliveries since the just-in-time system was installed reveals that 22 deliveries were late. Use this sample information to test whether the proportion of late deliveries was reduced significantly. Let a = 0.05. a) Show the verification that the sampling distribution of the proportion is normal. b) Follow the procedure for tests of hypothesis and use a 5% level of significance. Use the following hypotheses... Họ: p = 0.32 %3D H1: p < 0.32

Tardigrades, or water bears, are a type of micro-animal famous for their resilience. In examining the effects of radiation on arganisms, an expert claimed that the amount of gamma radiation needed to sterilize a colany of tardigrades no longer has a mean of 1200 Gy (grays). (For comparison, humans cannot withstand more than 10 Gy.) A study was conducted on a sample of 21 randomly selected tardigrade colonies, finding that the amount of gamma radiation needed to sterilize a colony had a sample mean af 1225 Gy, with a sample standard deviation of 65 Gy. Assume that the population of amounts of gamma radiation needed to sterilize a colony of tardigrades is approximately normally distributed. Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level af significance, to support the claim that H, the mean amount of gamma radiation needed to sterilize a colony of tardigrades, is nat equal to 1200 Gy. (a) State the null hypothesis and the…

In a study comparing banks in two countries, a sample of 142 matched pairs of banks was formed. Each pair contained one bank from country x and one from country y. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (country x - country y) was 0.0579, and the sample standard deviation of the differences was 0.3086. Test, against a two-sided alternative, the null hypothesis that the two population means are equal. (Assume α = 0.01.) Click the icon to view the critical values of the Student's t distribution. The test statistic is t = 2.236. (Round to three decimal places as needed.) The critical value(s) is(are) (Round to three decimal places as needed. Use a comma to separate answers as needed.)

A cell phone manufacturer tests the battery lifetimes of its cell phones by recording the time it takes for the battery charges to run out while testers are streaming videos on the phones continuously. The manufacturer claims that the population mean of the battery lifetimes of all phones of their latest model is 5.26 hours. As a researcher for a consumer information service, you want to test that claim. To do so, you select a random sample of 45 cell phones of the Español manufacturer's latest model and record their battery lifetimes. Assume it is known that the population standard deviation of the battery lifetimes for that cell phone model is 2.42 hours. Based on your sample, follow the steps below to construct a 99% confidence interval for the population mean of the battery lifetimes for all phones of the manufacturer's latest model. Then state whether the confidence interval you construct contradicts the manufacturer's claim. (If necessary, consult a list of formulas.) 00 (a)…

It has been estimated that the 1991 G-car obtains a mean of 35 miles per gallon on the highway, and the company that manufactures the car claims that it exceeds the estimate in highway driving. To support its assertion, the company randomly selects 49 1991 G-cars and records the mileage obtained for each car over a driving course similar to that used to obtain the estimate. The following data resulted: x = 36.5 miles per gallon, s = 7 miles per gallon. %3D a. Find the observed significance level for testing Ho: u= 35 vs. Ha: u> 35. %3D b. For what value of significance level a, you will reject the null hypothesis.

c and d please

1. The data in Table 7–6 were collected in a clinical trial to evaluate a new compound designed to improve wound healing in trauma patients. The new compound was com- pared against a placebo. After treatment for 5 days with the new compound or placebo, the extent of wound healing was measured. Is there a difference in the extent of wound healing between the treatments? (Hint: Are treatment and the percent wound healing independent?) Run the appropriate test at a 5% level of significance. Use the data in Problem 1. Pool the data across the treatments into one sample of size n = 250. Use the pooled data to test whether the distribution of the percent wound healing is approximately normal. Specifically, use the following distribution 30%, 40%, 20%, and 10% and a = 0.05 to run the appropriate test.   Percent Wound Healing   Treatment 0-25% 26-50% 51-75% 76-100% New compound (n=125) 15 37 32 41 Placebo (n=125) 36 45 34 10

A father is concerned that his teenage son is watching too much television each day, since his son watches an average of 2 hours per day His son says that his TV habits are no different than those of his friends. Since this father has taken a stats class, he knows that can actually test to see whether or not his son is watching more TV than his peers The father collects a random sample of watching times from boys at his son's high school and gets the following data: 1.9, 2.3, 2.2, 1.9, 1.6, 2.6, 1.4, 2.0, 2.0, 2.2 Find a 95% confidence interval for the true mean number of hours teenagers at that school watch TV. a. Confidence Interval b. Conclusion

Clothes dried outside smell better than those dried in a dryer because of a process called photolysis. Sunlight breaks down compounds that cause odor. A trained smeller is asked to smell a random sample of 20 towels. The smeller does not know it, but all of the 20 towels have been dried in the sun. The smeller is blindfolded and is not allowed to touch the towels but leans forward to smell the towel then rates the towel on a smell-scale of 1 to 10, where 1 is the an extremely unpleasant smell and 10 is a very good smell. A smell rating of 8 is typically attributed to a towel that is dried in the dryer. The mean smell rating for these 20 towels is approximately normal with a mean of 8.45 and the standard deviation is 0.887. A statistician performs a test of Ho μ = 8 versus H₁ μ> 8 where is the true mean smell rating for all towels dried in the sun. This test yields a P-value of 0.0176. What conclusion would you make at the significance level a = 0.05? Because the P-value of 0.0176 < α =…

The Wall Street Journal asked Concur Technologies, Inc., an expense management company, to examine data from 8.3 million expense reports to provide insights regarding business travel expenses. Their analysis of the data showed that New York was the most expensive city. The following table shows the average daily hotel room rate (X) and the average amount spent on entertainment (Y) for a random sample of 9 of the 25 most-visited U.S. cities. These data lead to the estimated regression equation y = 17.49 + 1.0334x. For these data SSE = 1541.4. Use Table 1 of Appendix B.  a. Predict the amount spent on entertainment for a particular city that has a daily room rate of $89 (to 2 decimals). b. Develop a 95% confidence interval for the mean amount spent on entertainment for all cities that have a daily room rate of $89 (to 2 decimals). c. The average room rate in Chicago is $128. Develop a 95% prediction interval for the amount spent on entertainment in Chicago (to 2 decimals).