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In this assignment you will practice building and interpreting confidence intervals. Confidence intervals are used to give a range of plausible values for the population mean or proportion under consideration. If you have ever read the results of a poll, you might recall that a percentage is given followed by the statement, "...plus or minus..." some value. For example, the presidential approval rating is 49% plus or minus 1.73%. The 49% is the sample proportion with the 1.73% noting the margin of error. Understanding how confidence intervals are built and interpreted will help you become a better consumer of information. Confidence intervals are used in a variety of applications besides analyzing polling data. For instance, manufacturing clothing (sizes are based on an average plus or minus an acceptable error), volume of food items (the 12 oz cereal box holds on average 12 oz plus or minus a margin of error), lifetimes of items (a lightbulb will last on average 400 hours plus or minus an acceptable error), etc. Task Based on USDA suggestion, a U.S. adult should be spending anywhere from 165 to 345 per month on food, depending on age and gender. A random sample of 120 adults in Washington State is polled regarding the amount of money they spend each month on food. The sample has an average of 273.61 with a standard deviation of 61.59. Based on this information, complete the following questions. a) (5pts)Show that the requirements to build a confidence interval are met. b) (5 pts) State the critical value for a 90% confidence interval c) (5pts)Compute the 90% confidence interval. State values to two decimal places. d) (5pts) Interpret the 90% confidence interval using complete sentences.
The TTC has set a bus mechanical reliability goal of 3,900 bus miles. Bus mechanical reliability is measured specifically as the number of bus miles between mechanical road calls. Suppose a sample of 100 buses resulted in a sample mean of 3,950 bus miles and a sample standard deviation of 175 bus miles. Complete parts (a) and (b) below. a. Is there evidence that the population mean bus miles is more than 3,900 bus miles? (Use a 0.05 level of significance.) State the null and alternative hypotheses. H0: μ = T H1: μ ≠ T (Type integers) Find the test statistic for this hypothesis test. The test statistic t= (Round to two decimal places as needed.) The critical value for the test statistic is(are) . (Round to two decimal places as needed) Is there sufficient evidence to reject the null hypothesis using α=0.05 ? A. Do not reject the null hypothesis. There is insufficient evidence at the 0.05 level of significance that the population mean bus miles is less than 3,900 bus miles. B. Reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance that the population mean bus miles is greater than 3,900 bus miles. C. Reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance that the population mean bus miles is less than 3,900 bus miles. D. Do not reject the null hypothesis. There is insufficient evidence at the 0.05 level of significance that the population mean bus miles is greater than 3,900 bus miles. b. The p-value is . (Round to 3 decimal places as needed.) What does this p-value mean given the results of part (a)? A. The p-value is the probability of getting a sample mean of 3,950 bus miles or greater if the actual mean is 3,900 bus miles. B. The p-value is the probability that the actual mean is 3,950 bus miles or less. C. The p-value is the probability that the actual mean is 3,900 bus miles or greater given the sample mean is 3,950 bus miles.