Question 3: a. A point estimate of the (population) average cost per person of automobile crashes.b.

Question 3: a. A point estimate of the (population) average cost per person of automobile crashes.b. The margin of error (using 95% confidence) for your point estimate.c. How large a sample would have to be taken to make the margin of error one-half thesize that you calculated in part (b) above?Question 4.Here you are with the Department of Weights and Measures again. Suppose that you areinterested in estimating the mean amount of soda that is placed in 2-liter bottles at the localbottling plant of a large nationally known soda company. The bottling plant has informed theinspection division that the standard deviation for 2-liter bottles is .07 liter. A random sampleof fifty 2-liter bottles obtained from this bottling plant indicated a sample mean of 2.02 liters.a. What is the point estimate of the population mean?b. Calculate the 95% margin of error for your point estimate in part (a).c. How large a sample would you need to make the margin of error 1/3 as large as it is in (b)?Question 5.A Sears department store manager would like to estimate the average (mean) amount ofpurchases per month by its customers with charge accounts to within $25 (the desiredmargin of error) with 90 percent confidence. How many accounts should be selected fromthe store’s records if the sample standard deviation of monthly account balances is estimatedto be $90?Question 6: a. Calculate a point estimate of the population mean.b. Calculate a point estimate of the population standard deviation.c. Calculate the 95% margin of error for the point estimate in part (a).d. Calculate a 95% confidence interval for the population mean. Question 8: a. A point estimate of ?.b. The 99% margin of error.c. A 99% confidence interval estimate of ?.Question 9: – a. A point estimate (p ) of the population proportion (p).b. The 95% margin of error.c. The 95% confidence interval estimate of the population proportion.