(A) What price do farmers get for their watermelon crops? In the third week of July, a random sample of44farming regions gave a sample mean of= $6.88per 100 pounds of watermelon. Assume thatis known to be$1.98per 100 pounds.

(I)Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.)

lower limit$=?

upper limit$=?

margin of error$=?

(II)Find the sample size necessary for a 90% confidence level with maximal error of estimateE=0.31for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.)

farming regions

(III)A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)?Hint:1 ton is 2000 pounds. (For each answer, enter a number. Round your answers to two decimal places.)

lower limit$

upper limit$

margin of error$

(B) Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl. Recently, the patient's total calcium tests gave the following readings (in mg/dl). Assume that the population ofxvalues has an approximately normal distribution.9.98.610.58.59.49.810.09.911.212.1

(I)Use a calculator with mean and sample standard deviation keys to find the sample mean readingand the sample standard deviations. (in mg/dl; round your answers to two decimal places.)

=mg/dl

s=mg/dl

(II)Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (in mg/dl; round your answer to two decimal places.)

lower limitmg/dl

upper limitmg/dl

(C) For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a random sample of62professional actors, it was found that43were extroverts.

(I)Letprepresent the proportion of all actors who are extroverts.Find a point estimate forp. (Round your answer to four decimal places.)

(II)Find a 95% confidence interval forp. (Round your answers to two decimal places.)

lower limit

upper limit

(III)For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a marketing survey, a random sample of980supermarket shoppers revealed that262always stock up on an item when they find that item at a real bargain price.

(I)Letprepresent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain.Find a point estimate forp. (Enter a number. Round your answer to four decimal places.)

(II)Find a 95% confidence interval forp. (For each answer, enter a number. Round your answers to three decimal places.)

lower limit

upper limit