# Statistics MA132

**Question description**

**(1) Pinworm:** In a random sample of 810 adults in the U.S.A., it was found that 80 of those had a pinworm infestation. You want to find the 99% confidence interval for the proportion of all U.S. adults with pinworm.

(a) What is the point estimate for the proportion of all U.S. adults with pinworm? **Round your answer to 3 decimal places. **

.................

(b) What is the critical value of *z* (denoted *z*_{α/2}) for a 99% confidence interval? **Use the value from the table or, if using software, round to 2 decimal places. ***z*_{α/2} = ..................

(c) What is the margin of error (*E*) for a 99% confidence interval? **Round your answer to 3 decimal places. ***E* = ...................

(d) Construct the 99% confidence interval for the proportion of all U.S. adults with pinworm. **Round your answers to 3 decimal places. **

................ < *p* < ..............

(e) Based on your answer to part (d), are you 99% confident that more than 5% of all U.S. adults have pinworm?

(f) In Sludge County, the proportion of adults with pinworm is found to be 0.17. Based on your answer to (d), does Sludge County's pinworm infestation rate appear to be greater than the national average?

**(2) Pinworm:**In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 8% of the people in this sample had pinworm but the margin of error for the population estimate was too large. They want an estimate that is in error by no more than 2.5 percentage points at the 90% confidence level.

**Enter your answers as whole numbers.**

(a) What is the minimum sample size required to obtain this type of accuracy? Use the prior sample proportion in your calculation.

(b) What is the minimum sample size required to obtain this type of accuracy when you assume no prior knowledge of the sample proportion?

The minimum sample size is ................ U.S. adults.

**(3) Diet Methods:**Here we study the effectiveness of two dieting methods. Method 1 is a diet high in protein, fiber, and fat but low in carbohydrates. Method 2 is a diet low in fat but high in fiber and carbohydrates. After one year, a sample of participants from each diet method were checked for the amount of weight lost. The 99% confidence intervals for the mean weight loss of each plan are given below.

**Method 1:** 8.25 < μ < 9.15**Method 2:** 6.25 < μ < 7.75

**Round your answer to 1 decimal place.**

x

_{1}= .............. pounds

(b) What is the sample mean weight loss for participants in the Method 2 diet plan?

**Round your answer to 1 decimal place.**

x_{2} = ............... pounds

(c) Are you 99% confident that the mean weight loss for all those on the Method 1 diet plan is greater than the mean weight loss for all on the Method 2 diet plan and why?

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