Question

1. You sample 16 students in your school, and they average 13 hours of TV a week. Assume​ σ​ =3. Find a 99% Confidence Interval

2. A hardware manufacturer produces bolts used to assemble various machines. Assume that the diameter of bolts produced by this manufacturer has an unknown population mean and the standard deviation is 0.1 mm. Suppose the average diameter of a simple random sample of 75 bolts is 5.12 mm. Calculate the margin of error of a 95% confidence interval for the mean then determine the 95% confidence interval

3. You want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a simple random sample of 30 apartments advertised in the local newspaper is $1,450. The standard deviation of the population is $220. Find a 99% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.

4. An appliance manufacturer stockpiles washers and dryers in a large warehouse for shipment to retail stores. Sometimes, handling them the appliances get damaged. Even though the damage may be minor, the company must sell those machines at drastically reduced prices. One day an inspector randomly checks 50 washers and finds that 5 of them have scratches or dents. Compute a 95% confidence interval for the proportion of appliances from this manufacturer that get damaged during shipment.

5. You are testing chocolate chip cookies to estimate the mean number of chips per cookie.You sample 20 cookies and you find a sample mean of 10 chips per cookie. Assume ​σ​ = 2 .Find a 99% confidence interval.

6. If 64% of a sample of 550 people leaving a shopping mall claims to have spent over $25, determine a 99% confidence interval estimate for the proportion of shopping mall customers who spend over $25.

7. In a random sample of machine parts, 18 out of 225 were damaged in a shipment. Establish a 95% confidence interval estimate for the proportion of damaged machine parts in shipment.

8. A travel agent wants to estimate the proportion of vacationers who plan to travel outside the United States in the next 12 months. A random sample of 150 vacationers revealed that 45 had plans for foreign travel in that time frame.

a) Suppose (at the 95% con fidence level) you need to have a margin of error no more than 4 percentage points. How many vacationers would you have to sample? (Use the sample proportion you calculated in part (a) as an estimate of ^p. )

b) Suppose (at the 95% con dence level) you need to have a margin of error no more than 4 percentage points, but you have no estimate of ^p. How many vacationers would you have to sample?

Answer 1

1) At 99% confidence interval the critical value is z^* = 2.58

The 99% confidence interval for population mean is

+/- z^* *

= 13 +/- 2.58 * 3/

= 13 +/- 1.935

= 11.065, 14.935

2) At 95% confidence interval the critical value is z^* = 1.96

Margin of error =  z^* *

= 1.96 * 0.1/ = 0.0226

So the 95% confidence interval for population mean is

+/- E

= 5.12 +/- 0.0226

= 5.0974, 5.1426

3) The 99% confidence interval for population mean is

+/- z^* * 

= 1450 +/- 2.58 * 220/sqrt(30)

= 1450 +/- 103.6291

= 1346.3709, 1553.6291

4) = 5/50 = 0.1

The 95% confidence interval for population proportion is

+/- z^* * sqrt((1 - )/n)

= 0.1 +/- 1.96 * sqrt(0.1 * (1 - 0.1)/50)

= 0.1 +/- 0.0832

= 0.0168, 0.1832

5) The 99% confidence interval for population mean is

+/- z^* *

= 10 +/- 2.58 * 2/sqrt(20)

= 10 +/- 1.1538

= 8.8462, 11.1538

6) The 99% confidence interval for population proportion is

+/- z^* * sqrt((1 - )/n)

= 0.64 +/- 2.58 * sqrt(0.64 * (1 - 0.64)/550)

= 0.64 +/- 0.0548

= 0.5852, 0.6948

7)   = 18/225 = 0.08

The 95% confidence interval for population proportion is

+/- z^* * sqrt((1 - )/n)

= 0.08 +/- 1.96 * sqrt(0.08 * (1 - 0.08)/225)

= 0.08 +/- 0.0354

= 0.0446, 0.1154

8)a) = 45/150 = 0.3

Margin of error = 0.04

Or, z^* * sqrt(p(1 - p)/n) = 0.04

Or, 1.96 * sqrt(0.3 * 0.7/n) = 0.04

Or, n = (1.96 * sqrt(0.3 * 0.7)/0.04)^2

Or, n = 505

b) Margin of error = 0.04

Or, z^* * sqrt(p(1 - p)/n) = 0.04

Or, 1.96 * sqrt(0.5 * 0.5/n) = 0.04

Or, n = (1.96 * sqrt(0.5 * 0.5)/0.04)^2

Or, n = 601