Unit 4 Assignment 1 PDF

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Unit 4 Assignment 1...

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1. Find the critical value zc necessary to form a confidence interval at the level of confidence shown below. C = 0.81 The level of confidence c is the area under the standard normal curve between the critical values -Z c and Zc The area outside the critical values is 1 – c, so the area in each tail is ½(1 – C) Substitute the value of c to solve for the area in each tail of the confidence region. ½ (1 – C) = ½ (1 – 0.81) = 0.095 While you can use a standard normal table or technology to find the critical value, for this problem we will use a standard normal table. Use a standard normal table to find the positive critical value that corresponds to a tail area of 0.095 Zc = 1.31 2. Use the values on the number line to find the sampling error.

The difference between the point estimate and actual parameter value is called the sampling error. The formula for the sampling error is shown below.

3.7-4.53 = -0.83 3. Construct the confidence interval for the population mean µ

Substitute the value of c and evaluate to find the area in each tail. ½ (1 – 0.90) = 0.05 While you can use a standard normal table or technology to find the critical value, for this problem we will use a standard normal table. Zc = 1.645 Substitute the values zc = 1.645 o = 0.7 and n=57 to find E

E = 1.65

0.7 √ 57

E = .15 Use the margin of error to find the left endpoint.

= 7.6 – 0.15 = 7.45 Now find the right endpoint.

= 7.6 + 0.15 = 7.75 Therefore, a 90% confidence interval for u is (7.45, 7.75) 4. Use the confidence interval to find the margin of error and the sample mean. (1.66,2.04)

Divide the width of the confidence interval by 2 to find the margin of error. = 0.19 0.19 is the margin of error.

2.04 −1.66 / 2

1.66 + 0.19 = 1.85 5. You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 35 home theater systems has a mean price of $124.00. Assume the population standard deviation is $19.20.

First find the 90% confidence interval. The level of confidence c is the probability that the interval estimate contains the population parameter. For a 90% confidence interval, c is .90

The population standard deviation σ is given. Substitute the values zc = 1.645, σ = 19.20, and n = 35 to find E, rounding to two decimal places.

= 1.645

19.20 √ 35

= 5.34

124 – 5.34 = 118.66

124 + 5.34 = 129.34 Therefore, the 90% confidence interval is (118.66,129.34) Construct a 95% confidence interval for the population mean. Find the 95% confidence interval. The level of confidence c is the probability that the interval estimate contains the population parameter. For a 95% confidence interval, c is 0.95

Substitute the values zc = 1.960 σ = 19.20, and n = 35 to find E, rounding to two decimal places.

E = 1.960

19.20 √ 35

E = 6.36

= 124 – 6.36 = 117.64

124 + 6.36 = 130.36 Therefore, the 95% confidence interval is (102.64,115.36)

6. Find the critical value tc for the confidence level c = 0.80 and sample size n = 14 Use a t-distribution table to look up the t-value using the level of confidence c and the degrees of freedom n – 1 The level of confidence c = 0.80 is given. The degrees of freedom are n – 1 = 14 – 1 = 13 Use the column with a level of confidence c = 0.80 and the row with 13 degrees of freedom to determine the critical value Tc = 1.350 7. Find the margin of error for the given values of c, s, and n. C = 0.99, s = 3.6, n = 18

While either a table or technology can be used, for this example, use a t-distribution table to find the tvalue tc using the confidence c and the degrees of freedom n – 1. The level of confidence c = 0.99 is given. The degrees of freedom are n – 1 = 18 – 1 = 17. Use the column with level of confidence c = 0.99 and the row with 17 degrees of freedom to determine the critical value. Tc = 2.898 Substitute tc = 2.898 s = 3.6, and n = 18 into the formula to find the margin of error, rounding to one decimal place.

E = 2.898

3.6 √ 18

E = 2.5 8. Construct the indicated confidence interval for the population mean µ using the t-distribution. Assume the population is normally distributed.

N = 10 X = 14.6 s = 2.0

Df = 10 – 1 = 9 Identify the confidence level c. C = 0.99 Use technology to find the critical value t c with d.f = 9, rounding to three decimal places. 3.250

E = 3.250

2.0 √ 10

E = 2.1

Substitute the known values into the formula for the confidence interval and simplify. 14.6 – 2.1 < µ < 14.6 + 2.1 12.5 < µ < 16.7

12.5, 16.7 9. Use the given confidence interval to find the margin of error and the sample mean. (4.26,8.40) The sample mean is the value at the center of the confidence interval. It is found by taking the average of the endpoints of the interval. Calculate the average of the endpoints.

4.26 + 8.40 2

= 6.33

The margin of error is the distance from the sample mean to either end of the confidence interval. Subtract the sample mean, 6.33, from the upper endpoint of the confidence interval, 8.40 8.40 – 6.33 = 2.07 Therefore, the sample mean is 6.33 and the margin of error is 2.07...