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Exhibit: Department Store.

A random sample of 64 credit sales in a department store showed an average sale of $75.00. From past data, it is known that the standard deviation of the population of sales is $24.00.

(For this exhibit, avoid rounding intermediate steps and round your final solutions to 4 digits)

1. Construct the 96% confidence interval for the population mean and provide the lower and upper bound of the confidence interval below.

2. Assume that the information about average sales was lost. You decided to repeat the study and collect information for another sample. This time you obtain a random sample of 400 sales which accidentally resulted in the same average sale of $75.00. Construct the 96% confidence interval for the population mean and provide the lower and upper bound of the confidence interval below.

3. A population has a standard deviation of 80. A random sample of 400 items from this population is selected. The sample mean is determined to be 200. Find the margin of error at 94% confidence level assuming that the population is large. (Round your solution to 4 digits)

User Rikky
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1 Answer

4 votes

Answer:

Explanation:

Considering the central limit theorem, the distribution is normal since the number of samples is large.

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Confidence interval = mean ± z × σ/√n

Where

σ = population standard Deviation

σ/√n = sample standard deviation

Confidence interval = x ± z × σ/√n

1) x = $75

σ = $24

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.96 = 0.04

α/2 = 0.04/2 = 0.02

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.02 = 0.98

The z score corresponding to the area on the z table is 2.05. Thus, confidence level of 96% is 2.05

Confidence interval = 75 ± 2.05 × 24/√64

= 75 ± 2.05 × 3

= 75 ± 6.15

The lower end of the confidence interval is

75 - 6.15 = 68.85

The upper end of the confidence interval is

75 + 6.15 = 81.15

2) n = 400

x = $75

σ = $24

z = 2.05

Confidence interval = 75 ± 2.05 × 24/√400

= 75 ± 2.05 × 1.2

= 75 ± 2.46

The lower bound of the confidence interval is

75 - 2.46 = 72.54

The upper bound of the confidence interval is

75 + 2.46 = 77.46

3) n = 400

x = $200

σ = $80

The z score corresponding to the confidence level of 94% is 1.88

z = 1.88

Confidence interval = 200 ± 1.88 × 80/√400

= 200 ± 1.88 × 4

= 200 ± 7.52

Margin of error = 7.52

User MothOnMars
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