The average selling price of a smartphone purchased by a random sample of 37 customers was...

The average selling price of a smartphone purchased by a random sample of 37 customers was $314. Assume the population standard deviation was $35.

a. Construct a 95% confidence interval to estimate the average selling price in the population with this sample.

b. What is the margin of error for this interval?

a. The 95% confidence interval has a lower limit of $ and an upper limit of $ . (Round to the nearest cent as needed.)

b. The margin of error is $ . (Round to the nearest cent as needed.)

Solutions

Expert Solution

a. Construct a 95% confidence interval to estimate the average selling price in the population with this sample.

We need to construct the 95% confidence interval for the population mean . The following information is provided:

Sample Mean	314
Population Standard Deviation	35
Sample Size	37

The critical value for α=0.05 is z_c = 1.96 . The corresponding confidence interval is computed as shown below:

b. What is the margin of error for this interval?

Margin of error = Length of confidence interval / 2

Length of confidence interval = 325.278 - 302.722 = 22.556

Margin of error = 22.556/2 = 11.278

a. The 95% confidence interval has a lower limit of $ and an upper limit of $ . (Round to the nearest cent as needed.)

b. The margin of error is $ . (Round to the nearest cent asneeded.)

Margin of error = Length of confidence interval / 2

Length of confidence interval = 325 - 303 = 22

Margin of error = 22/2 = 11

orchestra answered 1 year ago

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