Question

In a random sample of cell​ phones, the mean full retail price was ​$ and the standard deviation was ​$. assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a ​% confidence interval for the population mean . interpret the results.

Answer 1

Using the​ t-distribution the margin of error  is 128.04 and 95% confidence interval for the population mean is:315.46; 571.54.

Confidence Interval

Using the margin of error for a confidence interval formula

Margin of error(m) = t × (s/√n)

Where:

t=critical value for a 2 tailed test at a 5% level of significance

Degree of freedom  = 10 - 1 = 9

Level of significance α = 5%

Value of t using t distribution table= 2.262

Hence:

Margin of error = 2.262 × (179/√10)

Margin of error = 128.04

95% confidence level for the population mean (u):

Confidence level =Mean ± Margin of error

Confidence level = 443.5 ± 128.04

Confidence level=(443.5 - 128.04); (443.5 + 128.04)

Confidence level= (315.46; 571.54)

Inconclusion the margin of error  is 128.04 and 95% confidence interval for the population mean is:315.46; 571.54.

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