Question

In a random sample of ten cell​ phones, the mean full retail price was ​$443.50 and the standard deviation was ​$179.00. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean mu. Interpret the results.

Answer 1

Answer: Margin of error = 128.04, confidence level = (315.46, 571.54)

Step-by-step explanation: The marginal error for a confidence interval is given as

Marginal error = tα/2 ×s/√n

Where s = sample standard deviation = $179.00

n = sample size = 10

tα/2 = this is the critical value for a 2 tailed test at a 5% level of significance (note level of significance + confidence level = 100, since our question is about a 95% confidence level, then the level of significance is 5%).

The value of tα/2 is gotten using a t distribution table with respect to the degree of freedom = sample size - 1 (df = 10 - 1 = 9) and the level of significance α = 5%

From the t distribution table, it is known that tα/2 = 2.262

Hence the margin of error = 2.262 × 179/√10

Margin of error = 2.262 × 56.605

Margin of error = 128.04

To construct a 95 % confidence level for the population mean (u), we have that

u = x ± tα/2 ×s/√n

Where tα/2 ×s/√n = margin of error = 128.04

Hence

u = x ± 128.04

But x = sample mean = 443.5

For upper limit of the interval

u = x + 128.04

u = 443.5 + 128.04

u = 571.54.

For lower limit of the interval

u = x - 128.04

u = 443.5 - 128.04

u = 315.46

This results of ours simply implies that the population mean is between 315.46 and 571.54. Comparing this interval value of population mean to the estimated sample mean, we can see that the values are not the same which implies that some errors have been introduced in the question, this error is the marginal error which is as a result of the sample size, it is given as 128.04