Question

In a random sample of five mobile​ devices, the mean repair cost was ​$90.00 and the standard deviation was ​$14.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results.The 95 % confidence interval for the population mean is ______(Round to two decimal places as needed.) The margin of error is ______ (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. A. If a large sample of microwaves is taken approximately 95 percent of them will have repair costs between the bounds of the confidence interval. B. With 95 percent confidence, it can be said that the repair cost is between the bounds of the confidence interval. C. It can be said that 95 percent of microwaves have a repair cost between the bounds of the confidence interval. D. With 95 percent confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval. E. The 95 percent confidence interval for the population mean repair cost is between the bounds of the confidence interval.

Answer 1

Answer:

The correct option is;

B. With 95 percent confidence, it can be said that the repair cost is between the bounds of the confidence interval

Step-by-step explanation:

Here we have the formula for confidence interval, CI given as follows;

$CI=\bar{x}\pm t\frac{s}{\sqrt{n}}$

Where:

$\bar x$ = Sample mean = $90.00

s = Sample standard deviation = $14.00

n = Sample size = 5

t = z value for the confidence level = 2.7764 from z-score relations

Plugging in the values, we have;

$CI=90.00\pm 2.7764 \frac{14.00}{\sqrt{5}}$

Which gives the CI as 72.6167 to 107.38.