Question

In a random sample of eleven cell phones, the mean full retail price was $553.00and the standard deviation was $179.00. Assume the population is normallydistributed and use the t-distribution to find the margin of error and construct a90% confidence interval for the population mean u. Interpret the results

Answer 1

We are given;

Sample size, n = 11

Sample mean, x' = $553

Sample standard deviation, s = $179

90% confidence interval

We need to calculate;

the margin of error, M

Formula for the margin of error, M for a confidence interval:

`M=t*(s)/(√(n))`

t is the critical value for a 2 tailed test at a 10% level of significance

degree of freedom = sample size - 1

df = 11 - 1 = 10 and the level of significance α = 10%

The value of t from a t distribution table is 1.8125

Thus;

the margin of error is, M = $97.82

`M=1.8125*(179)/(√(11))=97.82`

The 90% confidence interval for the population mean is:

`\begin{gathered} CI=x^(\prime)±M \\ CI=553\pm97.8 \\ CI=\left(455.2,650.8\right) \end{gathered}`