Question

In a random sample of eight cell phones, the mean full retail price was $575.00 and the standard deviation was $150.00. Assume the population iserror and construct a 90% confidence interval for the population mean H. Interpret the results.Identify the margin of error.(Round to one decimal place as needed.)Construct a 90% confidence interval for the population mean.(Round to one decimal place as needed.)Interpret the results. Select the correct choice below and fill in the answer box to complete your choice.(Type an integer or a decimal. Do not round.)OA. It can be said that% of the population of cell phones have full retail prices (in dollars) that are between the interval's endpoints.OB. With% confidence, it can be said that the population mean full retail price of cell phones (in dollars) is between the interval's endpoints.O c. With% confidence, it can be said that most cell phones in the population have full retail prices (in dollars) that are between the interval's endponOD.% of all random samples of eight people from the population of cell phones will have a mean full retail price in dollars) that is between the intervViJTMoreClick to select

Answer 1

Given the following parameters

$\begin{gathered} \bar{x}=\text{ \ 575.00}\Rightarrow\operatorname{mean} \\ \sigma=\text{ \ 150.00}\Rightarrow\text{ standard deviation} \\ n=8\Rightarrow\text{ sample size} \\ \text{confidence level=90\%} \end{gathered}$

Calculate the alpha value

$\begin{gathered} \alpha=1-confidence\text{ level} \\ \alpha=1-\frac{90}{100}=1-0.90=0.1 \end{gathered}$

Calculate the P-value and critical value

$\begin{gathered} P=\frac{\alpha}{2}=0.05 \\ Degree\text{ of freedom}=n-2=8-2=6 \\ CV=\text{ check 0.05 in the P-value for t-test under DF of 6} \\ CV=1.94 \end{gathered}$

Calculate the standard error

$\begin{gathered} SE=\frac{\sigma}{\sqrt[]{n}} \\ =\frac{150}{\sqrt[]{8}}=\frac{150}{2.828}=53.04 \end{gathered}$

Thus, the margin of error is given as

$\begin{gathered} MOE=CV\times SE \\ =1.94\times53.04=102.9 \end{gathered}$

Therefore the confidence interval will be

$\begin{gathered} CI=\bar{x}\pm MOE \\ =\text{ \ 575.00}\pm\text{ \ 102.9} \end{gathered}$

Hence, the final conclusion is option C