Question

In a random sample of eight cell​ phones, the mean full retail price was ​$496.00 and the standard deviation was ​$220.00. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 90​% confidence interval for the population mean μ. Interpret the results.

Answer 1

The 90% confidence interval for the population mean (μ) of cell phone full retail prices, based on a sample of eight phones with a mean of $496.00 and a standard deviation of $220.00, is approximately $348.60 to $643.40. This means we are 90% confident that the true population mean falls within this range.

Step-by-step explanation:

To find the margin of error and construct a 90% confidence interval for the population mean using the t-distribution, you need to follow these steps:

Identify the sample mean (X), sample standard deviation (s), and sample size (n).**
In this problem, we have:
- Sample mean (X) = $496.00
- Sample standard deviation (s) = $220.00
- Sample size (n) = 8

Determine the degrees of freedom (df).**
Degrees of freedom for the t-distribution is calculated as the sample size minus one.
df=n-1

=8-1

=7

Find the critical t-value (t*) for the t-distribution with df degrees of freedom at the desired confidence level (90%).**

For a 90% confidence interval, you need to find the t-value that corresponds to the middle 90% of the t-distribution.

The remaining 10% is split evenly between the two tails, which is 5% in each tail.

You can find this critical value, t*, using a t-table or a statistical calculator.

For df = 7, the t-value for a 90% confidence level and two-tailed test (which is equivalent to the one-tailed 95th percentile) is approximately:
t* ≈ 1.895 (This value might vary slightly depending on the t-table or tool you are using.)

Calculate the margin of error (ME).
The margin of error can be calculated using the formula:

`\[ \text{ME} = t* * \left((s)/(√(n))\right) \]`
Plug in our values:

`\[ \text{ME} = 1.895 * \left((220.00)/(√(8))\right) \]`
Compute the margin of error:

`\[ \text{ME} \approx 1.895 * 77.78100 \]`

`\[ \text{ME} \approx 147.43 \]`

Construct the 90% confidence interval (CI).**

To construct the confidence interval, subtract and add the margin of error from the sample mean:

`\[ \text{Lower limit} = X - \text{ME} \]`

`\[ \text{Upper limit} = X + \text{ME} \]`
So, the confidence interval is:

`\[ \text{CI} = (496.00 - 147.43, 496.00 + 147.43) \]`

`\[ \text{CI} = (348.57, 643.43) \]`

So the 90% confidence interval for the population mean is approximately between $348.57 and $643.43.

The results tell us that if we took many samples of eight cell phones and computed a 90% confidence interval for the mean full retail price from each sample, about 90% of those intervals would contain the true population mean of full retail prices for cell phones.

We are 90% confident that the interval from $348.57 to $643.43 contains the true mean full retail price for all cell phones.