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Answer :
Final answer:
To achieve a 98% confidence level that the weekly sales sample mean is within $400 of the population mean, with a standard deviation of $1400, a sampling period of 13 weeks is necessary.
Explanation:
To estimate the mean weekly sales of a new line of athletic footwear, you would need to calculate the size of your sample. Confidence intervals, standard deviation, and sample mean are essential to this calculation. Here, we've been given a 98% confidence level, a known population standard deviation of $1400, and we desire our estimate to be within $400 of the population mean.
The formula for a confidence interval can be expressed as follows: [sample mean - Z*(standard deviation/√n) , sample mean + Z*(standard deviation/√n)], where the 'Z' value depends on your desired confidence level. For 98% confidence level, 'Z' is approximately 2.33. The $400 represents the maximal amount of error acceptable (or standard deviation/√n), and rearranging this equation allows us to solve for 'n'.
($1400/√n) = $400 --> √n = $1400/$400 --> √n = 3.5
Squaring both sides to get 'n' yields n = 12.25. However, since we cannot gather a fraction of a week's data, we need to round up to the next whole week, resulting in 13 weeks required to gather enough data for a 98% confidence level that the sample mean is within $400 of the population mean.
Learn more about Sample Size Calculation here:
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Answer:
[tex]n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67[/tex]
So the answer for this case would be n=67 rounded up
Step-by-step explanation:
Information given
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean
[tex]\sigma=1400[/tex] represent the sample standard deviation
n represent the sample size
Solution to the problem
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (a)
And on this case we have that ME =400 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (b)
The critical value for 98% of confidence interval now can be founded using the normal distribution. And the critical value would be [tex]z_{\alpha/2}=2.33[/tex], replacing into formula (b) we got:
[tex]n=(\frac{2.33(1400)}{400})^2 =66.50 \approx 67[/tex]
So the answer for this case would be n=67 rounded up
