A research firm conducted a survey to determine the mean amount people spend at
a popular coffee shop during a week. They found the amounts spent per week
followed a normal distribution, with a population standard deviation of $4. A sample
of 49 customers revealed that the mean is $25.
a) What is the 95% confidence interval estimate of µ?
b) What is the 99% confidence interval estimate of µ?
23.88, 26.12
To find the confidence interval estimate of the population mean (µ), we can use the formula:
Confidence interval = sample mean ± (Z * standard deviation / √n)
Where:
- Sample mean: $25 (given)
- Z: Z-score (depends on the desired confidence level)
- Standard deviation: $4 (given)
- n: sample size = 49 (given)
a) For a 95% confidence interval estimate of µ:
Z-score for a 95% confidence level is 1.96.
Using the formula, the confidence interval is:
25 ± (1.96 * 4 / (√49)) = 25 ± (1.96 * 4 / 7) = 25 ± 1.117 (rounded to 3 decimal places)
Therefore, the 95% confidence interval estimate of µ is ($23.883, $26.117).
b) For a 99% confidence interval estimate of µ:
Z-score for a 99% confidence level is 2.576.
Using the formula, the confidence interval is:
25 ± (2.576 * 4 / (√49)) = 25 ± (2.576 * 4 / 7) = 25 ± 1.438 (rounded to 3 decimal places)
Therefore, the 99% confidence interval estimate of µ is ($23.562, $26.438).
Note: The Z-scores used to find the confidence intervals can be obtained from a Z-table or calculated using statistical software or online calculators.