Construct confidence interval at 90% confidence

Question

In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42 and standard deviation of $13. Construct a confidence interval at a 90% confidence level.

Give your answers to one decimal place.

______________ plus or minus ______________

Answer 1

To construct a confidence interval for the mean spending on a child's last birthday gift, we can use the formula:

Confidence interval = Sample mean ± Margin of error

First, we need to calculate the margin of error. The margin of error depends on the confidence level and the standard deviation of the sample. Since we have the standard deviation, we can use the following formula:

Margin of error = Z * (Standard deviation / √n)

Where:

Z = Z-score corresponding to the desired confidence level (for a 90% confidence level, Z ≈ 1.645)

Standard deviation = $13

n = sample size = 11

Calculating the margin of error:

Margin of error = 1.645 * (13 / √11)

Next, we calculate the confidence interval by subtracting and adding the margin of error to the sample mean:

Confidence interval = $42 ± Margin of error

Now we can substitute the values and calculate the confidence interval:

Confidence interval = $42 ± (1.645 * (13 / √11))

Calculating the values:

Confidence interval ≈ $42 ± (1.645 * 3.92) ≈ $42 ± 6.45

Therefore, the confidence interval at a 90% confidence level is approximately $35.55 to $48.45.