Question

PLS HELP (statistics) pleasethis is my last chance before 11:59 to get this right.

In a survey, 15 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $34 and standard deviation of $3. Construct a confidence interval at a 98% confidence level.

Give your answers to one decimal place.

Answer 1

Sure, I can help you with that! To construct a confidence interval at a 98% confidence level for the mean spending on birthday gifts, we can use the formula:

Confidence Interval = (Mean ± Z * (Standard Deviation / √n))

where Z is the critical value for the desired confidence level. For a 98% confidence level, Z is approximately 2.33.

Given the values you provided:

Mean = $34

Standard Deviation = $3

Sample Size (n) = 15

Z (for 98% confidence) ≈ 2.33

Now, let's calculate the confidence interval:

Confidence Interval = (34 ± 2.33 * (3 / √15))

Now, compute the upper and lower bounds:

Upper Bound = 34 + 2.33 * (3 / √15) ≈ 36.16

Lower Bound = 34 - 2.33 * (3 / √15) ≈ 31.84

The 98% confidence interval for the mean spending on birthday gifts is approximately $31.84 to $36.16. Good luck with your surveys