Question

A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $22.91,$17.13,$13.60 Construct the 99% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal.

Step 1 of 4: Calculate the sample mean for the given sample data. Round your answer to two decimal places.

Step 2 of 4: Calculate the sample standard deviation for the given sample data. Round your answer to two decimal places.

Step 3 of 4: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 4 of 4: Construct the 99% confidence interval. Round your answer to two decimal places.

Answer 1

1)

= X / n = 53.64 / 3 = 17.88

2)

S = Sqrt [ X2 - n 2 / ( n - 1) ] = sqrt [ 1003.265 - 3 * 17.882 / 2 ] = 4.70

3)

From T table

T critical value at 0.01 significance level with 2 df = 9.925

4)

99% CI is

- t * S / sqrt(n) < < + t * S / sqrt(n)

17.88 - 9.925 * 4.70 / sqrt(3) < < 17.88 + 9.925 * 4.70 / sqrt(3)

-9.05 < < 44.81

99% CI Is ( -9.05 , 44.81 )