A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $10.31,$17.22,$26.62,$22.84 Construct the 98% 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-by-step explanation:

From the information given,

Mean, μ = (10.31 + 17.22 + 26.62 + 22.84)/4 = 19.2475

Standard deviation, σ = √summation(x - mean)/n

Summation(x - mean) = (10.31 - 19.2475)^2+ (17.22 - 19.2475)^2 + (26.62 - 19.2475)^2 + (22.84 - 19.2475)^2 = 151.249475

σ = √(151.249475/4)

σ = 6.15

number of sample, n = 4

The z score for 98% confidence interval is 2.33

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

19.2475 ± 2.33 × 6.15/√4

= 19.2475 ± 2.33 × 3.075

= 19.2475 ± 7.16

The lower end of the confidence interval is 19.2475 - 7.16 = 12.09

The upper end of the confidence interval is 19.2475 + 7.16 = 26.41

i think the correct option is the second option.