Question

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.

Answer 1

To calculate the sample mean, all the amounts spent on a trip to the mall are summed up and then divided by the number of data points, resulting in a sample mean of $19.25.

Step-by-step explanation:

To calculate the sample mean for the amounts spent by 9 to 11 year olds at the mall, we need to add all the given amounts together and then divide by the number of amounts to find the average. The amounts given are $10.31, $17.22, $26.62, and $22.84.

The sum of these is:

$10.31 + $17.22 + $26.62 + $22.84 = $76.99

There are four amounts in total, so we divide the sum by four:

$76.99 / 4 = $19.2475

When rounded to two decimal places, the sample mean is $19.25.

Answer 2

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