A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mall: $18.31,$25.09,$26.96,$26.54,$21.84,$21.46 Construct the 98% confidence interval for the average amount spent by 10 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Step 3 of 4 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answers

Answer 1

The 98% confidence interval for the average amount spent by 10 to 11-year-olds on a trip to the mall is 23.36 ± 2.933.

How to calculate the confidence interval and its critical value?

The confidence interval for a given level of percentage is given by

C. I = μ ± Z(σ/√n)

Where,

μ - mean, σ - standard deviation, n - sample size, and z - critical value.

The critical value is calculated by

step 1: 100% - (the confidence level)

step 2: Converting the step 1 result into decimal value

step 3: dividing the step 2 result by 2

This is indicated by α/2. So, from the normal distribution table, the z-value at α/2 is said to be the required critical value and denoted by Z_(α/2) or Z.

Calculation:

It is given that,

A survey of several 10 to 11-year-olds recorded the following amounts spent on a trip to the mall: $18.31,$25.09,$26.96,$26.54,$21.84,$21.46

Sample size n = 6

Step 1: Finding the mean for the given amounts of the survey:

Mean μ = (18.31 + 25.09 + 26.96 + 26.54 + 21.84 + 21.46)/6

             = 23.36

Step 2: Finding the standard deviation:

Standard deviation σ = √summation(x - mean)²/n

On calculating, we get σ = 3.09

Step 3: Finding the critical value:

It is given that the confidence level is 98%

So, (100% - 98%) = 2%

Converting into decimal gives 0.02

So, α/2 = 0.02/2 = 0.01

Thus, at 0.01, the critical value Z = 2.326

Step 4: Constructing the confidence interval:

C.I = 23.36 ± (2.326) × (3.09/√6)

    = 23.36 ± (2.326 × 1.261)

    = 23.36 ± 2.933

So, the lower bound = 23.36 - 2.933 = 20.427

the upper bound = 23.36 + 2.933 =26.293

Therefore, the 98% confidence interval lies from 20.427 to 26.293.

Learn more about constructing the confidence interval here:

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