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.
<h3>How to calculate the confidence interval and its critical value?</h3>
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.
<h3>Calculation:</h3>
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:
brainly.com/question/15714113
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