A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mall:$19.17,$21.18,$20.38,$25.08Construct the 90\%90\% 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.

UkusakazaL

UkusakazaL

Answered question

2021-08-03

A survey of several 10 to 11 year olds recorded the following amounts spent on a trip to the mall:
$19.17,$21.18,$20.38,$25.08
Construct the 90% 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.
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Step 4 of 4 : Construct the 90% confidence interval. Round your answer to two decimal places.

Answer & Explanation

Falak Kinney

Falak Kinney

Beginner2021-08-07Added 1 answers

Step 1
The mean and standard deviation are obtained as follows:
xx219.17367.488921.18448.592420.38415.344425.08629.0064SUM85.811860.432
Mean =(x)=xn=85.814=21.45
Standard Deviation =x2(x)nn1
Standard Deviation =1860.43285.81243
Standard Deviation =2.55
Step 2
The 90% confidence interval is obtained as follows:
Confidence Interval =[x±t(n1,α2)×sn]
Confidence Intarval =[21.45±t(3,0.102)×2.554]
Confidence Interval =[21.45±2.353×2.554]
Comfidence Interval =[18.45,24.45]
Lower Limit =18.45
Upper Limit =24.45

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