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6.2.31-T Question Help The monthly incomes for 12 randomly selected people, each with a bachelors 4450.24 degree in economic

(a) find the sample mean

(b) find the sample standard deviation

(c) Construct a 99% confidence interval for the population mean μ.

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Answer #1

Solution :

Given data is 4450.24,4596.57,4366.61,4455.36,4151.27,3727.14,4283.33,4527.47,4407.49,3946.14,4023.48,4221.28

=> sample mean x-bar = sum of terms/number of terms

= 51156.38/12

= 4263.0317

= 4263.0

=> standard deviation s = 260.1469

=> df = n - 1 = 11

=> For 99% confidence interval, t = 3.106

=> A 99% confidence interval for the population mean μ is

=> x-bar +/- t*s/sqrt(n)

=> 4263.0 +/- 3.106*260.1469/sqrt(12)

=> (4029.7458 , 4496.2542)

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