Question

In a sample of 22 people, the average cost of a cup of coffee is $2.83. Assume the population standard deviation is $1.01. What is the 90% confidence interval for the cost of a cup of coffee?

Group of answer choices

($2.45, $3.21)

($2.54, $3.12)

($2.55, $3.11)

($2.48, $3.18)

Answer 1

Ans. Option d

Point estimate,c = average cost of a cup of coffee = $2.83

Standard deviation, s = $1.01

Sample size, n = 22

Standard error of c, se = s/n0.5 = 1.01/220.5 = 0.2153

Zcritical at level of significance 10%,z = 1.645

90% confidence interval for the cost of cup of coffee, CI = [c - z*se, c + z*se]

=> CI = [2.83 - 1.645*0.2153, 2.83 + 1.645*0.2153]

=> CI = [2.476, 3.184]

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