21villalobosjabez
24.06.2020 •
Mathematics
In a sample of 22 people, the average cost of a cup of coffee is $2.70. Assume the population standard deviation is $0.93. What is the 90% confidence interval for the cost of a cup of coffee
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Ответ:
$2.70+/-$0.33
= ( $2.37, $3.03)
Therefore, the 90% confidence interval (a,b) = ( $2.37, $3.03)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean x = $2.70
Standard deviation r = $0.93
Number of samples n = 22
Confidence interval = 90%
z(at 90% confidence) = 1.645
Substituting the values we have;
$2.70+/-1.645($0.93/√22)
$2.70+/-1.645($0.198276666210)
$2.70+/-$0.326165115916
$2.70+/-$0.33
= ( $2.37, $3.03)
Therefore, the 90% confidence interval (a,b) = ( $2.37, $3.03)
Ответ: