Answer :
The confidence interval at a 90% is 29.533 ≤ μ ≤ 40.467
Given that
mean = 35
standard deviation = 15
We compute the confidence interval as follows:
μ =¯ˣ± ME
Where:
- μ = is the population mean,
- ¯x = 35 is the sample mean and,
- ME = is the margin of error.
To calculate the margin of error, we will use the t-distribution since the sample is very small n<30. Thus, our margin of error will be calculated as:
ME = t ×[tex]\frac{s}{\sqrt{n} }[/tex]
Where:
t = is the t-score at the 90% confidence level
degree of freedom df = 24 -1 =23
s = 15 is the standard deviation and,
n =24 is the sample size.
From the t-distribution tables, we will read the values of t at a critical value [tex]\alpha[/tex] = 0.10, and the degree of freedom is 23. This will give t= 1.714.
Thus, the margin of error is:
ME = 1.714 ×[tex]\frac{15}{\sqrt{23} }[/tex]
ME = 1.714 ×3.19
ME = 5.467
the confidence interval for the population mean is equal to:
μ = 35 ± 5.467
29.533 ≤ μ ≤ 40.467
To learn more about t- distribution table:
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