Answer :
Using the t-distribution, the 99% confidence intervals for the given standard deviation are given as follows: (5053.39 , 4126.61)
Given that,
Number of college students = 23
A simple random sample with a mean = $4590
A simple random sample with a standard deviation = $860
What is a t-distribution confidence interval?
The t distributions is wide (has thicker tailed) for smaller sample sizes, reflecting that s can be smaller than σ. The thick tails ensure that the 80%, 95%, 99% confidence intervals are wider than those of a standard normal distribution (so are better for capturing the population mean).
The confidence interval is:
= μ ± [tex]t\frac{s}{\sqrt{n} }[/tex]
In which:
μ is the sample mean.
t is the critical value.
n is the sample size.
s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 23 - 1 = 22 df,
Upon using a t -table or a calculator, we see that the critical t -value for this 99% confidence interval is t0.005=2.581
The parameters are:
μ = $4590
n = 23
Hence, with s = 860, the bounds of the interval are:
= μ ± [tex]t\frac{s}{\sqrt{n} }[/tex]
= 4590 ± 2.581 [tex]\frac{860}{\sqrt{23} }[/tex]
= 4590 ± 2.581 [tex]\frac{860}{4.79}[/tex]
= 4590 ± 2.581 (179.54)
= 4590 ± 463.39
Then,
= 4590 + 463.39 , 4590 - 463.39
= 5053.39 , 4126.61
Therefore,
Using the t-distribution, the 99% confidence intervals for the given standard deviation are given as follows: (5053.39 , 4126.61)
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