Answer:
[tex]19.22\:<\:\mu\:<\:44.78[/tex]
Step-by-step explanation:
Tthe population is normally distributed and the sample size is [tex]n=22\:<\:30[/tex].
Since the population standard deviation [tex]\sigma[/tex] is unknown and the sample standard deviation [tex]s[/tex], must replace it, the t distribution must be used for the confidence interval.
Hence with degrees of freedom of 21, [tex]t_{\frac{\alpha}{2} }=3.527[/tex].(Read from the t distribution table)
The 98% confidence interval can be constructed using the formula:
[tex]\bar X-t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )\:<\:\mu\:<\:\bar X+t_{\frac{\alpha}{2}}(\frac{s}{\sqrt{n} } )[/tex].
From the question the sample mean is [tex]\bar X=32[/tex]dollars and the sample standard deviation is [tex]s=17[/tex] dollars.
We substitute the values into the formula to get
[tex]32-3.527(\frac{17}{\sqrt{22} } )\:<\:\mu \:<\:32+3.527(\frac{17}{\sqrt{22} } )[/tex]
[tex]19.22\:<\:\mu\:<\:44.78[/tex]
Therefore, we can be 98% confident that the population mean is between is between 19.22 and 44.78 dollars.
