# A researcher is interested in finding a 90% confidence interval for the mean number minutes students are concentrating on their professor during a one

A researcher is interested in finding a $90\mathrm{%}$ confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 117 students who averaged 40.9 minutes concentrating on their professor during the hour lecture. The standard deviation was 11.8 minutes. Round answers to 3 decimal places where possible.
a.
To compute the confidence interval use a ? distribution.
b.
With $90\mathrm{%}$ confidence the population mean minutes of concentration is between ____ and ____ minutes.
c.
If many groups of 117 randomly selected students are studied, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population mean minutes of concentration and about ____ percent will not contain the true population mean number of minutes of concentration.
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Step 1
Given: $\stackrel{―}{x}=40.9,s=11.8,n=117$
a)
The sampling distribution follows a t-distribution.
Reason:
Since the population standard deviation is unknown, so the t-distribution is appropriate
b)The degree of freedom is,
$df=n-1$
$=117-1$
$=116$
Critical value:
By using the t-tables, the critical value at $10\mathrm{%}$ level of significance for a two tailed t -distribution is,
${t}_{0.10\text{/}2.116}=±1.658$
Step 2
The $90\mathrm{%}$ confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture is,
$90\mathrm{%}CI=\stackrel{―}{x}-{t}_{\alpha \text{/}2}\left(\frac{s}{\sqrt{n}}\right)<\mu <\stackrel{―}{x}+{t}_{\alpha \text{/}2}\left(\frac{s}{\sqrt{n}}\right)$
$=40.9-1.658\left(\frac{11.8}{\sqrt{117}}\right)<\mu <40.9+1.658\left(\frac{11.8}{\sqrt{117}}\right)$
$=40.9-1.658\left(1.090910386\right)<\mu <40.9+1.658\left(1.090910386\right)$
$=40.9-1.80872942<\mu <40.9+1.80872942$
$=39.09127058<\mu <42.70872942$
$=39.091<\mu <42.709$
With $95\mathrm{%}$ confidence the population the mean minutes of concentration is between 39.091 and 42.709.
c)
If many groups of 117 randomly selected students are studied, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.