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A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students te

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Answer #1

Answer:-

Given that:-

(a) t-distribution

here t-distribution should be used as the population standard deviation(\sigma) is not given

(b)

Given

n=126 X = 34.8 s=18.7

98% confidence interval for POPulation Mean is

(\overline{X}-t*\frac{s}{\sqrt{n}}<\mu<\overline{X}+t*\frac{s}{\sqrt{n}})

ME=margin of error

ME=t*\frac{s}{\sqrt{n}}

t=critical value

\alpha =1-0.98=0.02

Z_{critical}=Z_{0.02/2}=2.3658

ME=2.3658*\frac{18.7}{\sqrt{126}}

ME=3.95

Now, 98% Confidence interval for Population Mean is

\overline{x}\pm Z_{critical}\times \frac{s}{\sqrt{n}}

=34.8\pm 2.3658\times \frac{18.7}{\sqrt{126}}

=34.8\pm 3.95

=30.85,38.75

98% confidence interval for population mean is (34.8-3.95<\mu<34.8+3.95)

(30.85 <\mu< 38.75) is the 98% confidence interval for population mean

#98% confidence interval for population mean number of text per day is between 30.85 and 38.75

c) If many groups of 126 randomly selected days were observed, then a different confidence interval would be produced from each group. About 98 percent of these confidence intervals will contain the true population mean number of hours of sunlight per day that the solar panel receives and about 2 percent will not contain the true population mean number of hours of sunlight that the solar panel receives

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