Stacie Worsley

2022-01-09

Ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours, and the population variance was 0.6084 hour. Find the 99% confidence interval of the mean time. Assume the variable is normally distributed. In connection with the given data, explain the concept of the statement ‘the sample mean is significant or not significant at the given significance level’.

### Answer & Explanation

Lynne Trussell

Step 1
Given information-
Sample mean, $x-\stackrel{―}{=}7.1$
Population variance, $\sigma 2=0.6084$
So, Standard deviation $=0.78$
Confidence level, $c=99\mathrm{%}$
Sample size, $n=10$
Significance level $\left(\alpha \right)=1-0.99=0.01$
Step 2
Since here, population variance is given, therefore using z-test.
99% Confidence interval is given by the following formula
$C.I=\stackrel{―}{x}±{z}_{\frac{0.1}{2}}×\frac{\sigma }{\sqrt{n}}$
$C.I=7.1±2.57583×\frac{0.78}{\sqrt{10}}=\left(6.4646,7.7353\right)$
${z}_{\frac{0.01}{2}}=±2.57583$ (Using excel =NORM.S.INV (0.005))
Since sample mean is in the range of confidence interval, therefore the sample mean is significant at the given significance level.

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