 Reeves

2021-02-11

I have a question:
With the given sample, you want to calculate the population mean $\mu$.
54.7
48.9
56.3
43.5
41.5
41.2
54.2
You think the population is dispersed normally. Find the 99.99 percent confidence level range. Put your response in parenthesis as an open-interval that is accurate to two decimal places (because the sample data are reported accurate to one decimal place).
$99.9\mathrm{%}C.I.=\left(35.68,61.54\right)$Incorrect
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. Delorenzoz

Step 1
Given information-
Sample size, $n=7$
Excel is used to compute the sample mean and sample standard deviation.
$=AVERAGE\left(54.7:54.2\right)$
And standard deviation-
$=STDEV.S\left(54.7:54.2\right)$
Sample mean, $x-\stackrel{―}{=}48.61429$
Sample standard deviation, $s=6.571765$
Confidence level, $c=0.999$
So, significance level, $\alpha =0.001$
We shall employ the t-distribution because the population's standard deviation is unknown here.

Now, degree of freedom is given by-
Degree of freedom, $df=n-1$
Degree of freedom, $df=7-1=6$
Step 2
Now, $99.9\mathrm{%}$ confidence interval is given by
Confidence interval $=\stackrel{―}{x}±{t}_{\frac{\alpha }{2},n-1}×\frac{s}{\sqrt{n}}$
So, t-critical is given by
${t}_{\frac{0.001}{2},7-1}=5.959\left(\text{From excel using formula}=T.INV.2T\left(0.001,6\right)\right)$
$C.I.=48.61429±5.959×\frac{6.571765}{\sqrt{7}}$
$C.I=\left(33.81,63.42\right)$
So, $99.9\mathrm{%}$ confidence interval is 33.81 to 63.42

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