 # I have a question: You intend to estimate a population mean displaystylemu with the following sample. 54.7 48.9 56.3 43.5 41.5 41.2 54.2 You believe t Reeves 2021-02-11 Answered
I have a question:
You intend to estimate a population mean $\mu$ with the following sample.
54.7
48.9
56.3
43.5
41.5
41.2
54.2
You believe the population is normally distributed. Find the $99.9\mathrm{%}$ confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp 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.
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Step 1
Given information-
Sample size, $n=7$
Calculating sample mean and sample standard deviation using excel.
$=AVERAGE\left(54.7:54.2\right)$
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, $\alpha =0.001$
Since here population standard deviation is unknown so we will use t-distribution.
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