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).

$99.9\mathrm{\%}C.I.=(35.68,61.54)$Incorrect

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Delorenzoz

Skilled2021-02-12Added 91 answers

Step 1

Given information-

Sample size, $n=7$

Excel is used to compute the sample mean and sample standard deviation.

$=AVERAGE(54.7:54.2)$

And standard deviation-

$=STDEV.S(54.7:54.2)$

Sample mean, $x-\stackrel{\u2015}{=}48.61429$

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 $=\stackrel{\u2015}{x}\pm {t}_{\frac{\alpha}{2},n-1}\times \frac{s}{\sqrt{n}}$

So, t-critical is given by

${t}_{\frac{0.001}{2},7-1}=5.959(\text{From excel using formula}=T.INV.2T(0.001,6))$

$C.I.=48.61429\pm 5.959\times \frac{6.571765}{\sqrt{7}}$

$C.I=(33.81,63.42)$

So, $99.9\mathrm{\%}$ confidence interval is 33.81 to 63.42

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