Question

A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Construct a 90% confidence interval estimate of the standard deviation of body temperature of all healthy humans. Is it safe to conclude that the population standard deviation is less than 1.40°F​?

_____°F


Is it safe to conclude that the population standard deviation is less than 1.40°F​?


A. This conclusion is not safe because 1.40 °F is in the confidence interval.


B. This conclusion is safe because 1.40 °F is in the confidence interval.


C. This conclusion is not safe because 1.40°F is outside the confidence interval.


D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Answer 1

`0.609 \leq \sigma \leq 0.772`

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Explanation:

1) Data given and notation

s=0.68 represent the sample standard deviation

`\bar x =98.90` represent the sample mean

n=98 the sample size

Confidence=90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

`df=n-1=98-1=97`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:

`\chi^2_(\alpha/2)=120.990`

`\chi^2_(1- \alpha/2)=75.282`

And replacing into the formula for the interval we got:

`((97)(0.68)^2)/(120.990) \leq \sigma \leq ((97)(0.68)^2)/(75.282)`

`0.371 \leq \sigma^2 \leq 0.596`

Now we just take square root on both sides of the interval and we got:

`0.609 \leq \sigma \leq 0.772`

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.