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

Answer:

$0.609 \leq \sigma \leq 0.772$  

And the best conclusion would be:

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

Step-by-step 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:  

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$  

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:  

$\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}$  

$0.371 \leq \sigma^2 \leq 0.596$  

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.