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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.

1 Answer

3 votes

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.

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.

User Akayh
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