Q1-16BSC
Question
Test Statistics. In Exercises 13–16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)
16. Exercise 8 “Pulse Rates”
Step-by-Step Solution
VerifiedThe value of the chi-square test statistic is equal to 160.404.
The pulse rates of a sample of 153 adult males have a standard deviation equal to 11.3 bpm. The claim is that the standard deviation of the pulse rates of adult males is more than 11bpm.
In Correspondence with the given claim, the following hypotheses are set up:
Null Hypothesis: The standard deviation of the pulse rates of adult males is equal to 11bpm. Mathematically,
\({H_0}:\sigma = 11\;{\rm{bpm}}\)
Alternative Hypothesis: The standard deviation of the pulse rates of adult males is more than 11bpm. Mathematically,
\({H_1}:\sigma > 11\;{\rm{bpm}}\)
Since the claim involves testing the population's standard deviation's equality with a hypothesized value, the test statistic used will be the Chi-square statistic.
The chi-square test statistic has the following expression:
\({\chi ^2} = \frac{{\left( {n - 1} \right){s^2}}}{{{\sigma ^2}}}\)
Where
n is the sample size
\({s^2}\) is the sample variance
\({\sigma ^2}\) is the population variance
Here, the sample size (n) is equal to 153.
The value of the sample variance is computed below:
\(\begin{array}{c}{s^2} = {\left( {11.3} \right)^2}\\ = 127.69\end{array}\)
The population variance is computed below:
\(\begin{array}{c}{\sigma ^2} = {\left( {11} \right)^2}\\ = 121\end{array}\)
Thus, the value of the test statistic is as follows:
\(\begin{array}{c}{\chi ^2} = \frac{{\left( {n - 1} \right){s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {153 - 1} \right)127.69}}{{121}}\\ = 160.404\end{array}\)
Therefore, the value of the test statistic is equal to 160.404.