A study was to be undertaken to determine if a particular training program would improve physical fitness.An SRS of 31 university students was selected to be enrolled in the fitness program.One important measure of fitness is maximum oxygen uptake.Measurements of oxygen uptake in untrained individuals are known to follow a Normal distribution with a mean of $\mu$= 45 mL/kg/min.The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects.The measurements made on the subjects coming into this study produced a mean of = 47.4 mL/kg/min with a standard deviation of s = 5.3 mL/kg/min.What is the approximate P-value for the test? A)P

Question

A study was to be undertaken to determine if a particular training program would improve physical fitness.An SRS of 31 university students was selected to be enrolled in the fitness program.One important measure of fitness is maximum oxygen uptake.Measurements of oxygen uptake in untrained individuals are known to follow a Normal distribution with a mean of $\mu$= 45 mL/kg/min.The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects.The measurements made on the subjects coming into this study produced a mean of  = 47.4 mL/kg/min with a standard deviation of s = 5.3 mL/kg/min.What is the approximate P-value for the test?
A)P < 0.005 
B)0.005 < P < 0.01 
C)0.01 < P < 0.02 
D)P > 0.02 
E)P = 0.0118

Quizplus · Accepted Answer

The null hypothesis is that the sample of students does not differ from the general population of untrained subjects, i.e. the mean of the population is still 45 mL/kg/min. The alternative hypothesis is that the mean of the sample is significantly different from 45 mL/kg/min.

We will use a two-tailed t-test with significance level alpha = 0.05. We calculate the t-statistic as:

t = (47.4 - 45) / (5.3 / sqrt(31)) = 2.421

Using a t-distribution table with degrees of freedom 30, we find the p-value to be approximately 0.018.

Since this is a two-tailed test, we double the p-value to get 0.036. This value is between 0.01 and 0.02, so the answer is choice C: 0.01 < P < 0.02.

A Study Was to Be Undertaken to Determine If a Particular