A study was to be undertaken to determine if a particular training program would improve physical fitness. A simple random sample of 35 university students was selected to be enrolled in the training program. After the training program was completed, the maximum oxygen uptake of the students was measured to indicate their fitness level. The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects, whose maximum oxygen uptake is known to be 45 mL/kg/min on average. Based on the sample of 35 students, the maximum oxygen uptake was found to have slight skewness but no strong outliers. Why can we safely use a t procedure in this testing situation

The t-test for one mean can be safely used for this situation.

Step-by-step explanation:

A statistical experiment is conducted to determine if a particular training program would improve physical fitness.

The objective of the experiment was to determine if there is some evidence that the maximum oxygen uptake of the sample of students differed from the general population of untrained subjects, whose maximum oxygen uptake is known to be 45 ml/kg/min on average.

The hypothesis is defined as:

H₀: The population mean is 45 ml/kg/min, i.e. μ = 45.

Hₐ: The population mean is different from 45 ml/kg/min, i.e. μ ≠ 45.

The test for single mean can be either using the z-distribution or the t-distribution.

The z-distribution is used if it provided that:

The t-distribution is used if :

To use a t-distribution for a single mean test the following assumptions are to made:

In this case we will use a t-distribution since there is no information about the population standard deviation.

The sample of 35 students are randomly selected for the training program.

The maximum oxygen uptake of every student is independent of the others.

Thus, the t-test for one mean can be safely used for this situation.

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%

19

step-by-step explanation: