Question

A physician claims that joggers’ average oxygen uptake is greater than the average for all adults. A simple random sample of fifteen joggers has a mean of 40.6 ml/kg and a standard deviation of 6ml/kg. If the average of all adults is 36.7 ml/kg, is there enough evidence to support the physician’s claim at α = 0.05? Use the P-Value Method to solve this problem. Assume the population is normally distributed. The standard deviation for all adults is not known.

State the information given:

State the claim:

State the requirements (are they met?):

Step 1 (state the claim symbolically):

Step 2 (state the opposite if the claim is false):

Step 3 (state the null and alternative hypotheses):

Step 4 (state α):

Step 5 (state the test statistic):

Step 6a (calculate test statistic):

Step 6b (find the p-value):

Step 7 (compare p-value to alpha and accept or reject Ho):

Step 8 (state, as a sentence, the conclusion):

Answer 1

We have a mean and a standard deviation of fifteen joggers and a mean of all adults.

The claim is that joggers’ average oxygen uptake is greater than the average for all adults.

The one-sample t-test has four main assumptions:

• • The dependent variable must be continuous (interval/ratio).
• • The observations are independent of one another.
• • The dependent variable should be approximately normally distributed.
• • The dependent variable should not contain any outliers.

All the assumptions are met.

The hypothesis being tested is:

H0: µ = 36.7

Ha: µ > 36.7 (Claim)

α = 0.05

The test statistic, z = (x - µ)/σ/√n

z = (40.6 - 36.7)/6/√15 = 2.52

The p-value is 0.0059.

Since the p-value (0.0059) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that joggers’ average oxygen uptake is greater than the average for all adults.

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