Question

Every user of statistics should understand the distinction between statistical significance and practical importance. A sufficiently large sample will declare very small effects statistically significant. Consider the study of elite female Canadian athletes that investigated whether elite athletes are deficient in their nutritional intake. A total of n athletes from eight Canadian sports centers participated in the study. Female athletes were consuming an average of 2403.7 kcal/day with a standard deviation of 880 kcal/day. Suppose a nutritionist is brought in to implement a new health program for these athletes. This program should increase mean caloric intake but not change the standard deviation. Given the standard deviation and how caloric deficient these athletes are, a change in the mean of 50 kcal/day to 2453.7 is of little importance. However, with a large enough sample, this change can be significant. To see this, calculate theP-value for the test of

H0:μ=2403.7
Ha:μ>2403.7

in the following situation:

A sample of 100 athletes; their average caloric intake is x= 2453.7.

(Fill in the blanks below and roundzto to 2 decimal places and round P to 4 decimal places.)

For n=100, z=____ P=P(Z >___) =

Answer 1

The test statistic value is, z = 0.57.

The p-value of the test is 0.2843.

Step-by-step explanation:

A one-sided, single mean z-test can be performed to determine whether the mean caloric intake increased from 2403.7 kcal/day or not.

The hypothesis is defined as:

H₀: The mean caloric intake did not increase, i.e. μ = 2403.7.

Hₐ: The mean caloric intake did increase, i.e. μ > 2403.7.

The information provided is:

`\bar x = 2453.7\ kcal/day\\n=100\\\sigma = 880\ kcal/day`

The z-statistic is:

`z=(\bar x-\mu)/(\sigma/√(n))=(2453.7-2403.7)/(880/√(100))=0.5682\approx0.57`

The test statistic value is, z = 0.57.

Compute the p-value of the test statistic as follows:

`p-value=P(Z>0.57)=1-P(Z<0.57)=1-0.7157=0.2843`

*Use a standard normal table.

The p-value of the test is 0.2843.