Question

A researcher would like to examine the effect of labor unions on the average pay of workers. The investigator obtains information from the National Carpenters Union, which states that the average pay for union carpenters is μ=$7.80 per hour. The researcher then obtains a random sample of 8 carpenters who do not belong to a union and records the pay for each. The data from this sample are as follows: 6.25, 7.50, 6.75, 6.90, 7.10, 7.15, 8.00, 6.15 per hour. Does the average pay of nonunion carpenters differ from union members? Test at .01 level of significance and report effect size.

Answer 1

The given problem is to test whether the average pay of non union carpenters differ from union members. Thus we use "ONE SAMPLE T TEST" to test this hypothesis.

I have used excel function to calculate mean and standard deviation of given sample.

To calculate Mean: "=AVERAGE(select array of data values)"

To calculate Standard deviation: "=STDEV(select array of data values)"

GIVEN:

Sample size

Sample mean

Sample standard deviation

HYPOTHESIS:

(That is, the average pay of non union carpenters is not significantly different from union carpenters $7.80 per hour.)

(That is, the average pay of non union carpenters is significantly different from union carpenters $7.80 per hour.)

LEVEL OF SIGNIFICANCE:

TEST STATISTIC:

which follows t distribution with degrees of freedom.

CALCULATION:

CRITICAL VALUE:

The two tailed (since ) t critical value with degrees of freedom at is .

DECISION RULE:

INFERENCE:

Since the calculated t statistic value (-3.80) is less than the t critical value (-3.499), we reject null hypothesis and conclude that the average pay of non union carpenters is significantly different from union carpenters ($7.80 per hour).

EFFECT SIZE:

The formula of effect size for one sample is given by,

where is sample mean, is sample standard deviation and is hypothesized value .

Thus the effect size is,

Since the calculated effect size is less than , there is a small effect size.