A proofreader checked 160 ads for grammatical errors.The distribution obtained is shown: At α = .01,what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution?
A) Reject the null and conclude the distribution is not uniform.
B) Conclude that there is insufficient evidence to reject the null.
C) No conclusion can be made due to small expected frequencies.
D) No conclusion can be made due to inadequate sample size.

Question

A proofreader checked 160 ads for grammatical errors.The distribution obtained is shown:  At α = .01,what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution?
A) Reject the null and conclude the distribution is not uniform. 
B) Conclude that there is insufficient evidence to reject the null. 
C) No conclusion can be made due to small expected frequencies. 
D) No conclusion can be made due to inadequate sample size.

Quizplus · Accepted Answer

In a goodness-of-fit test, we test whether a sample fits a hypothesized distribution. In this case, the hypothesized distribution is uniform. We can use the chi-square test to test the goodness of fit.

The expected frequency for each interval in a uniform distribution is equal. Therefore, we can calculate the expected frequency for each interval by dividing the total frequency (160) by the number of intervals (6): 160/6 = 26.67.

We can then calculate the chi-square statistic using the formula:

χ2 = ∑(Observed frequency - Expected frequency)2/Expected frequency

Using the observed and expected frequencies from the distribution table, we get:

χ2 = (20-26.67)2/26.67 + (26-26.67)2/26.67 + (28-26.67)2/26.67 + (26-26.67)2/26.67 + (26-26.67)2/26.67 + (34-26.67)2/26.67 = 6.246

The degrees of freedom for a goodness-of-fit test with k intervals is (k-1). In this case, we have 6 intervals, so the degrees of freedom are 5.

Using a chi-square distribution table with 5 degrees of freedom and α = .01, the critical value is 15.086. The calculated χ2 value (6.246) is less than the critical value (15.086). Therefore, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the distribution is not uniform.

A Proofreader Checked 160 Ads for Grammatical Errors