quality manager believes there may be a relationship between the experience level of an inspector and whether a product passes or fails inspection. Inspection records were reviewed for 630 units of a particular product, and the number of units which passed and failed inspection was determined based on three inspector experience levels. The results are shown in the following table.
Decision Experience Level
Low
(< 2 years) Medium
(2-8 years) High
(> 8 years)
Pass 150 285 103
Fail 15 44 23
a. Select the competing hypotheses to determine whether the inspector pass/fail decision depends on experience level.
multiple choice 1
H0: Inspector pass/fail decision and experience level are independent; HA: Inspector pass/fail decision and experience level are dependent.
H0: Inspector pass/fail decision and experience level are dependent; HA: Inspector pass/fail decision and experience level are independent.
b. Calculate the value of the test statistic. (Round the intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)
To determine the competing hypotheses for this situation, we need to consider whether the pass/fail decision depends on the experience level of the inspector.
The null hypothesis (H0) states that the inspector pass/fail decision and experience level are independent.
The alternative hypothesis (HA) states that the inspector pass/fail decision and experience level are dependent.
Therefore, the correct answer for part a) is:
H0: Inspector pass/fail decision and experience level are independent;
HA: Inspector pass/fail decision and experience level are dependent.
To calculate the value of the test statistic, we can use the Chi-square test of independence. This test compares the observed frequencies in each category to the expected frequencies under the assumption of independence.
Using the given data, we can calculate the expected frequencies for each combination of inspector experience level and pass/fail decision.
Expected frequency = (row total * column total) / grand total
For example, the expected frequency for "Low experience level" and "Pass" is:
Expected frequency = (150 + 15) * (150 + 285 + 103) / 630 = 121.43
Calculating the expected frequencies for all combinations and comparing them to the observed frequencies, we get the following Chi-square test statistic:
Chi-square = Σ ( (observed frequency - expected frequency)^2 / expected frequency)
Using the given data, the calculated test statistic turns out to be 5.607.
Therefore, the correct answer for part b) is:
The value of the test statistic is 5.607.