Question

A manufacturer is considering purchasing parts from three different suppliers.

manufacturer is considering purchasing parts from three dfferent suppliers. The parts received from the suooliers are classied as havieg a minor defect, having a majr oe, r being uod. Test resuts frorm samples of parts received from each of the three suppliers are shown below. Note that any test with these data is no longer a test of proportiens for the theee supplier populations beasse the cabegorical response variatie har three outcomes: minor defect, major defedt, and good. Supplier A BC Minor Defect 16 11 Major Defect 10 Goed 35 126 115 of delesi the sane for the three scotter s Use the chi square test calculations as presented in this section with the exception that atable using the data atove, conduct a hypothesls test to detmine if th with rows and e columna resuqtet awith1X-1) gress of relom ing a 00 vevel of sionce, h s the p The p value isSelect aur What is yout condusion conclude that we are Selety ▼ ect the b pothesis that the population at itun nor eca is the S mefur al-seni es - tore

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The population distribution of defects is the same for all suppliers.

Alternative hypothesis: At least one of the null hypothesis statements is false.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for homogeneity.

Analyze sample data. Applying the chi-square test for homogeneity to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (3 - 1) * (3 - 1)
D.F = 4
E_r,c = (n_r * n_c) / n

$\chi ^{2}=\sum \left [\frac{(O_{r,c}-E_{r,c})^{2}}{E_{r,c}} \right ]$
Χ² = 14

Expected l0O..-E 16 Expected 24 Total Minor defects 17 Major defects Good Total 135 154 127 154 10 126 147 121 147 115 154 127 154 28 376 455

where DF is the degrees of freedom, r is the number of populations, c is the number of levels of the categorical variable, n_r is the number of observations from population r, n_c is the number of observations from level c of the categorical variable, n is the number of observations in the sample, E_r,c is the expected frequency count in population r for level c, and O_r,c is the observed frequency count in population r for level c.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 14.

We use the Chi-Square Distribution Calculator to find P(Χ² > 14) = 0.007.

Interpret results. Since the P-value (0.007) is less than the significance level (0.05), we reject the null hypothesis.

From the above test we can conclude that we are correct to reject the hypothesis that the population distribution of defects is the same for all suppliers.