A worker is supposed to manually weigh and mix apples, oranges, pineapple, andkiwi in a 7:5:2:2...

A worker is supposed to manually weigh and mix apples, oranges, pineapple, andkiwi in a 7:5:2:2 ratio by weight. A bag containing 80oz of these mixed fruits was found to have 39ozof apples, 26 oz of oranges, 12 oz of pineapple, and 3 oz of kiwi. At the 0.05 level of significance, testthe hypothesis that the worker is mixing the fruit in the 7:5:2:2 ratio.

Expert Solution

Chi-Square Goodness of Fit test

(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H0: p1 =0.4375, p2 =0.3125, p3=0.125, p4=0.125
Ha: Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit.

(2) Degrees of Freedom
The number of degrees of freedom is df=n-1=4-1=3

(3) Test Statistics
The Chi-Squared statistic is computed as follows:

(4)Critical Value and Rejection Region
Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 7.8147.
Then the rejection region for this test is R={χ2:χ2>7.8147}.

(5)P-value
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 7.8147.
The p-value is p=Pr(χ2>5.7971)=0.1219

(6) The decision about the null hypothesis
Since it is observed that χ2=5.7971≤χc2=7.8147, it is then concluded that the null hypothesis is not rejected.

(7) Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!

orchestra answered 2 years ago

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