Question

A large supermarket carries four qualities of ground beef. Customers are believed to purchase these four varieties with probabilities of 0.12, 0.28, 0.16, and 0.44, respectively, from the least to most expensive variety. A sample of 497 purchases resulted in sales of 42, 146, 90, and 219 of the respective qualities. Does this sample contradict the expected proportions? Use α = 0.05.

(a) Find the test statistic. (Round your answer to two decimal places.)


(ii) Find the p-value. (Round your answer to four decimal places.)


(b) State the appropriate conclusion.

Fail to reject the null hypothesis. There is significant evidence of a difference from the expected proportions. Fail to reject the null hypothesis. There is not significant evidence of a difference from the expected proportions.     Reject the null hypothesis. There is significant evidence of a difference from the expected proportions. Reject the null hypothesis. There is not significant evidence of a difference from the expected proportions.

Answer 1

here, we will do chi square goodness of fit test.

hypothesis:-

the necessary calculation table:-

variety	observed frequency (Oi)	test proportion	expected frequency(Ei)
1	42	0.12	497*0.12 = 59.64	5.21746
2	146	0.28	497*0.28 = 139.16	0.33620
3	90	0.16	497*0.16 = 79.52	1.38117
4	219	0.44	497*0.44 = 218.68	0.00047
Total	497			6.9353

a).the test statistic be:-

p value = 0.0740

[ df = (4-1)= 3

in any blank cell of excel type =CHISQ.DIST.RT(6.9353,3) press enter.

i had used 6.9353 in spite of 6.54, for getting more accurate answers.]

b).conclusion:-

Fail to reject the null hypothesis. There is not significant evidence of a difference from the expected proportions

[ p value = 0.0740 >0.05 (alpha)

so, we fail to reject the null hypothesis.]

*** if you have any doubt regarding the problem please write it in the comment box.if you are satisfied please give me a LIKE if possible..