Question

In: Statistics and Probability

A large supermarket carries four qualities of ground beef. Customers are believed to purchase these four...

A large supermarket carries four qualities of ground beef. Customers are believed to purchase these four varieties with probabilities of 0.13, 0.27, 0.14, and 0.46, respectively, from the least to most expensive variety. A sample of 480 purchases resulted in sales of 48, 148, 74, and 210 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.)

A program for generating random numbers on a computer is to be tested. The program is instructed to generate 100 single-digit integers between 0 and 9. The frequencies of the observed integers were as follows. At the 0.05 level of significance, is there sufficient reason to believe that the integers are not being generated uniformly?

Integer 0 1 2 3 4 5 6 7 8 9
Frequency 10 8 6 8 13 10 7 11 14 13

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


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

Solutions

Expert Solution

1)

           relative observed Expected Chi square
category frequency(p) Oi Ei=total*p R2i=(Oi-Ei)2/Ei
1 0.13 48.0 62.40 3.323
2 0.27 148.0 129.60 2.612
3 0.14 74.0 67.20 0.688
4 0.46 210.0 220.80 0.528
total 1.000 480 480 7.1518
test statistic X2 = 7.15

ii)

p value = 0.0672

2)

applying chi square goodness of fit test:
           relative observed Expected residual Chi square
category frequency(p) Oi Ei=total*p R2i=(Oi-Ei)/√Ei R2i=(Oi-Ei)2/Ei
0 0.10 10.0 10.00 0.00 0.000
1 0.10 8.0 10.00 -0.63 0.400
2 0.10 6.0 10.00 -1.26 1.600
3 0.10 8.0 10.00 -0.63 0.400
4 0.10 13.0 10.00 0.95 0.900
5 0.10 10.0 10.00 0.00 0.000
6 0.10 7.0 10.00 -0.95 0.900
7 0.10 11.0 10.00 0.32 0.100
8 0.10 14.0 10.00 1.26 1.600
9 0.10 13.0 10.00 0.95 0.900
total 1.000 100 100 6.8000
test statistic X2 = 6.80
ii) p value = 0.6579

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