Question

1A) A large supermarket carries four qualities of ground beef. Customers are believed to purchase these four varieties with probabilities of 0.13, 0.16, 0.13, and 0.58, respectively, from the least to most expensive variety. A sample of 482 purchases resulted in sales of 46, 151, 88, and 197 of the respective qualities. Does this sample contradict the expected proportions? Use α = .05.

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


(b) Find the p-value. (Give your answer bounds exactly.)
____< p < ____

1B) 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	12	6	8	8	14	10	6	12	13	11

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


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

Answer 1

Solution:

Question 1A) A large supermarket carries four qualities of ground beef.

Customers are believed to purchase these four varieties with probabilities of 0.13, 0.16, 0.13, and 0.58, respectively, from the least to most expensive variety.

Sample size = n = 482

and observed counts of purchases resulted in sales as 46, 151, 88, and 197

We have to test if this sample contradict the expected proportions.

Level of significance = 0.05

Thus hypothesis would be:

H0: p₁ = 0.13, p₂ =0.16, p₃= 0.13, p₄ = 0.58

H1: this sample contradict the expected proportions.

Part a) Find the test statistic

Since we have to test if sample or observed counts are same as expected counts, we use Chi-square test of goodness of fit test.

Thus Chi-square test statistic is given by:

$\chi ^{2}=\sum \frac{O_{i}^{2}}{E_{i}}-N$

Where Oi = Observed counts and Ei = Expected counts

Oi : Observed counts	Expected proportions	Calculations	Ei : Expected Counts	Oi^2/Ei
46	0.13	=482*0.13	62.66	33.770
151	0.16	=482*0.16	77.12	295.656
88	0.13	=482*0.13	62.66	123.588
197	0.58	=482*0.58	279.56	138.822
N = 482				$\sum \frac{O_{i}^{2}}{E_{i}}=591.835$

Thus

$\chi ^{2}=\sum \frac{O_{i}^{2}}{E_{i}}-N$

$\chi ^{2}=591.835 -482$

$\chi ^{2}=109.84$

Thus the test statistic is $\chi ^{2}=109.84$

Part b) Find the p-value

df = k - 1 = 4 - 1 = 3

look in Chi-square table for df = 3 row and find the interval in which 109.835 fall and then find corresponding right tail area.

10.000 0.000 0.00 0.004 0.016 2.706 3.841 5.024 6.635 7.879 20.010 0.020 000.103 4.605 5.991 7.378 9.210 10.597 8951 78150248 11 2412.838 4 0.207 0.297 0.484 0.711.064 7.779 9.488 11.143 13.27714.860 5 0.412 0.554 0.83 1.145 1.610 9.236 11.070 12.833 15.086 16.750

$\chi ^{2}=109.84$ > 12.838 ,

Area for 12.838 is 0.005

So Area corresponding to 109.84 would be less than 0.005

Thus bounds on p-value are:

0.000 < p-value < 0.005

Since p-value < 0.05, we reject H0 and thus we conclude that: this sample contradict the expected proportions.

Question 1B) 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.

Integer	0	1	2	3	4	5	6	7	8	9
Frequency	12	6	8	8	14	10	6	12	13	11

We have to test if there is sufficient reason to believe that the integers are not being generated uniformly .

Thus H0 and H1 are:

H0: the integers are being generated uniformly

Vs

H1: the integers are not being generated uniformly

Part a) Find the test statistic

$\chi ^{2}=\sum \frac{O_{i}^{2}}{E_{i}}-N$

Integer	Oi: Frequency	Calculations	Ei: Expected Counts	Oi^2/Ei
0	12	=100/10	10	14.4
1	6	=100/10	10	3.6
2	8	=100/10	10	6.4
3	8	=100/10	10	6.4
4	14	=100/10	10	19.6
5	10	=100/10	10	10
6	6	=100/10	10	3.6
7	12	=100/10	10	14.4
8	13	=100/10	10	16.9
9	11	=100/10	10	12.1
	N = 100			$\sum \frac{O_{i}^{2}}{E_{i}} =107.40$

Thus

$\chi ^{2}=\sum \frac{O_{i}^{2}}{E_{i}}-N$

$\chi ^{2}=107.40 -100$

$\chi ^{2}= 7.40$

Part b) Find the p-value

To find exact p-value, we use Excel command:

=CHISQ.DIST.RT( x ,df )

where x = $\chi ^{2}= 7.40$ and df = k - 1 = 10 - 1 = 9

=CHISQ.DIST.RT( 7.40 , 9 )

=0.5955

Thus p-value = 0.5955

Since p-value = 0.5955 > 0.05 level of significance , we fail to reject H0 and thus there is not sufficient reason to believe that the integers are not being generated uniformly .

That is: the integers are being generated uniformly.