Question

15.

Cookies: Following are the weights of 8 boxes of cookies, each of which is labeled as containing 16 ounces. Assume that the population of weights is normally distributed. 15.91    16.06    16.08    15.97    16.02    15.88    15.89    16.01 Can you conclude that the population standard deviation is less than 0.1? Use the α = 0.01 level of significance.

Answer 1

Solution:

Given:

Claim: the population standard deviation is less than 0.1

level of significance = α = 0.01

Sample size =n = 8

Step 1) State H0 and H1:

Ho: 0 = 0.1

Vs

H1:0 <0.1

( Left tailed test)

Step 2) Find test statistic:

Chi square test statistic for variance ( Standard deviation)

12 _ (n-1) * $

where

2 Σα - (Σα): /n η – 1

Thus we need to make following table:

x	x^2
15.91	253.1281
16.06	257.9236
16.08	258.5664
15.97	255.0409
16.02	256.6404
15.88	252.1744
15.89	252.4921
16.01	256.3201
Στ = 127.82	Σχ2 = 2042.286

Thus

2 Σα - (Σα): /n η – 1

| 2042.286-(127.8252/8 8-1

2042.286-2042.24405 s =

0.04195 S2 =

$2 = 0,005993

Thus

12 _ (n-1) * $

(8 - 1) < 0.005993 x² = ! 0.12

x =- 7 x 0.005993 0.01

= 4.195

Step 3) Find chi-square critical value;

df = n- 1 = 8 -1 = 7

level of significance = α = 0.01 but this is left tailed test , thus use Area= 1 - 0.01 = 0.99

X-995 x2950 x2000 O CON 0.000 0.010 0.072 0.207 X.990 0 000 0.020 0115 0297 0554 0.872 1.239 0.001 0.051 0.216 0.484 0.831 1.237 1.690 0.004 0.103 0.352 0.711 1.145 1.635 2.167 0.016 0.211 0.584 1.064 1.610 2.204 2.833 X2100 2.706 4.605 6.251 7.779 9.236 10.645 12.017 0.12 0.676 0.00

Chi-square critical value = 1.239

Step 4) Decision Rule:

Reject null hypothesis H0, if Chi square test statistic < Chi-square critical value = 1.239 , otherwise we fail to reject H0.

Since Chi square test statistic 4.195 > Chi-square critical value = 1.239 , we fail to reject H0.

Step 5) Conclusion:
There is not sufficient evidence to  conclude that the population standard deviation is less than 0.1