In: Statistics and Probability
A poultry producer is interested in which feed type will yield the poultry weight. He has two types of feed; the first made from soybeans and the second made from sunflower seeds. He has 71 chickens and randomly selects 26 of these chickens to use in his experiment. 14 of these chickens are randomly assigned to receive only the soybean product and the remaining 12 are to receive only the sunflower product. Is there a difference in these weights at a ten percent error rate? Estimate the mean difference in the weights of these chickens using 90% confidence. The data are below.
Soybean Feed 199.86 237.32 335.21 203.80 166.66 340.68 234.31 254.03 296.78 289.57 246.15 247.51 125.27 296.90
Sunflower Feed 279.04 247.33 362.32 262.55 305.91 295.22 335.41 292.62 336.80 307.60 307.67 317.60
Soybean Feed
n1 = 10
x̄1 = 0.022
s1 = 0.159
Sunflower Feed
n2 = 13
x̄2 = 0.008
s2 = 0.008
Soybean Feed ( X ) | Sunflower Feed ( Y ) | |||
199.86 | 2331.5764 | 279.04 | 631.6426 | |
237.32 | 117.2109 | 247.33 | 3231.0698 | |
335.21 | 7580.0704 | 362.32 | 3381.1318 | |
203.8 | 1966.6032 | 262.55 | 1732.4325 | |
166.66 | 6640.0334 | 305.91 | 3.0189 | |
340.68 | 8562.4671 | 295.22 | 80.1473 | |
234.31 | 191.446 | 335.41 | 975.7814 | |
254.03 | 34.6167 | 292.62 | 133.4603 | |
296.78 | 2365.227 | 336.8 | 1064.5538 | |
289.57 | 1715.9146 | 307.6 | 11.7478 | |
246.15 | 3.9856 | 307.67 | 12.2325 | |
247.51 | 0.405 | 317.6 | 180.2978 | |
125.3 | 15098.6097 | |||
296.9 | 2376.9135 | |||
Total | 3474.05 | 48985.0795 | 3650.07 | 11437.5165 |
Mean
Standard deviation
Mean
Standard deviation
To Test :-
H0 :-
H1 :-
Test Statistic :-
t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) +
(S_{2}^{2} / n2)}
t = ( 248.1464 - 304.1725 ) / \sqrt{ ( 61.3847^{2} / 14) + (
32.2455^{2} / 12)}
t = -2.9702
Test Criteria :-
Reject null hypothesis if
DF = 20
Result :- Reject Null Hypothesis
Decision based on P value
P - value = P ( t > 2.9702 ) = 0.0076
Reject null hypothesis if P value <
level of significance
P - value = 0.0076 < 0.1 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis
There is sufficient evidence to support the claim that there is difference in average poultry weight at 10% level of significance.
Confidence interval :-
Lower Limit =
Lower Limit = -88.5587
Upper Limit =
Upper Limit = -23.4935
90% Confidence interval is ( -88.5587 , -23.4935 )