In: Statistics and Probability
A random sample of communities in western Kansas gave the following information for people under 25 years of age. Rate of hay fever per 1000 population for people under 25 124 96 98 107 120 122 142 136 112 124 81 124 108 120 144 115 A random sample of regions in western Kansas gave the following information for people over 50 years old. : Rate of hay fever per 1000 population for people over 50 87 103 100 98 105 98 106 117 83 105 99 86 117 105 Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use State the null and alternate hypotheses.
Under 25 Years ( X ) | Over 50 years ( Y ) | |||
25 | 7507.7199 | 50 | 2234.1409 | |
124 | 152.5941 | 87 | 105.4051 | |
96 | 244.8317 | 103 | 32.8707 | |
98 | 186.2433 | 100 | 7.4709 | |
107 | 21.5955 | 98 | 0.5377 | |
120 | 69.7709 | 105 | 59.8039 | |
122 | 107.1825 | 98 | 0.5377 | |
142 | 921.2985 | 106 | 76.2705 | |
136 | 593.0637 | 117 | 389.4031 | |
112 | 0.1245 | 83 | 203.5387 | |
124 | 152.5941 | 105 | 59.8039 | |
81 | 939.2447 | 99 | 3.0043 | |
124.0 | 152.5941 | 86 | 126.9385 | |
108 | 13.3013 | 117 | 389.4031 | |
120 | 69.7709 | 105 | 59.8039 | |
144 | 1046.7101 | |||
115 | 11.2419 | |||
Total | 1898 | 12189.8817 | 1459 | 3748.93 |
Mean
Standard deviation
Mean
Standard deviation
To Test :-
H0 :-
H1 :-
Test Statistic :-
t = 1.8166
Test Criteria :-
Reject null hypothesis if
DF = 26
Result :- Reject Null Hypothesis
Decision based on P value
P - value = P ( t > 1.8166 ) = 0.0404
Reject null hypothesis if P value <
level of significance
P - value = 0.0404 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis
There is sufficient evidence to support the claim that the age group over 50 has a lower rate of hay fever at 5% level of significance.