If we test the null hypothesis at the 5% level of significance, then the decision is (c)reject null hypothesis and conclude the means are different .
In the question ,
it is given that ,
the results of a mathematics placement exam for two different campuses of college ,
we have to test the hypothesis that the mean score on campus 1 is higher than on campus 2 .
sample size (n₁) is = 330
the sample size (n₂) is = 310
the sample mean (μ₁) = 33
the sample mean (μ₂) = 31
the standard deviation , σ₁ = 8
the standard deviation , σ₂ = 7
the test statistic test z can be calculated using the formula ,
z = (μ₁ - μ₂)/√(σ₁²/n₁ + σ₂²/n₂)
Substituting the values of the sample size , sample mean and standard deviation ,
we get ,
z = (33 - 31)/√(8²/330 + 7²/310)
Simplifying further ,
we get ,
Z = 3.37
the test is for the null hypothesis at the 5% level of significance .
that means ,
Z₀.₉₅ = 1.645 ;
So , the p value = P(Z > 3.37)
= 1 - P(Z < 3.37)
= 1 - 0.9996
= 0.0004
since the p-value is (0.0004) < the level of significance (0.05) .
we can reject the null hypothesis .
There is significant evidence to conclude that the mean score on campus 1 is higher than campus 2 .
Therefore , the correct option is (c) .
The given question is incomplete the complete question is
The results of a mathematics placement exam at two different campuses of mercy college follow: ( the table is given below)
The college wants to test the hypothesis that the mean score on campus 1 is higher than on campus 2. if we test the null hypothesis at the 5% level of significance, what is the decision ?
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