As of 2012, the proportion of students who use a MacBook as their primary computer is 0.37. You believe that at your university the proportion is actually different from 0.37. The hypotheses for this test are Null Hypothesis: p = 0.37, Alternative Hypothesis: p ≠ 0.37. If you randomly select 29 students in a sample and 13 of them use a MacBook as their primary computer, what is your test statistic and p-value?
Solution :
The null and alternative hypothesis is
H0 : p = 0.37
Ha : p 0.37
= x / n = 13 / 29 = 0.45
P0 = 0.37
1 - P0 = 1 - 0.37 = 0.63
Test statistic = z =
= - P0 / [P0 * (1 - P0 ) / n]
= 0.45 - 0.37 / [(0.37 * 0.63) / 29]
Test statistic = z = 0.89
P(z > 0.89) = 1 - P(z < 0.89) = 1 - 0.8133 = 0.1867
This is the two tailed test .
P-value = 2 * P(z > 0.89)
P-value = 2 * 0.1867
P-value = 0.3734
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