Question

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?

Answer #1

Solution :

The null and alternative hypothesis is

H_{0} : p = 0.37

H_{a} : p
0.37

= x / n = 13 / 29 = 0.45

P_{0} = 0.37

1 - P_{0} = 1 - 0.37 = 0.63

Test statistic = z =

=
- P_{0} / [P_{0
*} (1 - P_{0} ) / 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

Question 11 (1 point)
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