Question

Suppose we want to test the null hypothesis H0 : p = 0.28 against the alternative hypothesis H1 : p ≠ 0.28. Suppose also that we observed 100 successes in a random sample of 400 subjects and the level of significance is 0.05. What is the p-value for this test?

a. 0.9563

b. 0.0901

c. 0.05

d. 0.1802

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p =0.28

H_{a} : p
0.28

n = 400

x = 100

= x / n = 100 / 400 = 0.25

P_{0} = 0.28

1 - P_{0} = 1 - 0.28 = 0.72

Test statistic = z

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

= 0.28 / [ 0.28 * / ]

= −1.336

Test statistic = z = −1.34

P(z < -1.34 ) = 0.0901 * 2 =

P-value = 0.1802

= 0.05

P-value ≥

0.1802 ≥ 0.05

Do not reject the null hypothesis .

There is insufficient evidence to suggest that

Option d ) is correct,

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