Question

In a recent Super Bowl, a TV network predicted that 33 % of the
audience would express an interest in seeing one of its forthcoming
television shows. The network ran commercials for these shows
during the Super Bowl. The day after the Super Bowl, and
Advertising Group sampled 150 people who saw the commercials and
found that 42 of them said they would watch one of the television
shows.

Suppose you are have the following null and alternative hypotheses
for a test you are running:

H0:p=0.33H0:p=0.33

Ha:p≠0.33Ha:p≠0.33

Calculate the test statistic, rounded to 3 decimal places

z=___

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p =0.33

H_{a} : p
0.33

n = 150

x =42

= x / n = 42 / 150 = 0.28

P_{0} = 0.33

1 - P_{0} = 1 - 0.33 = 0.67

Test statistic = z

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

= 0.28 - 0.33 / [0.33 * 0.67 ) / 150 ]

= −1.302

Test statistic = z = −1.302

P-value = 0.1928

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