Question

According to a reputable​ magazine,

34​%

of people in a certain large country have sleepwalked at least once in their lives. Suppose a random sample of

200

people showed that

44

reported sleepwalking. The null and alternative hypotheses for a test that would test whether the proportion of people who have sleepwalked is

0.34

are

H0:p=0.34

and

Ha​: p≠0.34

respectively. The conditions for such a test are satisfied. Use the technology output provided to determine the test statistic and​ p-value for the​ test, and provide a conclusion. Use a significance level of 0.05

Hypothesis test​ results:

p​ : Proportion of successes

Upper H 0H0​:

pequals=0.340.34

Upper H Subscript Upper AHA​:

pnot equals≠0.340.34

Proportion	Count	Total	Sample Prop.	Std. Err.	​Z-Stat	​P-value
p	44	200	0.22	0.03349627	3.5824886	0.0003

Determine the test statistic.

z=_____

​(Round to two decimal places as​ needed.)

Determine the​ p-value.

​p-value=____

​(Round to three decimal places as​ needed.)

What is the conclusion to this​ test?

▼

Upper H 0H0.

There

▼

is not

is

sufficient evidence to conclude that the proportion of people who have sleepwalked

▼

is less than

is not

is greater than

is

0.34

Answer 1

H0: p = 0.34

Ha: p 0.34

Sample proportion = 44 / 200 = 0.22

Test staistics

z = ( - p) / sqrt [ p ( 1- p) / n ]

= (0.22 - 0.34) / sqrt [ 0.34 ( 1 - 0.34) / 200 ]

= -3.58

For two tailed test,

p-value = 2 * P(Z < z)

= 2 * P(Z < -3.58)

= 2 * 0.0002

= 0.000

conclusion -

Reject H0 , There is sufficient evidence to conclude that the proportion of people who have sleepwalked

is not 0.34