Question

Question 1

In a sample of 75 female Facebook users in a certain group, 42 say that they are very active and make at least one post a day. A researcher wants to test if the proportion of Facebook posting by this group is different from the national percent of 45%. What would be the null hypothesis for this test?

Group of answer choices

H0:p=0.45

H0:p<0.45

Ha:p>0.45

Question 2

In test of hypothesis, we either accept or reject the null hypothesis.

Group of answer choices

True

False

Question 3

Let μ denote the average score on a standardized test for all children in a certain region of the United States. The average score for all children in the United States is 100. Regional education authorities are interested in testing H 0: μ = 100 versus H a: μ > 100 using a significance level of 0.01. A sample of 2500 children resulted in the values n = 2500, x ¯ = 101, s = 15.

What test would you use here?

Group of answer choices

t test

Z test

Question 4

In a sample of 75 female Facebook users in a certain group, 42 say that they are very active and make at least one post a day. A researcher wants to test if the proportion of Facebook posting by this group is different from the national percent of 45%. What is the test statistic value for this test?

Group of answer choices

1.042

1.915

2.561

Answer 1

Question 1)

Null hypothesis means no difference

So, null hypothesis Ho : P = 0.45

True

We either reject or accept the null hypothesis

Question 4)

N = 75

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 75*0.45 = 33.75

N*(1-p) = 41.25

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 42/75

Claimed P = 0.45

N = 75

After substitution

Test statistics z = 1.915