Question

According to a including 7730 adults, 4097 said they talk to their siblings monthly. Use these data to test the claim that more than 45% of adults talk to their siblings monthly, using alpha=0.01.

p-value=1.117 evidence support claim

p-value=0.530, evidence support claim

p-value=0.000, evidence not support claim

p-value=0.000, evidence support claim

p-value=0.470, evidence support claim

Answer 1

Solution:

Given:

Sample size = n = 7730

x = Number of adults said they talk to their siblings monthly = 4097

Claim: more than 45% of adults talk to their siblings monthly

Use following steps:

Step 1) State H0 and H1:

H0: p =0.45 Vs H1: p > 0.45

Step 2) Test statistic:

where

thus

Step 3) Find p-value:

For right tailed test , p-value is:

p-value = P(Z > z test statistic)

p-value = P(Z > 14.14)

p-value =1 - P(Z < 14.14)

Since z = 14.14 is greater than z = 3, area below z = 14.14 is approximately = 1

that is: P( Z< 14.14) = 1.000

thus

p-value =1 - P(Z < 14.14)

p-value =1 - 1.000

p-value = 0.000

We can use Excel command:

=1-NORM.S.DIST(z,cumulative)

=1-NORM.S.DIST(14.14,TRUE)

=0.000

Thus p-value = 0.000

Decision Rule:
Reject null hypothesis H0, if P-value < 0.01 level of significance, otherwise we fail to reject H0

Since p-value = 0.000 < 0.01 level of significance, we reject null hypothesis H0.

Thus at 0.01 level of significance, we have sufficient evidence to support the claim that  more than 45% of adults talk to their siblings monthly.

Thus correct answer is:

p-value=0.000, evidence support claim