Question

According to the Gallup survey, 23% of Americans reported eating less meat in the past year than they had previously. Results for this Gallup poll are based on telephone interviews conducted Sept. 16-30, 2019, with a random sample of 2,431 adults, aged 18 and older, living in all 50 U.S. states and the District of Columbia. Test that the proportion of Americans who reduced meat consumption last year is less than 0.25. Use α = 0.05. State the rejection region. Group of answer choices z > 1.65 z > 1.65 z < − 1.65 z < − 1.65 z > 1.96 z > 1.96

Answer 1

Answer:

Null hypothesis:$p\geq 0.25$  

Alternative hypothesis:$p < 0.25$  

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Now we need to find the critical value for the rejection zone of the null hypothesis. Since we have a left tailed test we need to find in the normal standard distirbution a value who accumulate 0.05 of the area in the left tail and we got:

$z_{crit}= -1.65$

And the best choice for this case would be:

z < − 1.65

Step-by-step explanation:

Information provided

n=2431 represent the random sample taken

$\hat p=$ estimated proportion of interest

$p_o=0.25$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

Hypothesis to test

We want to verify if the true proportion of Americans who reduced meat consumption last year is less than 0.25, then the system of hypothesis are :

Null hypothesis:$p\geq 0.25$  

Alternative hypothesis:$p < 0.25$  

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Now we need to find the critical value for the rejection zone of the null hypothesis. Since we have a left tailed test we need to find in the normal standard distirbution a value who accumulate 0.05 of the area in the left tail and we got:

$z_{crit}= -1.65$

And the best choice for this case would be:

z < − 1.65