Question

In a previous​ year, 58​% of females aged 15 and older lived alone. A sociologist tests whether this percentage is different today by conducting a random sample of 600 females aged 15 and older and finds that 339 are living alone. Is there sufficient evidence at the alphaequals0.01 level of significance to conclude the proportion has​ changed?

Answer 1

Answer:

$z=\frac{0.565 -0.58}{\sqrt{\frac{0.58(1-0.58)}{600}}}=-0.744$  

Since is a bilateral test the p value would be given by:  

$p_v =2*P(z$  

And since the p value is higher than the significance level we have enough evidence to conclude that the true proportion is not significantly different from 0.58

Step-by-step explanation:

Information given

n=600 represent the random sample selcted

X=339 represent the number of females aged 15 and older that living alone

$\hat p=\frac{339}{600}=0.565$ estimated proportion of females aged 15 and older that living alone

$p_o=0.58$ is the value that we want to check

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_$ represent the p value

Sytem of hypothesis

We want to check if the true proportion females aged 15 and older that living alone is significantly different from 0.58.:  

Null hypothesis:$p=0.58$  

Alternative hypothesis:$p \neq 0.58$  

The statistic is given by:

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

Replacing the info given we got:

$z=\frac{0.565 -0.58}{\sqrt{\frac{0.58(1-0.58)}{600}}}=-0.744$  

Since is a bilateral test the p value would be given by:  

$p_v =2*P(z$  

And since the p value is higher than the significance level we have enough evidence to conclude that the true proportion is not significantly different from 0.58