Respuesta :
Answer:
The p-value of the test is 0.1611>0.01, which means that there is not significant evidence at the 1% significance level that less than 34% of fathers take no responsibility for child care
Step-by-step explanation:
A journal article reports that 34% of fathers take no responsibility for child care. A group of fathers believes that this estimate is too high.
At the null hypothesis, we test if the proportion is of at least 34%, that is:
[tex]H_0: p \geq 0.34[/tex]
At the alternative hypothesis, we test if the proportion is less than that, that is:
[tex]H_1: p < 0.34[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.34 is tested at the null hypothesis:
This means that [tex]\mu = 0.34, \sigma = \sqrt{0.34*0.66}[/tex]
In a random sample of 80 fathers, it was found that 23 of those fathers take no responsibility for child care.
This means that [tex]n = 80, X = \frac{23}{80} = 0.2875[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.2875 - 0.34}{\frac{\sqrt{0.34*0.66}}{\sqrt{80}}}[/tex]
[tex]Z = -0.99[/tex]
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion below 0.2875, which is the p-value of Z = -0.99.
Looking at the Z-table, Z = -0.99 has a p-value of 0.1611.
The p-value of the test is 0.1611>0.01, which means that there is not significant evidence at the 1% significance level that less than 34% of fathers take no responsibility for child care
