Respuesta :
Answer:
There is enough statistical evidence to support the claim that the survey is not accurate.
Step-by-step explanation:
We have to perform a test of hypothesis on the proportion.
The claim is that the proportion of families that own stock differs from 48.8%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.488\\\\H_a:\pi\neq0.488[/tex]
The significance level is 0.05.
The sample. of size n=250, has a proportion of p=0.568.
[tex]p=X/n=142/250=0.568[/tex]
The standard error of the proportion is
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.488*0.512}{250}}=\sqrt{0.00099}=0.032[/tex]
The z-statistic can now be calculated:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.568-0.488-0.5/250}{0.032}=\dfrac{0.078}{0.032}=2.4375[/tex]
The P-value for this two-tailed test is then:
[tex]P-value=2*P(z>2.4375)=0.015[/tex]
As the P-value is smaller than the significance level, the effect is significant. The null hypothesis is rejected.
There is enough evidence to support the claim that the survey is not accurate.
The suffecient eveidence to insure that the percentage of families who won stock is different from 48.8%.
Null Hypothesis:
The hypothesis that there is no significant difference between specified populations, any observed difference being due to sampling or experimental error.
The null hypothesis can be stated as shown below:
[tex]H_{0} :P=48.8[/tex]%
That is, the percentage of families who own stock is [tex]48.8[/tex]%.
The alternative hypothesis can be stated as shown below:
[tex]H_{a} :p\neq 48.8[/tex]%
That is, the percentage of families who own stock is different from [tex]48.8[/tex]%
The [tex]z[/tex] test statistic is given below:
[tex]z=\frac{\hat{p-p}}{\sqrt{\frac{p\left ( 1-p \right )}{n}}}[/tex]
Here, [tex]\hat{p}[/tex] is the observed proportion then, [tex]\hat{p}[/tex] is calculated by,
[tex]\hat{p}=\frac{x}{n} \\=\frac{142}{250} \\=0.568[/tex]
Therefore,
[tex]z=\frac{\hat{p}-p}{\sqrt{\frac{p\left ( 1-p \right )}{n}}} \\=\frac{0.08}{0.0316}\\ =2.532[/tex]
Calculating the p-value be excel as attached below then we get,
The p-value is 0.01.
This p-value is called the actual level of significance.
The value of alpha is given as [tex]0.05[/tex]
Decision: The rejection of the null hypothesis [tex]H_0[/tex] because [tex]p-value < \alpha[/tex]
So, reject the null hypothesis.
Conclusion: The p-value is less than considered level of significance 5%.
Therefore the null hypothesis get rejected while the alternative hypotheisis is accepted.
Hence, there is suffecient eveidence to insure that the percentage of families who won stock is different from 48.8%.
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