Respuesta :
Answer:
The probability that the sample proportion will differ from the population proportion by less than 6% is 0.992.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p\\\\[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
The information provided is:
[tex]p=0.22\\n=276[/tex]
As the sample size is large, i.e. n = 276 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion.
Compute the value of [tex]P(\hat p-p<0.06)[/tex] as follows:
[tex]P(\hat p-p<0.06)=P(\frac{\hat p-p}{\sigma_{\hat p}}<\frac{0.06}{\sqrt{\frac{0.22(1-0.22)}{276}}})\\\\=P(Z<2.41)\\\\=0.99202\\\\\approx 0.992[/tex]
Thus, the probability that the sample proportion will differ from the population proportion by less than 6% is 0.992.
