Answer :
Answer:
The probability that the sample proportion will be with 0.04 of the population proportion is 0.8904.
Step-by-step explanation:
Let p = proportion of advertisers who claim they have been a victim of click fraud.
The population proportion is, p = 0.40.
A random sample of 380 advertisers is selected.
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 sample selected is too large, i.e. n = 380 > 30.
Then the proportion of advertisers who claim they have been a victim of click fraud can be approximated by the normal distribution.
The mean and standard deviation of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p=0.40\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{380}}=0.025[/tex]
Compute the probability that the sample proportion will be with 0.04 of the population proportion as follows:
[tex]P(p-0.04<\hat p<p+0.04)=P(\frac{-0.04}{0.025}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.04}{0.025})[/tex]
[tex]=P(-1.60<Z<1.60)\\=P(Z<1.60)-P(Z<-1.60)\\=0.94520-0.05480\\=0.8904[/tex]
Thus, the probability that the sample proportion will be with 0.04 of the population proportion is 0.8904.