Question

Goofy's fast food center wishes to estimate the proportion of people in its city that will purchase its products. Suppose the true proportion is 0.050.05. If 254254 are sampled, what is the probability that the sample proportion will differ from the population proportion by greater than 0.030.03? Round your answer to four decimal places.

Answer 1

Answer:

The probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.0143.

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:

 $\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

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

The sample selected consists of n = 254 individuals. The sample is quite large, i.e. n = 254 > 30. So the central limit theorem can be applied to approximate the distribution of sample proportion of  people in the city that will purchase the products of Goofy's fast food.

The mean is:

$\mu_{\hat p}=p=0.05$

And the standard deviation is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.05(1-0.05)}{254}}=0.0137$

Now, we need to compute the probability that the sample proportion will differ from the population proportion by greater than 0.03.

That is:

$\hat p-p>0.03\\\hat p -0.05>0.03\\\hat p>0.08$

Compute the value of $P(\hat p>0.08)$ as follows:

$P(\hat p>0.08)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.08-0.05}{0.0137})$

                   $=P(Z>2.19)\\=1-P(Z$

*Use a z-tale for the probability.

Thus, the probability that the sample proportion will differ from the population proportion by greater than 0.03 is 0.0143.