Question

A researcher believes that 9% of females smoke cigarettes. If the researcher is correct, what is the probability that the proportion of smokers in a sample of 703 females would differ from the population proportion by more than 3%? Round your answer to four decimal places.

Answer 1

Answer:

The probability that the sample proportion would differ from the population proportion by more than 3% is 0.0027.

Step-by-step explanation:

Let X = number of females who smoke cigarettes.

The probability that a female smoke cigarettes is, p = 0.09.

A random sample of size, n = 703 females were selected to determine the sample proportion of females who smoke cigarettes.

A female smoking cigarettes is independent from others.

The random variable X follows a Binomial distribution with parameters n = 703 and p = 0.09.

Now, as the sample size is quite large, i.e. n = 703 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be described by the Normal distribution.

The mean of this sampling distribution is same as the population proportion, i.e. $\mu_{\hat p}=p=0.09$.

And the standard deviation of the sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.09(1-0.09)}{703}}=0.0108$

Compute the probability that the sample proportion would differ from the population proportion by more than 3% as follows:

$\hat p-p>0.03\\\hat p>0.03+p\\\hat p>0.03+0.09\\\hat p>0.12$

Compute the value of $P(\hat p>0.12)$ as follows:

$P(\hat p>0.12)=P(\frac{\hat p-p}{\sigma_{\hat p}}>\frac{0.03}{0.0108})\\=P(Z>2.78)\\=1-P(Z$

Thus, the probability that the sample proportion would differ from the population proportion by more than 3% is 0.0027.