Chapter 7: Q. 16 (page 463)

According to the U.S. census, the proportion of adults in a certain county who owned their own home was $0.71$ . An SRS of $100$ adults in a certain section of the county found that $65$ owned their home. Which one
of the following represents the approximate probability of obtaining a sample of $100$ adults in which fewer than $65$ own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?
(a) $(\begin{matrix} 100 \\ 65 \end{matrix}) (0.71)^{65} (0.29)^{35}$
(b) $(\begin{matrix} 100 \\ 65 \end{matrix}) (0.29)^{65} (0.71)^{35}$
(c) $P (z < \frac{0.65 - 0.71}{\sqrt{\frac{(0.71) (0.29)}{100}}})$
(d) $P (z < \frac{0.65 - 0.71}{\sqrt{\frac{(0.65) (0.35)}{100}}})$
(e) $P (z < \frac{0.65 - 0.71}{\frac{(0.71) (0.29)}{\sqrt{100}})}$

Short Answer

Expert verified

The correct answer is option (c) $P (z < \frac{0.65 - 0.71}{\sqrt{\frac{(0.71) (0.29)}{100}}})$ .

Step by step solution

Given information

The proportion of adults who owned their own home was $0.71 .$ An SRS of $100$ adults of the county found that $65$ owned their home.

Explanation

Let, sample size $n = 100$
Number of successes $x = 65$
Population proportion $p = 0.71$
Requirements for a normal approximation of the binomial distribution: $n p \geq 10$ and $n q \geq 10$ .
$n p = 100 (0.71) = 71 \geq 10 n q = n (1 - p) = 100 (1 - 0.71) = 29 \geq 10$

The proportion is:
$\hat{p} = \frac{x}{n} = \frac{65}{100} = 0.65$
Then the mean is:
$μ_{\hat{p}} = p = 0.71$

Calculation

The standard deviation is:
$σ_{\hat{p}} = \sqrt{\frac{p (1 - p)}{n}} = \sqrt{\frac{0.71 (1 - 0.71)}{100}} = \sqrt{\frac{0.71 (0.29)}{100}}$

The z-score is:

$z = \frac{x - μ}{σ} = \frac{0.65 - 0.71}{\sqrt{\frac{0.71 (0.29)}{100}}}$

To determine the probability that the sample proportion is less than $0.65$ .

$P (\hat{p} < 0.65) = P (Z < \frac{0.65 - 0.71}{\sqrt{\frac{0.71 (0.29)}{100}}})$

Hence, option (c) $P (z < \frac{0.65 - 0.71}{\sqrt{\frac{(0.71) (0.29)}{100}}})$ is correct.

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