Question

Most married couples have two or three personality preferences in common. Myers-

Answer 1

Answer:

The 90% confidence interval for the proportion of married couples with three personality preferences in common compared  with the proportion of couples with two preferences in common is (-0.081, 0.025).

Step-by-step explanation:

The (1 - α)% confidence interval for difference in proportion formula is,

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

The given information is:

n₁ = 375,

n₂ = 571,

X₁ = 132,

X₂ = 217.

Compute the sample proportion as follows:

$\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{132}{375}=0.352\\\\\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{217}{571}=0.38$

For the 90% confidence level, the z-value is,

z₀.₀₅ = 1.645

*Use a z-table.

Compute the 90% confidence interval for the difference in proportion as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

     $=(0.352-0.38)\pm 1.645\sqrt{\frac{0.352(1-0.352)}{375}+\frac{0.38(1-0.38)}{571}}\\=-0.028\pm 0.05256\\=(-0.08056, 0.02456)\\\approx (-0.081, 0.025)$

The 90% confidence interval for the proportion of married couples with three personality preferences in common compared  with the proportion of couples with two preferences in common is (-0.081, 0.025).

This confidence interval implies that the true difference between the proportions lies in this interval with probability 0.90.