Question

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


Briggs used a random sample of 375 married couples and found that 132 had three


preferences in common. Another random sample of 571 couples showed that 217 had


two personality preferences in common. Find a 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. (Show your confidence


in interval notation and show your interpretation, which was larger?

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.