Question

For a recent report on Sleep deprivation, CA vs. OR. For a recent report on sleep deprivation, the Centers for Disease Control and Prevention interviewed 11508 residents of California and 4967 residents of Oregon. In California, 921 respondents reported getting insufficient rest or sleep during each of the preceding 30 days, while 452 of the respondents from Oregon reported the same. Round each calculation to 4 decimal places.

Answer 1

Answer:

The 95% confidence interval for difference between population proportion is (-0.0194, -0.0006).

Step-by-step explanation:

The confidence interval for difference in population proportion formula is,

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

The given information is:

n₁ = 11508

X₁ = 921

n₂ = 4967

X₂ = 452.

Compute the sample proportion values as follows:

$\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{921}{11508}=0.08\\\\\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{452}{4967}=0.09$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the standard normal table.

Compute the 95% confidence interval for difference between proportions as follows:

 $CI=(\hat p_{2}-\hat p_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}\\=(0.08-0.09)\pm1.96\times \sqrt{\frac{0.08(1-0.08)}{11508}+\frac{0.09(1-0.09)}{4967}}\\=-0.01\pm0.0094\\=(-0.0194, -0.0006)$

Thus, the 95% confidence interval for difference between population proportion is (-0.0194, -0.0006).