Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.

In this case we must use the formula to calculate the confidence interval for the difference between two proportions given by inferential statistics. The formula to calculate both limits of the interval is as it follows:

[tex](p_{1}-p_{2} )+\sqrt[]{\frac{p_{1}(1-p_{1})}{n_{1} }+\frac{p_{2}(1-p_{2})}{n_{2}} } \\(p_{1}-p_{2} )-\sqrt[]{\frac{p_{1}(1-p_{1})}{n_{1} }+\frac{p_{2}(1-p_{2})}{n_{2}} }[/tex]

Where:

[tex]p_{1}[/tex]: proportion of population one (in our case: Oregon residents who reported insufficient rest or sleep)

[tex]p_{2}[/tex]: proportion of population two (in our case: California residents who reported insufficient rest or sleep)

[tex]z_{(∝/2)}[/tex]: quantile of the normal distribution with α/2 probability (in our case, from the standard normal table we have 1.96 for a confidence level of 95%)

[tex]n_{1}[/tex]: sample size for population one (in our case, sample of Oregon residents)

[tex]n_{2}[/tex]: sample size for population two (in our case, sample of California residents)

Now, with our data we have:

[tex]p_{1}[/tex]=0.088

[tex]p_{2}[/tex]=0.08

[tex]z_{(∝/2)}[/tex]=1.96

[tex]n_{1}[/tex]=4,691

[tex]n_{2}[/tex]=11,545

Therefore, we obtain:

[tex](0.088-0.08)+1.96\sqrt{\frac{0.088(1-0.088)}{4,691}+\frac{0.08(1-0.08)}{11,545} } \\(0.088-0.08)-1.96\sqrt{\frac{0.088(1-0.088)}{4,691}+\frac{0.08(1-0.08)}{11,545} }[/tex]

Finally, the result for our interval:

0.008+0.0095=(-0.0015,0.0175)

According to the result we can say with a 95% confidence, that the proportion of Oregon residents with sleep deprivation issues is the same as the proportion of California residents. The reason for this is that 0 (zero) is contained within the interval.

joaobezerra joaobezerra · Answer 2 · 2021-11-03T15:06:24+01:00

Applying the Central Limit Theorem, it is found that the 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is (-0.0174, 0.0014). This means that we are 95% that the true difference for the two populations is in this interval.

The Central Limit Theorem states that when two proportions are subtracted.

The mean is the subtraction of the proportions.
The standard error is the square root of the sum of the variances.

For California:

[tex]p_C = 0.08[/tex]

[tex]v_C = \frac{0.08(0.92)}{11545}[/tex]

For Oregon:

[tex]p_O = 0.088[/tex]

[tex]v_O = \frac{0.088(0.912)}{4691}[/tex]

Thus, the distribution of the difference has:

[tex]p = p_C - p_O = 0.08 - 0.088 = -0.008[/tex]

[tex]s = \sqrt{v_C + v_O} = \sqrt{\frac{0.08(0.92)}{11545} + \frac{0.088(0.912)}{4691}} = 0.0048[/tex]

The confidence interval is:

[tex]p \pm zs[/tex]

We have to find the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.

In this problem, [tex]\alpha = 0.95[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], which means that it is z = 1.96.

Then

[tex]p - zs = -0.008 - 1.96(0.0048) = -0.0174[/tex]

[tex]p + zs = -0.008 + 1.96(0.0048) = 0.0014[/tex]

The 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is (-0.0174, 0.0014). This means that we are 95% that the true difference for the two populations is in this interval.

A similar problem is given at https://brainly.com/question/23089456