Question

According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insucient 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% con dence interval for the di erence between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.

Answer 1

`(0.08-0.0888) - 1.96 \sqrt{(0.08(1-0.08))/(11545) +(0.088(1-0.088))/(4691)}= -0.0175`

`(0.08-0.0888) + 1.96 \sqrt{(0.08(1-0.08))/(11545) +(0.088(1-0.088))/(4691)}= 0.0015`

And the 95% confidence interval would be given (-0.0175;0.0015).

We are confident at 95% that the difference between the two proportions is between
`-0.0175 \leq p_A -p_B \leq 0.0015`

And since the confidence interval contains the 0 we have enough evidence to conclude that the population proportions are not significantly different at 5% of significance.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`p_A` represent the real population proportion for California

`\hat p_A =0.08` represent the estimated proportion for California

`n_A=11545` is the sample size required for California

`p_B` represent the real population proportion for Oregon

`\hat p_B =0.088` represent the estimated proportion for Brand B

`n_B=4691` is the sample size required for Oregon

`z` represent the critical value for the margin of error

Solution to the problem

The sample proportion have the following distribution

`\hat p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`(0.08-0.0888) - 1.96 \sqrt{(0.08(1-0.08))/(11545) +(0.088(1-0.088))/(4691)}= -0.0175`

`(0.08-0.0888) + 1.96 \sqrt{(0.08(1-0.08))/(11545) +(0.088(1-0.088))/(4691)}= 0.0015`

And the 95% confidence interval would be given (-0.0175;0.0015).

We are confident at 95% that the difference between the two proportions is between
`-0.0175 \leq p_A -p_B \leq 0.0015`

And since the confidence interval contains the 0 we have enough evidence to conclude that the population proportions are not significantly different at 5% of significance.