Question

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.1. -.4401 ≤ p1 - p2 ≤ -.13802. -.4401 ≤ p1 - p2 ≤ .13803. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Answer 1

`(0.411-0.7) - 1.96 \sqrt{(0.411(1-0.411))/(73) +(0.7(1-0.7))/(80)}=-0.4401`

`(0.411-0.7) + 1.96 \sqrt{(0.411(1-0.411))/(73) +(0.7(1-0.7))/(80)}=-0.1380`

We are confident at 95% that the difference between the two proportions is between
`-0.4401 \leq p_B -p_A \leq -0.1380`

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

Explanation:

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.

1. -.4401 ≤ p1 - p2 ≤ -.1380

2. -.4401 ≤ p1 - p2 ≤ .1380

3. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Solution to the problem

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_1` represent the real population proportion for San Jose

`\hat p_1 =(30)/(73)=0.411` represent the estimated proportion for San Jos

`n_1=73` is the sample size required for San Jose

`p_2` represent the real population proportion for San Francisco

`\hat p_2 =(56)/(80)=0.7` represent the estimated proportion for San Francisco

`n_2=80` is the sample size required for San Francisco

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

The population proportion have the following distribution

`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_1 -\hat p_1) \pm z_(\alpha/2) \sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2 (1-\hat p_2))/(n_2)}`

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:

`(0.411-0.7) - 1.96 \sqrt{(0.411(1-0.411))/(73) +(0.7(1-0.7))/(80)}=-0.4401`

`(0.411-0.7) + 1.96 \sqrt{(0.411(1-0.411))/(73) +(0.7(1-0.7))/(80)}=-0.1380`

We are confident at 95% that the difference between the two proportions is between
`-0.4401 \leq p_B -p_A \leq -0.1380`

Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.

Based on this the correct options are:

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%