Respuesta :
Answer:
[tex](0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401[/tex]
[tex](0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380[/tex]
We are confident at 95% that the difference between the two proportions is between [tex]-0.4401 \leq p_B -p_A \leq -0.1380[/tex]
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%
Step-by-step 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".
[tex]p_1[/tex] represent the real population proportion for San Jose
[tex]\hat p_1 =\frac{30}{73}=0.411[/tex] represent the estimated proportion for San Jos
[tex]n_1=73[/tex] is the sample size required for San Jose
[tex]p_2[/tex] represent the real population proportion for San Francisco
[tex]\hat p_2 =\frac{56}{80}=0.7[/tex] represent the estimated proportion for San Francisco
[tex]n_2=80[/tex] is the sample size required for San Francisco
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_1 -\hat p_1) \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}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex](0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401[/tex]
[tex](0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380[/tex]
We are confident at 95% that the difference between the two proportions is between [tex]-0.4401 \leq p_B -p_A \leq -0.1380[/tex]
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%
