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lyudmila [28]
3 years ago
9

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 samp

le 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%
Mathematics
1 answer:
dsp733 years ago
8 0

Answer:

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

(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{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%

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".  

p_1 represent the real population proportion for San Jose

\hat p_1 =\frac{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 =\frac{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{\frac{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{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\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{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401  

(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{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%

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Nadusha1986 [10]

Answer:

16 years old

Step-by-step explanation:

Let's use the letters s, b, and c for Sushil, Brian, and Caroline's ages. Sushil is the oldest, followed by Brian, and then Caroline, the youngest. From the problem description, we can set up three equations:

s = b + 6 <em>(Sushil is 6 years older than Brian)</em>

c = b - 5 <em>(Caroline is 5 years younger than Brian)</em>

s + b + c = 64 <em>(The total of their ages is 64)</em>

Since s and c are already in terms of b, we can substitute them into the last equation and solve to find Brian's age:

(b + 6) + b + (b - 5) = 64

3b + 1 = 64

3b = 63

b = 21

Now that we know Brian's age, we can simply subtract 5 to find Caroline's:

c = b - 5 = 21 - 5 = 16 years old.

8 0
2 years ago
Kaylee drove her scooter at an average speed of 15 miles per hour. Sophia drove her scooter at an average speed of 10 miles per
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Answer:

Kaylee drove her scooter for 2h and Sophia drove her scooter for 3h.

Step-by-step explanation:

From the graph shown below, we can see that the intersection point is (2,3). When using the graphing method to solve a system of equations, the intersection point is the answer. In this case, x = 2 and y = 3.

Thus, Kaylee drove her scooter for 2h and Sophia drove her scooter for 3h.

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Yo plz help me I don’t know this<br><br> Which number line shows the solution to -6 - (-4)?
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Answer:

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Step-by-step explanation:

To solve -6-(-4) you need to break it down.

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Then add 4 to -6 (-6+4)

This equation is being expressed through option B

Hope this helps! :)

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Georg invests $5000 for 14 years at a rate of 2% per year compound interest.
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Answer:

\frac{1}{50} or 0.02

Step-by-step explanation:

2% x 1 to give the answer

or

reduce it to faction like

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Maria has three children. there is two years age difference between each child. the total ages of all three children is 36 years
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T= Tina's age
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t+4= third child
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Add the ages of all children equal to 36 years. Add 2 to each age because there is 2 years age difference between each child.

t + (t+2) + (t+4)= 36
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