Mathematics, 04.06.2021 01:00 trint4

Duncan is investigating if residents of a city support the construction of a new school. He is curious about the difference of opinion between residents in north and south parts of the city. He obtained separate random samples of voter from each region. Here are results: Support Construction North South
Yes 274 240
No 726 520
Total 1000 760
Duncan wants to use these results to construct a 95% confidence of interval to estimate the difference in the proportion of the residents in these regions who support the construction (PN-Ps). Assume that all of the condition for inference have been met.

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Answer from: rama64

The 95% confidence of interval to estimate the difference in the proportion of the residents in these regions who support the construction is (-0.0849, 0.0013).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

North:

274 out of 1000. So

$p_N = \frac{274}{1000} = 0.274, s_N = \sqrt{\frac{0.274*0.726}{1000}} = 0.0141$

South

240 out of 760. So

$p_S = \frac{240}{760} = 0.3158, s_S = \sqrt{\frac{0.3158*0.6842}{760}} = 0.0169$

Distribution of the difference:

p = p_N - p_S = 0.274 - 0.3158 = -0.0418

$s = \sqrt{s_N^2+s_S^2} = \sqrt{0.0141^2+0.0169^2} = 0.022$

Confidence interval:

The confidence interval is:

$p \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so Z = 1.96 .

Lower bound:

p - zs = -0.0418 - 1.96*0.022 = -0.0849

Upper bound:

p + zs = -0.0418 + 1.96*0.022 = 0.0013

The 95% confidence of interval to estimate the difference in the proportion of the residents in these regions who support the construction is (-0.0849, 0.0013).

Answer from: Quest

this is a very good question

step-by-step explanation:

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Answer from: Quest

This is a robot. try again lol. maybe you attached the wrong image.
Can some find the area and perimeter for this triangle !