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Sagot :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
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
[tex]p_N = \frac{274}{1000} = 0.274, s_N = \sqrt{\frac{0.274*0.726}{1000}} = 0.0141[/tex]
South
240 out of 760. So
[tex]p_S = \frac{240}{760} = 0.3158, s_S = \sqrt{\frac{0.3158*0.6842}{760}} = 0.0169[/tex]
Distribution of the difference:
[tex]p = p_N - p_S = 0.274 - 0.3158 = -0.0418[/tex]
[tex]s = \sqrt{s_N^2+s_S^2} = \sqrt{0.0141^2+0.0169^2} = 0.022[/tex]
Confidence interval:
The confidence interval is:
[tex]p \pm zs[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
Lower bound:
[tex]p - zs = -0.0418 - 1.96*0.022 = -0.0849[/tex]
Upper bound:
[tex]p + zs = -0.0418 + 1.96*0.022 = 0.0013[/tex]
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).
Given the proportion and number of residents in the North and South of
0.274, 1000, and approximately 0.316, 760, we have;
- The 95% Confidence interval for the difference in proportion of those in support of the construction is; C.I. = (-0.085, 0.00109)
How can the confidence interval be found?
The given parameter presented in a tabular form are;
[tex]\begin{array}{|c|c|c|}Support \ Constructtion&North&South\\Yes&274&240\\No&726&520\\Total & 1000& 760\end{array}\right][/tex]
Which gives;
Proportion of the North resident that support, [tex]\mathbf{\hat p_N}[/tex] = 274 ÷ 1000 = 0.274
The number of people sampled in the north, n₁ = 1,000
Proportion of the South resident that support, [tex]\mathbf{\hat p_S}[/tex] = 240 ÷ 760 ≈ 0.316
The number of people sampled in the south, n₂ = 760
The confidence interval for the difference of two proportions are given
as follows;
[tex]\hat{p}_N-\hat{p}_S\pm \mathbf{ z^{*}\sqrt{\dfrac{\hat{p}_N \cdot \left (1-\hat{p}_N \right )}{n_{1}}+\dfrac{\hat{p}_S \cdot \left (1-\hat{p}_S \right )}{n_{2}}}}[/tex]
Where;
The z-score at 95% confidence level is 1.96
Which gives;
[tex]C.I. = \left(0.274-0.316 \right)\pm 1.96 \times \sqrt{\dfrac{0.274 \times \left (1-0.274 \right )}{1000}+\dfrac{0.316 \times \left (1-0.316 \right )}{760}}[/tex]
C.I. ≈ (-0.085, 0.00109)
The 95% confidence interval, C. I. of the difference in the proportion of
the residents in the regions who support the construction, [tex]\hat p_N[/tex] - [tex]\hat p_S[/tex] is
therefore;
- [tex]\underline{C.I. = (-0.085, \, 0.00109)}[/tex]
Learn more calculating the confidence interval here:
https://brainly.com/question/15308274
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