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3 years ago

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Use the sample data and confidence level to construct the confidence interval estimate of the population proportion p. n equals

550 comma x equals 440 comma 95 % confidence nothingless thanpless than nothing (Round to three decimal places asneeded.)

1 answer:

Alekssandra [29.7K]3 years ago

3 0

Answer:

(0.767,0.833)

Step-by-step explanation:

The 95% confidence interval for population proportion p can be computed as

$p-z_{\frac{\alpha }{2} } \sqrt{\frac{pq}{n} }$

The z-value associated with 95% confidence level is 1.96.

whereas p=x/n

We are given that x=440 and n=550.

p=440/550=0.8

$0.8-1.96\sqrt{\frac{0.8(0.2)}{550} }$

$0.8-1.96\sqrt{\frac{0.16}{550} }$

$0.8-1.96\sqrt{0.00029 }$

$0.8-1.96(0.01706)$

$0.8-0.03343$

$0.76657$

Thus, the required confidence interval is

0.767<P<0.833 (rounded to 3 decimal places)

Hence, we are 95% confident that our true population proportion will lie in the interval (0.767,0.833)

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Answer:

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Step-by-step explanation:

To find the exact solution, find the equation for each line. And solve for x and y.

To do this, represent each equation in the slope-intercept form, y = mx + b. Where m is the slope, and b is the y-intercept.

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Substitute m = 2 and b = 7 in y = mx + b

Equation 1 would be:

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✍️Equation 2 for the line that slopes downwards from left to your right:

Slope = $m = \frac{rise}{run} = -\frac{3}{1} = -3$

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Substitute m = -3 and b = 1 in y = mx + b

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y = -3x + 1

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Collect like terms

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<h3>Further explanation </h3>

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<h3>Learn more</h3>

the multiplicative identity property

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Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

3 years ago