The equation of a circle in general form is x^2+y^2+22x+14y-55=0

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

4 years ago

Justin's car used 12 gallons to travel 252 miles. At what rate does the car use gas, in

s2008m [1.1K]

Answer:

All you have to do is divide the miles by the gallons... 252 miles ÷ 12 gallons = 21 miles/gallons

Your final answer is 21 miles/gallon!!

7 0

3 years ago

Slope = 1/5 y-intercept =-5

Novay_Z [31]

Answer:

what do you need help with??

6 0

3 years ago

The table shows the steps for solving the given inequality 4-3(x-5)>6x-17

mina [271]

Step-by-step explanation:

(=) 1 ( x-5 ) > 6x-17

(=) x-5 > 6x -17

(=) -5x > -12

=> x < 12/5

7 0

3 years ago

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a

Vikki [24]

Answer:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

Step-by-step explanation:

Information given:

$n=131$ represent the sample size

$\hat p=0.81$ represent the estimated proportion

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.326$

And replacing into the confidence interval formula we got:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

6 0

4 years ago