Please solve the question

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

Media experts claim that daily print newspapers are declining because of Internet access. Listed below, from left to right and

yKpoI14uk [10]

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic $t_{\alpha/2}$ is 6.678155

(d) The p value is 1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean, $\bar{x}_{1}$ = 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean, $\bar{x}_{2}}$ = 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The formula for t test is given by;

$t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic $t_{\alpha/2}$ = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.

5 0

3 years ago

Tristan has some dimes and some quarters. He has at least 20 coins worth at most $4.25 combined. If Tristan has 10 dimes, determ

Dovator [93]

Answer:

The maximum number of quarters that he could have is 13

Step-by-step explanation:

Let x represent the number of quarters he could have. If he has 10 dimes and has at least 20 coins, then

→ x + 10 >= 20

→ x >= 10 (1)

His coins worth at most $4.25. Also, it is known that 1 dollar is equal to 100 cents, 1 dime is equal to 10 cents and 1 quarter is equal to 25 cents.

→ (x * 25) + (10 * 10) <= (4.25 * 100)

→ 25x + 100 <= 425

→ 25x <= 325

→ x <= 13 (2)

If we combine the equation 1 and 2:

10 <= x <= 13

The maximum value of x is 13

7 0

3 years ago

Plz answer the below question u will be awarded brainliest and will get points too..

brilliants [131]

Answer:

x=0.6241

Step-by-step explanation:

Square both sides and end up with

75.24+x=75.8641

isolate x

x=75.8641 - 75.24

x=0.6241

6 0

3 years ago

Match the width with the given area and length of a rectangle.

torisob [31]

Answer:

5 cm A = 44cm; w = 8.8 cm

7.5 cm A = 48.75 cm2; w = 6.5 cm

5 cm A = 31.25; w = 6.25

9.25 cm A = 55.5 cm; w = 6 cm

Step-by-step explanation:

What you have to do to find the answer is to take the length which for the first problem is 5 cm and then multiply the 5 by each width until you get the area. Example; Length x Width = Area

Basically you use the length times all of the widths until you match up the area.

Hope I could help

3 0

3 years ago