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VashaNatasha [74]
2 years ago
7

Answer these questions please please

Mathematics
1 answer:
Natalka [10]2 years ago
4 0

Answer:

Step-by-step

i) a² - 2ab + b² = (a + b)²

a² = p² ;     a  = p

b² = 16 = 4²  ;        b = 4

2ab = 2*p*4 = 8p

p² - 8p + 16 = (p - 4)²

ii) a² + 2ab + b² = (a +b)²

a² = 121x² = (11x)²                 ;  b² = 4y² = (2y)²

2ab = 2 * 11x * 2y = 44xy

121x² + 44xy + 4y² = (11x + 2y)²

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Is education related to programming preference when watching TV? From a poll of 80 television viewers, the following data have b
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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.

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