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To investigate whether there is a difference in opinion on a certain proposal between two voting districts, A and B, two independent random samples were taken. From district A, 35 of the 50 voters selected were in favor of the proposal, and from district B, 36 of the 60 voters selected were in favor of the proposal. Which of the following is the test statistic for the appropriate test to investigate whether there is a difference in the proportion of voters who are in favor of the proposal between the two districts (district A minus district B)? A. 35-36 A 50 B. 35-36 0.7 0.6 V 50 0 + 60 C. 0.7-0.6 (0..) (.x) (+) 0.7-0.6 D. V(0.7)(0.6) (Hot 50+60 0.7 -0.6 E. (0.7) (0.6) VE 60

Respuesta :

akiran007

Answer:

C:      t= x1`-x2`/√p^q^(1/n1 +1/n2)

0.7−0.6/(0.65)(0.35)√(1/50+1/60)

Step-by-step explanation:

The options given are

A

35−36/ 35/√50+3660

B

35−36/0.7/√50+0.660

C

0.7−0.6/(0.65)(0.35)√(1/50+1/60)

D

0.7−0.6/(0.7)(0.6)/√(1/50+1/60)

E

0.7−0.6/(0.7)(0.6)/√150+160

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This is a test hypothesis about difference of  means between two proportions.

The test statistic used is

t= x1`-x2`/√p^q^(1/n1 +1/n2)

x1`= 35/50= 0.7

x2`= 36/60= 0.6

so

x1`-x2`

= 35/50-36/60

= 0.7-0.6

p^= 35+36/50+60= 71/110= 0.645= 0.65

q^= 1-p^= 1-0.65= 0.35

p^q^= (0.65) (0.35)

n1= 50 and n2= 60

therefore √1/n1 +1/n2= √1/50+1/60

Now Putting the values in the test statistics we get

t= x1`-x2`/√p^q^(1/n1 +1/n2)

0.7−0.6/√(0.65)(0.35)(1/50+1/60)

Hence C is the correct option

Using the z-distribution, it is found that the test statistic for the appropriate test to investigate whether there is a difference in the proportion of voters who are in favor of the proposal between the two districts is given by:

[tex]z = \frac{0.1 - 0}{0.0906}[/tex]

What are the hypothesis tested?

At the null hypothesis, it is tested if there is no difference between the proportions, that is, the subtraction is zero, hence:

[tex]H_0: p_A - p_B = 0[/tex]

At the alternative hypothesis, it is tested if there is a difference between the proportions, that is, the subtraction is not zero, hence:

[tex]H_1: p_A - p_B \neq 0[/tex]

What is the mean and the standard error for the distribution of the difference of proportions?

First, we find the mean and the standard error for each proportion:

[tex]p_A = \frac{35}{50} = 0.7[/tex]

[tex]s_A = \sqrt{\frac{0.7(0.3)}{50}} = 0.0648[/tex]

[tex]p_B = \frac{36}{60} = 0.6[/tex]

[tex]s_B = \sqrt{\frac{0.6(0.4)}{60}} = 0.0633[/tex]

Then, for the difference:

[tex]\overline{p} = p_A - p_B = 0.7 - 0.6 = 0.1[/tex]

[tex]s = \sqrt{s_A^2 + s_B^2} = \sqrt{0.0648^2 + 0.0633^2} = 0.0906[/tex]

What is the test statistic?

It is given by:

[tex]z = \frac{\overline{p} - \mu}{s}[/tex]

In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.

Hence:

[tex]z = \frac{0.1 - 0}{0.0906}[/tex]

To learn more about the z-distribution, you can take a look at https://brainly.com/question/16313918

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