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In: Statistics and Probability

Use the information below to answer questions 11-15. A random sample of 1000 voters registered in...

Use the information below to answer questions 11-15. A random sample of 1000 voters registered in the state of Montana showed that 490 voted in the last general election. A random sample of 800 voters in the state of Arizona showed that 368 voted in the most recent general election. Do these data indicate that the population percentage of voter turnout in Arizona is lower than that in Montana? 11. Choose the appropriate null and alternate hypotheses. a. Null: The percentage of people who voted in Arizona and Montana is the same Alternate: The percentage of people who voted in Arizona is more than in Montana b. Null: The percentage of people who voted in Arizona is more than in Montana Alternate: The percentage of people who voted in Arizona and Montana is the same c. Null: The percentage of people who voted in Arizona and Montana is the same Alternate: The percentage of people who voted in Arizona is less than in Montana d. Null: : The percentage of people who voted in Arizona is less than in Montana Alternate: The percentage of people who voted in Arizona and Montana is the same 12. What is the standard error for the difference (SEdiff)? a. 0.03% b. 0.71% c. 3.0% d. 2.37% 13. What is the test statistic? a. 1.27 b. 0.01 c. 4.23 d. 100 14. What is the p-value? a. 10.2% b. 0% c. 49.48% d. 0.65% 15. What is your conclusion? a. Reject the null hypothesis b. Accept the null hypothesis c. Do not reject the null hypothesis d. Do not accept the alternate hypothesis

Expert Solution

(11) we have to check whether the population percentage of voter turnout in Arizona is lower than that in Montana.

Null: The percentage of people who voted in Arizona and Montana is the same

Alternate: The percentage of people who voted in Arizona is less than in Montana

therefore, option C is correct

(12) we have x1 = 490, x2 =368, n1 = 100 and n2 = 800

Pooled proportion = (x1+x2)/(n1+n2) = (490+368)/(1000+800) = 0.477

using the formula

Standard error =

option D

(13) Using TI 84 calculator

press STAT then TESTS then 2-PropZTest

enter the data

we get

z statistic = 1.27 (option A)

(14) p value = 0.1027 or 10.27% (using TI 84 result)

option A

(15) p value is 0.1027, which is too large. So, we failed to reject the null hypothesis as p value > 0.10 level

c. Do not reject the null hypothesis

orchestra answered 1 year ago

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