QUESTION 9: You hear on the local news that for the city of Kalamazoo, the propo

Question

QUESTION 9: You hear on the local news that for the city of Kalamazoo, the proportion of people who support President Obama is 0.43. However, you think it is different from 0.43. The hypotheses for this test are Null Hypothesis: p = 0.43, Alternative Hypothesis: p ? 0.43. If you randomly sample 21 people and 10 of them support President Obama, what is your test statistic and p-value?

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QUESTION 10: Consumers Energy states that the average electric bill across the state is $52.96. You want to test the claim that the average bill amount is actually greater than $52.96. The hypotheses for this situation are as follows: Null Hypothesis: ? ? 52.96, Alternative Hypothesis: ? > 52.96. A random sample of 40 customer's bills shows an average cost of $54.555 with a standard deviation of $9.4125. What is the test statistic and p-value for this test?

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QUESTION 11: You are looking for a way to incentivize the sales reps that you are in charge of. You design an incentive plan as a way to help increase in their sales. To evaluate this innovative plan, you take a random sample of your reps, and their weekly incomes before and after the plan were recorded. You calculate the difference in income as (after incentive plan - before incentive plan). You perform a paired samples t-test with the following hypotheses: Null Hypothesis: ?D? 0, Alternative Hypothesis: ?D < 0. You calculate a p-value of 0.5545. What is the appropriate conclusion of your test?

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QUESTION 12: Consumers Energy states that the average electric bill across the state is $50.43. You want to test the claim that the average bill amount is actually different from $50.43. The hypotheses for this situation are as follows: Null Hypothesis: ? = 50.43, Alternative Hypothesis: ? ? 50.43. If the true statewide average bill is $91.48 and the null hypothesis is not rejected, did a type I, type II, or no error occur?

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Test Statistic: -0.428, P-Value: 0.669

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Test Statistic: 0.428, P-Value: 1.666

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Test Statistic: 0.428, P-Value: 0.334

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Test Statistic: 0.428, P-Value: 0.669

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Test Statistic: 0.428, P-Value: 0.666

Explanation / Answer

Solution

Q9. sample p = 10/21 = 0.476

p = 0.43

test staistic = (0.476 - 0.43)/(0.43*0.57/210.5) = 0.428

p value two tailed = 0.669

option 4

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