In: Statistics and Probability
Consumers Energy states that the average electric bill across the state is $41.553. You want to test the claim that the average bill amount is actually greater than $41.553. The hypotheses for this situation are as follows: Null Hypothesis: μ ≤ 41.553, Alternative Hypothesis: μ > 41.553. A random sample of 47 customer's bills shows an average cost of $43.307 with a standard deviation of $8.0202. What is the test statistic and p-value for this test?
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It is reported in USA Today that the average flight cost nationwide is $396.81. You have never paid close to that amount and you want to perform a hypothesis test that the true average is actually less than $396.81. The hypotheses for this situation are as follows: Null Hypothesis: μ ≥ 396.81, Alternative Hypothesis: μ < 396.81. A random sample of 32 flights shows an average cost of $407.477 with a standard deviation of $59.2227. What is the test statistic and p-value for this test?
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Suppose the national average dollar amount for an automobile insurance claim is $973.04. You work for an agency in Michigan and you are interested in whether or not the state average is greater than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≤ 973.04, Alternative Hypothesis: μ > 973.04. You take a random sample of claims and calculate a p-value of 0.3431 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance.
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Suppose the national average dollar amount for an automobile insurance claim is $823.38. You work for an agency in Michigan and you are interested in whether or not the state average is greater than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≤ 823.38, Alternative Hypothesis: μ > 823.38. You take a random sample of claims and calculate a p-value of 0.0177 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance.
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(a) right choice is (2) Test Statistic: 1.499, P-Value: 0.0703
test statistic t==(43.307-41.553)/(8.0202/sqrt(47))=1.499
with n-1=47-1=46 df
p-value=P(t>1.499)=0.0703 ( using
ms-excel=tdist(1.499,46,1))
this is right tailed test so p-value is one tailed
(b) right choice is (4)Test Statistic: 1.019, P-Value: 0.8419
test statistic t==(407.477-396.81)/(59.2227/sqrt(32))=1.019
with n-1=32-1=31 df
p-value=P(t<1.019)=0.8419
(c) right choice is (5) We did not find enough evidence to say the true average claim amount is higher than $973.04.
since the p-value=0.3431 is more than alpha=0.05, so we fail to reject null hypothesis and conclude that national average dollar amount for an automobile insurance claim is μ ≤ 973.04 ( or not greater than $973.04 )
(d) right choice is (3)The true average claim amount is significantly higher than $823.38.
since the p-value=0.0177 is less than alpha=0.05, so we reject null hypothesis in favor of alternate hypothesis Ha: μ > 823.38 and conclude that true average claim amount is significantly higher than $823.38