Question

# Will eating oatmeal promote healthy levels of cholesterol? A consumer reports analyst took a sample of...

Will eating oatmeal promote healthy levels of cholesterol? A consumer reports analyst took a sample of 44 people with high cholesterol and asked them to eat oatmeal once a day for 3 months. Measurements were taken of their cholesterol levels before and after the 3 months in mg/dl. The analyst is testing whether the cholesterol levels after the diet are greater than the cholesterol levels before the diet. The hypotheses for this test are as follows: Null Hypothesis: μD ≤ 0, Alternative Hypothesis: μD > 0. If the analyst calculated the mean difference in cholesterol levels (after - before) to be -1.59 mg/dL with a standard deviation of 12.95 md/dL, what is the test statistic and p-value for the paired hypothesis t-test?

Question 11 options:

 1) Test Statistic: -0.814, P-Value: 1.58
 2) Test Statistic: -0.814, P-Value: 0.21
 3) Test Statistic: 0.814, P-Value: 0.79
 4) Test Statistic: -0.814, P-Value: 0.79
 5) Test Statistic: 0.814, P-Value: 0.21

Question 12 (1 point)

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: μD ≤ 0, Alternative Hypothesis: μD > 0. You calculate a p-value of 0.8586. What is the appropriate conclusion of your test?

Question 12 options:

 1) The average difference in weekly income is significantly larger than 0. The average weekly income was higher after the incentive plan.
 2) We did not find enough evidence to say there was a significantly negative average difference in weekly income. The incentive plan does not appear to have been effective.
 3) The average difference in weekly income is less than or equal to 0.
 4) We did not find enough evidence to say there was a significantly positive average difference in weekly income. The incentive plan does not appear to have been effective.
 5) We did not find enough evidence to say the average difference in weekly income was not 0. The incentive plan does not appear to have been effective.

Question 13 (1 point)

Suppose the national average dollar amount for an automobile insurance claim is \$721.2. You work for an agency in Michigan and you are interested in whether or not the state average is less than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 721.2, Alternative Hypothesis: μ < 721.2. Suppose the true state average is \$707.23 and the null hypothesis is rejected, did a type I, type II, or no error occur?

Question 13 options:

 1) We do not know the p-value, so we cannot determine if an error has occurred.
 2) Type I Error has occurred.
 3) Type II Error has occurred
 4) We do not know the degrees of freedom, so we cannot determine if an error has occurred.
 5) No error has occurred.

Question 14 (1 point)

Consumers Energy states that the average electric bill across the state is \$60.85. You want to test the claim that the average bill amount is actually less than \$60.85. The hypotheses for this situation are as follows: Null Hypothesis: μ ≥ 60.85, Alternative Hypothesis: μ < 60.85. If the true statewide average bill is \$94.02 and the null hypothesis is not rejected, did a type I, type II, or no error occur?

Question 14 options:

 1) No error has occurred.
 2) Type I Error has occurred.
 3) We do not know the degrees of freedom, so we cannot determine if an error has occurred.
 4) We do not know the p-value, so we cannot determine if an error has occurred.
 5) Type II Error has occurred

Question 11)

Test statistics t = (mean difference)/(s.d/√n)

t = (-1.59)/(12.95/√44) = -0.814

Degrees of freedom is = n-1 = 43

For 43 dof and -0.814 test statistics, P-value from t distribution is = 0.21

12)

Null hypothesis Ho : ud = 0

Alternate hypothesis Ha : ud > 0

P-value = 0.8586

As the p-value is extremely large we fail to reject the null hypothesis Ho

We did not find enough evidence to say there was a significantly positive average difference in weekly income. The incentive plan does not appear to have been effective.

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