Explain how the confidence interval provided in the output could be used to obta
ID: 3173444 • Letter: E
Question
Explain how the confidence interval provided in the output could be used to obtain the same conclusion as in Part b. For each of the following scenarios, determine whether the sample size is large enough to run a hypothesis test based on the normal distribution. H_0:p = .85, n = 60 H_0:p_1 = p_2, p_1 =6^79, p_2 = .83, n_1 = 65, n_2 = 62 A local car dealership is interested in determining how successful their salespeople are in turning selling a car. Specifically, they are interested in the average percentage of the price markup earned on the following table contains the percentages for a random sample of car sales by two of the dealer's.Explanation / Answer
6) Solution:-
a)
p = 0.85, n = 60
The sample includes at least 10 successes and 10 failures.
Success = n × p = 51
Failure = n × (1 - p) = 9
The sample size is not adequate to run a normal hypothesis.
b)
p1 = 0.79, n1 = 65
Success = n1 × p1 = 51.35
Failure = n1 × (1- p1) = 13.65
p2 = 0.83, n2 = 62
Success = n1 × p1 = 51.46
Failure = n1 × (1- p1) = 10.54
The sample sizes are adequate to run a normal hypothesis.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.8095
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.0697
z = (p1 - p2) / SE
z = - 0.57
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.57 or greater than 0.57.
Thus, the P-value = 0.5686
Interpret results. Since the P-value (0.5686) is greater than the significance level (0.05), we have to accept the null hypothesis.
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