i need help with 12, 13,14 please A manufacture guarantees that its new drug red
ID: 3176774 • Letter: I
Question
i need help with 12, 13,14 please
A manufacture guarantees that its new drug reduces systolic blood pressure (SBP in millimeters of mercury). The test results are shown in the table below. At significance level of 0.05, can you reject the manufacture's claim? Find critical F-values for right- and left-tailed test using the indicated level of significance alpha and degrees of freedom dfN and dflD: alpha = 0.05, dfN = 9, dfD = 16; alpha = 0.01, dfN = 5, dfD = 11; alpha = 0.025, dfN = 7, dfD = 5. A manager is designing a system that is intended to decrease the standard deviation of the time customers waiting for service. Under old system, a random sample of 10 customers had the standard deviation of 20min. Under new system, a random sample of 21 customers had the standard deviation of 16min. At alpha = 0.10, is there enough evidence to convince the manager to switch to the new system? Assume both populations are normally distributed. A study of 150 randomly selected occupants in passenger cars and 200 randomly selected occupants in pickup trucks shows that 86% of occupants in passenger cars and 74% of occupants in pickup trucks wear seat belts. At alpha 0.10, can you reject the claim that the proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks?Explanation / Answer
Solution:-
14)
p1 = 0.86, n1 = 150
p2 = 0.74, n2 = 200
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 = 0.7914
S.E = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
S.E = 0.0439
z = (p1 - p2) / SE
z = 2.7335
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 - 2.73 or greater than 2.73.
Thus, the P-value = 0.006267
Interpret results. Since the P-value (0.006267) is less than the significance level (0.05), we cannot accept the null hypothesis.
From the above test we have sufficient evidence to reject the claim that proportion of occupants who wear seat belts is the same for passenger cars and pickup trucks.
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