1.6 Problem 6 (8 points) V records indicate that of all vehicles undergoing emis
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Question
1.6 Problem 6 (8 points) V records indicate that of all vehicles undergoing emissions testing during the previous year, 80% passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 165 that passed on the initial test. a. Does this suggest that the true proportion for this county during the current year is different from the previous statewide proportion? Test the relevant hypotheses using (a),-0.10, (d) a 005, (as) a 0.01. b. Does this suggest that the true proportion for this county during the current year is greater than the previous statewide proportion? Test the relevant hypotheses using (b1) a 0.10, (b2) a0.05, (b3) a 0.01Explanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.80
Alternative hypothesis: P 0.80
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion 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, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.02828
z = (p - P) /
z = 0.88
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.88 or greater than 0.88.
Thus, the P-value = 0.3788
Interpret results. Since the P-value (0.3788) is greater than the significance level (0.10), we cannot reject the null hypothesis.
Interpret results. Since the P-value (0.3788) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Interpret results. Since the P-value (0.3788) is greater than the significance level (0.01), we cannot reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that there is significance in proportion from the previous year proportion.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.80
Alternative hypothesis: P > 0.80
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.02828
z = (p - P) /
z = 0.88
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.88.
Thus, the P-value = 0.1894.
Interpret results. Since the P-value (0.1894) is greater than the significance level (0.10), we cannot reject the null hypothesis.
Interpret results. Since the P-value (0.1894) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Interpret results. Since the P-value (0.1894) is greater than the significance level (0.01), we cannot reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that proportion of current year is greater than the previous year proportion.
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