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× O welcome tocxyDweeworK :KkXY One-sample: R Classes vebwork.certerville.k12.ohus/webwork2/Klein-ApStatistics/10.3-Large-Sample-Hypotheses_Testf." e A random sample of 130 observations is selected from a binomial population with unknown probability of success P. The computed value of pis 0.65. (1) Test H0 : p 0.65 against H, : p > 0.65 Use = 0.05 test statistic : critical: score 1.6449 The final conclustion is .A. There is not sufficient evidence to reject the null hypothesis that p 0.65 B. We can reject the null hypothesis that p 0.65 and accept that p > 0.65 (2) Test Ho : p 0.5 against HI : pExplanation / Answer
Solution:-
2)
phat =0.65
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P > 0.50
Alternative hypothesis: P < 0.50
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 ]
= 0.04385
z = (p - P) /
z = 3.42
zcritical = - 1.645
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 less than 3.42. Thus, the P-value = 0.9997
Interpret results. Since the P-value (0.9997) is less than the significance level (0.05), we have to accept the null hypothesis.
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