Consider the following null and alternative hypotheses:Ho: p = 0.85 versus H;: p # 0.85.Arandom sample of 575 observations taken from this population produced a sample proportion of 0.91.a. If this test is made at a 5% significance level, would you reject the null hypothesis? Use the critical-value approach.Round your answer for z to two decimal places.ZobservedZaritical leftZeritical rightWethe null hypothesis.b. What is the probability of making a Type l error in part a?P(Type l error) =c. Calculate the p-value for the test. Based on this p-value, would you reject the null hypothesis if a = 0.04? What if a = 0.1?Round your answer for the p-value to four decimal places.p-value =For a = 0.04, wev Họ.For a = 0.1, weHo.eTextbook an fail to rejectreject

a. Denote p as the true proportion. The null and alternative hypotheses are: H0: p = 0.85, vs. H1: p ≠ 0.85. The direction of the test is both sided. Hence, it is two-tailed test. Consider that, the test must is performed at 5% or 0.05 level of significance.If x denotes the number of cases favourable to an event in a sample of size n, then the sample proportion is, p̂ = x/n. Here, the sample size is, n = 575 and the sample proportion is p̂ = 0.91. The standardized sample test statistic is, z = (p̂ – p)/ √ [p (1 – p)/n], where p̂ is the point estimate of the population proportion, p is the population proportion and n is the sample size. The test statistic is: z = (p̂ – p)/ √ [p (1 – p)/n] [Where p is the proportion value suggested in H0] = (0.91 – 0.85) / √ [(0.85) (1 – 0.85)/575] = 0.06/0.014890907 ≈ 4.03. The test statistic is , zobserved = 4.03. The confidence level is 0.95. If 95% of the observations must lie within an interval, the remaining 5% must lie outside the interval. Further, due to symmetry, 2.25% will lie above the upper limit and the remaining 2.25% will lie below the lower limit of the interval. Thus, the upper limit of the interval is such that, 97.5% lie below it. As a result, zα/2 = z0.025 ≈ 1.96 [Using Excel formula: =NORM.S.INV(0.975)]. As it is both tailed test, the values of critical z tabular value are ±1.96. Thus, zcritical left = –1.96 and , zcritical right = 1.96. Here, the test statistic of 4.03 is greater than zcritical right of 1.96, reject the null hypothesis.b. The probability of type I error is the level of significance. That is, P(Type I error) = α = 0.05.c. Using the Excel formula =2*(1-NORM.S.DIST(4.03,TRUE)), the P-value is 0. Decision rule: If P-value is less than or equal to the level of significance, then reject the null hypothesis . Otherwise, fail to reject the null hypothesis. Conclusion: Here, the P-value is 0, which is less than the level of significance of 0.04. Hence, reject the null hypothesis H0. Here, the P-value is 0, which is less than the level of significance of 0.1. Hence, reject the null hypothesis H0.

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Consider the following null and alternative hypotheses: Ho:p = 0.85 versus H;: p + 0.85. Arandom sample of 575 observations taken from this population produced a sample proportion of 0.91. a. If this test is made at a 5% significance level, would you reject the null hypothesis? Use the critical-value approach. Round your answer for z to two decimal places. Zebserved Zaritical left = i Zaritical right" i We the null hypothesis. b. What is the probability of making a Type lerror in part a? P(Type l error) = c. Calculate the p-value for the test. Based on this p-value, would you reject the null hypothesis if a = 0.04? What if a = 0.1? Round your answer for the p-value to four decimal places. p-value - i For a = 0.04, we Но- For a = 0.1, we v Họ.

Consider the following null and alternative hypotheses: Ho:p = 0.85 versus H;: p + 0.85. Arandom sample of 575 observations taken from this population produced a sample proportion of 0.91. a. If this test is made at a 5% significance level, would you reject the null hypothesis? Use the critical-value approach. Round your answer for z to two decimal places. Zebserved Zaritical left = i Zaritical right" i We the null hypothesis. b. What is the probability of making a Type lerror in part a? P(Type l error) = c. Calculate the p-value for the test. Based on this p-value, would you reject the null hypothesis if a = 0.04? What if a = 0.1? Round your answer for the p-value to four decimal places. p-value - i For a = 0.04, we Но- For a = 0.1, we v Họ.

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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