You wish to test the following claim (Ha) at a significance level of a = 0.01.Ho:p = 0.76Ha:p> 0.76You obtain a sample of size n = 470 in which there are 380 successful observations. For this test, youshould use the (cumulative) binomial distribution to obtain an exact p-value. (Do not use the normaldistribution as an approximation for the binomial distribution.)The p-value for this test is (assuming Ho is true) the probability of observing...O at most 380 successful observationsat least 380 successful observationsWhat is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =

Answered: Image /qna-images/answer/9652993a-b961-4e1d-a64f-a96806708c03.jpg

Answered: You wish to test the following claim…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The test statistic of z= 1.65 is obtained when testing the claim that p # 0.857. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the P-value. c. Using a significance level of α = 0.01, should we reject Ho or should we fail to reject Ho? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. This is a test. b. P-value = (Round to three decimal places as needed.) c. Choose the correct conclusion below. A. Fail to reject Ho. There is sufficient evidence to support the claim that p *0.857. B. Reject Ho. There is sufficient evidence to support the claim that p # 0.857. C. Fail to reject Ho. There is not sufficient evidence to support the claim that p‡ 0.857. D. Reject Ho. There is not sufficient evidence to support the claim that p # 0.857.
A poll of 2,179 randomly selected adults showed that 88% of them own cell phones. The technology display below results from a test of the claim that 90% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e). Test of p=0.9 vs p≠0.9 Sample X N Sample p 95% CI Z-Value P-Value 1 1908 2,179 0.875631 (0.861775,0.889487) −3.79 0.000 a. Is the test two-tailed, left-tailed, or right-tailed? Right tailed test Left-tailed test Two-tailed test Your answer is correct. b. What is the test statistic? The test statistic is negative 3.79−3.79. (Round to two decimal places as needed.) c. What is the P-value? The P-value is 0.0020.002. (Round to three decimal places as needed.) d. What is the null hypothesis and what do you conclude about it? Identify the null…
This assignment is worth 1 points. The extra point will be added to your overall course grade. For example: if you receive an 88% in the course you can receive up to 1 point giving you a new score of 89%. The following rubric will be used. 0.25 point for drawing the normal distribution curve with the mean value labeled on the curve and the appropriate area shaded. 0.25 point for determining the value of the standard deviation of the sample mean. 0.5 point for finding the correct probability. All work must be shown in order to receive any credit. Please upload your completed assignment here. The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.
You wish to test the following claim (Ha) at a significance level of a = 0.002. Ho: p= 0.6 Ha:p> 0.6 You obtain a sample of size n = 176 in which there are 117 successful observations. For this test, you should use the (cumulative) binomial distribution to obtain an exact p-value. (Do not use the normal distribution as an approximation for the binomial distribution.) The p-value for this test is (assuming Ho is true) the probability of observing... at most 117 successful observations at least 117 successful observations What is the p-value for this sample? (Report answer accurate to four decimal places.) P-value= The p-value is... less than (or equal to) a Ogreater than a This test statistic leads to a decision to... O reject the null O accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence warrant rejection of the claim that the population proportion is greater than 0.6. There is not sufficient evidence to warrant rejection of the…
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A poll of 2,047 randomly selected adults showed that 89% of them own cell phones. The technology display below results from a test of the claim that 90% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.9 vs p#0.9 Sample X N Sample p 95% CI Z-Value P-Value 1 1822 2,047 0.890083 (0.872275,0.907891) - 1.50 0.135 a. Is the test two-tailed, left-tailed, or right-tailed? Right tailed test Two-tailed test Left-tailed test
You wish to test the following at a significance level of = 0.01. Ho:p = 0.1 H1:p > 0.1 You obtain a sample of sizen = 158 in which there are 18 successful observations. For this test, we use the normal distribution as an approximation for the binomial distribution. For this sample... a. The test statistic (z) for the data = (Please show your answer to three decimal places.) b. The p-value for the sample (Please show your answer to four decimal places.) c. The p-value is... greater than a O less than (or equal to) a d. (Recall that when p(-value) is low, the null must go; when p(-value) is high, the null must fly) Base on this, we should .... fail to reject the null hypothesis accept the null hypothesis O reject the null hypothesis the final conclusion is that... e. As suc The sample data suggest that the population proportion is significantly greater than 0.1 at the significant level of 0.01. a = O The sample data suggest that the population proportion is not significantly greater…
A poll of 2,030 randomly selected adults showed that 91% of them own cell phones. The technology display below results from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e). Test of p=0.92 vs p≠0.92 Sample X N Sample p 95% CI Z-Value P-Value 1 1855 2,030 0.913793 (0.901584,0.926003) −1.03 0.303 a. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Two-tailed test Right tailed test b. What is the test statistic? The test statistic is nothing. (Round to two decimal places as needed.) c. What is the P-value? The P-value is . (Round to three decimal places as needed.) d. What is the null hypothesis and what do you conclude about it? Identify the null hypothesis. A. H0: p>0.92 B. H0: p<0.92 C. H0: p=0.92 D. H0:…

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