You wish to test the following claim (\( H_a \)) at a significance level of \( \alpha = 0.005 \).\[ H_0: p = 0.78 \]\[ H_a: p > 0.78 \]You obtain a sample of size \( n = 621 \) in which there are 508 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 \( H_0 \) is true) the probability of observing...- \( \circ \) at most 508 successful observations- \( \circ \) at least 508 successful observationsWhat is the p-value for this sample? (Report answer accurate to four decimal places.)p-value = \[ \_\_\_\_\_\_\_\_\_ \]The p-value is...- \( \circ \) less than (or equal to) \( \alpha \)- \( \circ \) greater than \( \alpha \)This test statistic leads to a decision to...- \( \circ \) reject the null- \( \circ \) accept the null- \( \circ \) fail to reject the null

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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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A poll of 2,094 randomly selected adults showed that 95% 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 1 Sample p 95% CI Z-Value P-Value 1995 2,094 0.952722 (0.943632,0.961812) 5.52 0.000 ot toilod oE ciaht tailedO
A random sample of 20 binomial trials resulted in 9 successes. Test claim that the population proportion of successes does not equal 0.50. Use level of significance of 0.05. Can a normal distribution be used for the p distribution? State the hypotheses. Compute p and the corresponding standardized sample test statistic. Find the p value of the test. Do you reject or fail to reject Ho? What do the results tell?
A poll of 2,094 randomly selected adults showed that 95% 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 1 Sample p 95% CI Z-Value P-Value 1995 2,094 0.952722 (0.943632,0.961812) 5.52 0.000 ot toilod oE ciaht tailedO
J 1
A poll of 2,040 randomly selected adults showed that 89% 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 95% CI N 2,040 Sample p 0.893137 Z-Value P.Value 0.000 1 1822 (0.879731,0.906543) -4.47 Choose the correct answer below. O A. Reject the null hypothesis because the P-value is less than or equal to the significance level, a. OB. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. OC. Reject the null hypothesis because the P-value is greater than the significance level, a. OD. Fail to reject the null hypothesis because the P-value is greater than the significance level, a e. What is the final conclusion? OA. There is sufficient evidence to warrant rejection of the claim that…
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
The prices of a sample of books at University A were obtained by two statistics students. Then the cost of books for the same subjects (at the same level) were obtained for University B. Assume that the distribution of differences is Normal enough to proceed, and assume that the sampling was random. 1. Find the test statistic for this test. t = ___ (Round to two decimal places as needed.) 2. Find the p-value for this test. p-value = ____ (Round to three decimal places as needed.) 3. What is the conclusion for this test?
A poll of 1004 teens aged 13 to 17 showed that 57% of them have made new friends online. Use a 0.01 significance level to test the claim that half of all teens have made new friends online. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. C Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online. Identify the null and alternative hypotheses. Ho: P H₁: p (Type integers or decimals. Do not round.) Identify the test statistic. 2=0 (Round to two decimal places as needed.) Identify the P-value. P-value= (Round to three decimal places as needed.) State the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. the null hypothesis. There ▼sufficient evidence to the claim that half of all teens have made new friends online.
A poll of 2,115 randomly selected adults showed that 87% of them own cell phones. The technology display below results from a test of the claim that 94% 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 = 0.94 vs p *0.94 Sample X N 1 1844 2,115 Sample p 0.871868 95% CI (0.853147,0.890588) a. Is the test two-tailed, left-tailed, or right-tailed? Z-Value - 13.19 P-Value 0.000
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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:…
You want to test whether or not the following sample of 30 observations follows a normal distribution. The mean the sample equals 11.83, and the standard deviation equals 4.53. of 2 3 5 5 7 8 8 9 9 10 11 11 12 12 12 12 13 13 13 14 15 15 15 16 16 17 17 18 18 19 (A) The number of intervals or categories used to test the hypothesis for this problem is 6 (B) The expected frequency in the 2nd interval is 10 (C) The calculated value for the test statistic is 5 (D) The p-value is greater than 0.1 (E) a=0.05, the conclusion of the test is that the data follows a normal distribution

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