A poll of 2,084 randomly selected adults showed that 88% 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 0.05 significance level to complete parts (a) through (e). Test of p=0.94 vs p *0.94 Sample X N 1 1829 2,084 Sample p 0.877639 95% CI Z-Value (0.863570,0.891709) -11.99 a. Is the test two-tailed, left-tailed, or right-tailed? OTwo-tailed test O Left-tailed test O Right tailed test b. What is the test statistic? The test statistic is (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. OA. Ho: p*0.94 OB. Ho: p=0.94 OC. Ho: p<0.94 O D. Ho:p> 0.94 P.Value 0.000 Choose the correct answer below. O A. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. OB. Reject the null hypothesis because the P-value is less than or equal to the significance level, c. O C. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, c. O D. Reject the null hypothesis because the P-value is greater than the significance level, a. e. What is the final conclusion? O A. There is not sufficient evidence to support the claim that 94% of adults own a cell phone. OB. There is sufficient evidence to warrant rejection of the claim that 94% of adults own a cell phone. O C. There is sufficient evidence to support the claim that 94% of adults own cell phone. O D. There is not sufficient evidence to warrant rejection of the claim that 94% of adults own a cell phone.
A poll of 2,084 randomly selected adults showed that 88% 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 0.05 significance level to complete parts (a) through (e). Test of p=0.94 vs p *0.94 Sample X N 1 1829 2,084 Sample p 0.877639 95% CI Z-Value (0.863570,0.891709) -11.99 a. Is the test two-tailed, left-tailed, or right-tailed? OTwo-tailed test O Left-tailed test O Right tailed test b. What is the test statistic? The test statistic is (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. OA. Ho: p*0.94 OB. Ho: p=0.94 OC. Ho: p<0.94 O D. Ho:p> 0.94 P.Value 0.000 Choose the correct answer below. O A. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. OB. Reject the null hypothesis because the P-value is less than or equal to the significance level, c. O C. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, c. O D. Reject the null hypothesis because the P-value is greater than the significance level, a. e. What is the final conclusion? O A. There is not sufficient evidence to support the claim that 94% of adults own a cell phone. OB. There is sufficient evidence to warrant rejection of the claim that 94% of adults own a cell phone. O C. There is sufficient evidence to support the claim that 94% of adults own cell phone. O D. There is not sufficient evidence to warrant rejection of the claim that 94% of adults own a cell phone.
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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Question
A poll of 2,084 randomly selected adults showed that 88% 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
significance level to complete parts (a) through (e).
Test of:
p=0.94 vs p≠ 0.94
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