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A poll of 2,096 randomly selected adults showed that 91% of them own cell phones. The technology display below results from a test of the claim that 93% 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=0.93 vs p *0.93 Sample XN 1 1903 2,096 Sample p 0.907920 95% CI (0.895542,0.920298) Z-Value -3.96 P.Value 0.000

A poll of 2,096 randomly selected adults showed that 91% of them own cell phones. The technology display below results from a test of the claim that 93% 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=0.93 vs p *0.93 Sample XN 1 1903 2,096 Sample p 0.907920 95% CI (0.895542,0.920298) Z-Value -3.96 P.Value 0.000

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.3: Special Probability Density Functions
Problem 16E
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A poll of 2,096 randomly selected adults showed that 91% of them own cell phones. The technology display below results from a test of the claim that 93% 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 = 0.93 vs p *0.93 Sample X N 1 2,096 1903 Sample p 0.907920 95% CI (0.895542,0.920298) Z-Value -3.96 P-Value 0.000
Transcribed Image Text:A poll of 2,096 randomly selected adults showed that 91% of them own cell phones. The technology display below results from a test of the claim that 93% 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 = 0.93 vs p *0.93 Sample X N 1 2,096 1903 Sample p 0.907920 95% CI (0.895542,0.920298) Z-Value -3.96 P-Value 0.000
b. What is the test statistic? Z= (Round to two decimal places as needed.) c. What is the P-value? P-value= (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. OA. Ho: p *0.08 OB. Ho: p=0.08 Oc. Ho:p> 0.08 OD. Ho: p<0.08 Decide whether to reject the null hypothesis. Choose the correct answer below. OA. 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, a. Oc. Fail to 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 less than or equal to the significance level, a.
Transcribed Image Text:b. What is the test statistic? Z= (Round to two decimal places as needed.) c. What is the P-value? P-value= (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. OA. Ho: p *0.08 OB. Ho: p=0.08 Oc. Ho:p> 0.08 OD. Ho: p<0.08 Decide whether to reject the null hypothesis. Choose the correct answer below. OA. 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, a. Oc. Fail to 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 less than or equal to the significance level, a.
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