A poll of 2,079 randomly selected adults showed that 90% 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 0.93Sample X1N2,0791876Sample p0.90235795% CIZ-Value(089597,0.915116) - 4.94a. is the test two-talled, left-talled, or rignt-talled?O Right tailed testOLeft-tailed testOTwo-tailed testb. 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?13--66 LLLLLP-Value0.000…... A poll of 2,079 randomly selected adults showed that 90% 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 *0.93Sample X1N2,0791876Sample p0.90235795% CI(0.889597,0.915116)Z-Value-4.94P-Value0.000RECENChoose the correct answer below.O A. Reject the null hypothesis because the P-value is less than or equal to the significance level, aOB. Fail to reject the null hypothesis because the P-value is greater than the significance level, aOC. Reject the null hypothesis because the P-value is greater than the significance level, c.OD. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, ae. What is the final conclusion?OA. There is sufficient evidence to support the claim that 93% of adults own a cell phone.OB. There is sufficient evidence to warrant rejection of the claim that 93% of adults own a cell phone.OC. There is not sufficient evidence to support the claim that 93% of adults own a cell phone.O D. There is not sufficient evidence to warrant rejection of the claim that 93% of adults own a cell phone.4

Test of p=0.93 vs p≠0.93 Sample X N Sample p 95% CI Z-Value P-Value 1 1876 2079 0.902357…

Answered: A poll of 2,079 randomly selected…

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 accompanying table gives results from a study of words spoken in a day by men and women. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. n X S Men Women 185 211 15,667.8 16,215.3 8,632.7 7,301.6 What are the null and alternative hypotheses? OA. Ho₂:0² #0²₂2 B. H₂:0² = 0²/2 H₁:0² > 0² H₁ : 0 ²₁2 = 0²/2 OC. H₂ : 0 ²₁2 = 0²/2 OD. H₂:0² = 0²/22 H₁:0² <0² H₁: : 0² / ± 0²/2 Identify the test statistic. F= (Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is. (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? O A. Reject Ho. There is insufficient evidence to support the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. B.…
A certain drug is used to treat asthma. In a clinical trial of the drug, 22 of 277 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 8% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to complete parts (a) through (e) below. b. What is the test statistic? c. What is the P-value?
The accompanying table gives results from a study of words spoken in a day by men and women. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that the numbers of words spoken in a day by men vary more than the numbers of words spoken in a day by women. n x s Men 185 15,667.8 8,632.5 Women 212 16,215.5 7,301.2 What are the null and alternative hypotheses? A. H0: σ21≠σ22 H1: σ21=σ22 B. H0: σ21=σ22 H1: σ21>σ22 C. H0: σ21=σ22 H1: σ21<σ22 D. H0: σ21=σ22 H1: σ21≠σ22 Identify the test statistic. F= (Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? A.Reject H0. There is insufficient evidence to support the claim that…
A poll of 2,188 randomly selected adults showed that 90% 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 2,188 1969 Sample p 0.899909 b. What is the test statistic? a. Is the test two-tailed, left-tailed, or right-tailed? Two-tailed test Right tailed test Left-tailed test The test statistic is -7.90 . (Round to two decimal places as needed.) c. What is the P-value? 95% CI The P-value is. (Round to three decimal places as needed.) (0.887333,0.912484) Z-Value - 7.90 P-Value 0.000
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A poll of 2,163 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) A.) Is the test two-tailed, left-tailed, or right-tailed? B.)What is the test statistic? C.)What is the P-value? D.) What is the null hypothesis and what do you conclude about it? E.) Choose the correct answer below. A. Reject the null hypothesis because the P-value is less than or equal to the significance level, alphaα. B. Fail to rejectFail to reject the null hypothesis because the P-value is greater than the significance level, alphaα. C.Fail to rejectFail to reject the null hypothesis because the P-value is less than or equal to the significance level, alphaα. D.Reject the null hypothesis because the P-value is greater than the significance…
A recent national report states the marital status distribution of the male population age 18 or older is as follows: Never Married (31.7%), Married (54.7%), Widowed (2.8%), Divorced (10.8%). The table below shows the results of a random sample of 1888 adult men from California. Test the claim that the distribution from California is as expected at the a 0.10 significance level. = a. Complete the table by filling in the expected frequencies. Round to the nearest whole number: Frequencies of Marital Status Outcome Frequency Expected Frequency Never Married 624 Married 1015 Widowed 36 Divorced 213 b. What is the correct statistical test to use? Goodness-of-Fit c. What are the null and alternative hypotheses? Ho: The distribution of marital status in California is not the same as it is nationally. Marital status and residency are dependent. The distribution of marital status in California is the same as it is nationally. Marital status and residency are independent. H₁: Marital status and…
A study reported that finger rings increase the growth of bacteria on health-care workers’ hands. Research suggests that 31 percent of health-care workers who wear rings have bacteria on one or both hands. Suppose that independent random samples of 100 health-care workers wearing rings is selected. What is the standard deviation of the sampling distribution of the sample proportions of health-care workers having bacteria on one or both hands? A. 100 B. 0.0462 C. 0.69
1. We want to investigate whether traffic accidents are more likely on certain days of the week on certain road sections. Which of the following tests best fits the situation?Choose one:a. Two-way analysis of variance OR b. Compatibility test OR c. Paired t-test 2. In which of the following can it be assumed that the sample comes from a normal distribution?Choose one:a. In the Runs test (random test)b. In the compatibility testc. In a two-way analysis of variance
6
A certain drug is used to treat asthma. In a clinical trial of the drug, 15 of 287 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. e) Whats the final conclusion?
A random sample of 50 pieces of candy from a bag of assorted types is examined. Which of the following types of candy contributes the most to the chi-square statistics and why?

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