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 Sample p 95% CI Z-Value P-Value 1 1995 2,094 0.952722 (0.943632,0.961812) 5.52 0.000

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 Sample p 95% CI Z-Value P-Value 1 1995 2,094 0.952722 (0.943632,0.961812) 5.52 0.000

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A poll of 2,017 randomly selected adults showed that 92% 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 0.01 significance level to complete the equation.
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…
A poll of 2,009 randomly selected adults showed that 95% 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 Sample p 95% CI Z-Value P-Value 1 1912 2,009 0.951717 (0.942344,0.961091) 2.21 0.027 a. Is the test two-tailed, left-tailed, or right-tailed? O Right tailed test O Left-tailed test O Two-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. O A. Ho: p=0.94 Trash Click to select your answer(s).
The average retirement age in America is 63 years old. Do small business owners retire at a different average age? The data below shows the results of a survey of small business owners who have recently retired. Assume that the distribution of the population is normal. 66, 60, 62, 71, 58, 73, 60, 72, 70, 61, 57 What can be concluded at the the a 0.05 level of significance level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: uv = 63 H: uv> c. The test statistic ? = (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? a f. Based on this, we should Select an answerv the null hypothesis. g. Thus, the final conclusion is that O The data suggest the population mean retirement age for small business owners is not significantly different from 63 at a = 0.05, so there is insufficient evidence to conclude that the population mean…
A poll of 2,097 randomly selected adults showed that 94% 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 Sample p 95% CI Z-Value P-Value 1 1977 2,097 0.942775 (0.932834,0.952717) 6.53 0.000 a. Is the test two-tailed, left-tailed, or right-tailed? Two-tailed test Left-tailed test 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. О А. Но: р> 0.9 В. Но: р3D0.9
A poll of 1086 Americans showed that 47.1% of the respondents prefer to watch the news rather than read or listen to it. Use those results with a 0.01 significance level to test the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution.
A poll of 1188 Americans showed that 46.8% of the respondents prefer to watch the news rather than read or listen to it. Use those results with a 0.10 significance level to test the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all Americans who prefer to watch the news rather than read or listen to it. Identify the null and alternative hypotheses. Ho: P = 0.5 H₁: p < 0.5 (Type integers or decimals. Do not round.) Identify the test statistic. z = (Round to two decimal places as needed.)
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 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
Start A sales manager for a large department store believes that customer spending per visit with a sale is lower than customer spending without a sale, and would like to test that claim. A simple random sample of customer spending is taken from without a sale and with a sale. The results are shown below. Without sale with sale 82.904 71.316 Mean Variance 1984.96 1826.41 Observations 150 Hypothesized Mean Difference 328 PT<=t) one-tail 0.0071 Critical one-tail -1.65 PT<=t) two-tail Critical two-tail -1.967 6 Confidence Level 3 Ex: 1.234 Samples from without sale: n₁ = Ex 9 Samples from with sale: n₂ Point estimate for spending without sale: 1 Point estimate for spending with sale: 2 = 0 J. Ex: 1.234 EN P -7 T P= = t= DE UNDE IS-
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:…