A well-known brokerage firm executive claimed that 10% ofinvestors are currently confident of meeting theirinvestment goals. An XYZ Investor Optimism Survey,conducted over a two week period, found that in a sampleof 200 people, 7% of them said they are confident ofmeeting their goals.Test the claim that the proportion of people who areconfident is smaller than 10% at the 0.05 significance level.The null and alternative hypothesis would be:Ο H: μ -0.1Ha:µ # 0.1Ho : u < 0.1На: и > 0.1Но:р > 0.1На:р < 0.1Ho: u > 0.1На: р < 0.1О Но:р< 0.1Ha:p > 0.1Ho:p = 0.1Ha:p + 0.1The test is:O left-tailedtwo-tailedright-tailed

Answered: Image /qna-images/answer/3602961c-322e-4a75-945e-d2d4698b0be7.jpg

Answered: A well-known brokerage firm executive…

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 well-known brokerage firm executive claimed that 70% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 100 people, 77% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is larger than 70% at the 0.025 significance level. The null and alternative hypothesis would be: Ho:µ = 0.7 Ho : µ = 0.7 Ho:p= 0.7 Họ:p = 0.7 Họ:µ = 0.7 Ho:p= 0.7 H1: µ 0.7 H1:p > 0.7 %3D %3D %3D %3D The test is: right-tailed two-tailed left-tailed The test statistic is: (to 3 decimals) The p-value is: (to 4 decimals) Based on this we: Fail to reject the null hypothesis O Reject the null hypothesis Question Help: D Post to forum Submit Question ☆吧( NOV 20 W 3D ロ

A physical therapist is interested in studying the proportion of adults that receive a physical therapy treatment plan due to sciatica. The physical therapist states the proportion of adults that receive a physical therapy treatment plan due to sciatica is less than 0.30. If we would like to test the physical therapist's claim with a hypothesis test using a significance level of α=0.10 , which of the following choices are true? Select the correct answer below: There is a 10% chance we will conclude p=0.30, but is in fact p<0.30. There is a 10% chance of rejecting the null hypothesis. There is a 10% chance we will conclude p<0.30, but is in fact p=0.30. There is a 10% chance that p<0.30
A well-known brokerage firm executive claimed that 80% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 500 people, 75% of them said they are confident of meeting their goals.Test the claim that the proportion of people who are confident is smaller than 80% at the 0.10 significance level.The null and alternative hypothesis would be: H0:μ=0.8H0:μ=0.8H1:μ≠0.8H1:μ≠0.8 H0:p≤0.8H0:p≤0.8H1:p>0.8H1:p>0.8 H0:μ≤0.8H0:μ≤0.8H1:μ>0.8H1:μ>0.8 H0:μ≥0.8H0:μ≥0.8H1:μ<0.8H1:μ<0.8 H0:p≥0.8H0:p≥0.8H1:p<0.8H1:p<0.8 H0:p=0.8H0:p=0.8H1:p≠0.8H1:p≠0.8 The test is: right-tailed left-tailed two-tailed The test statistic is: (to 3 decimals)The p-value is: (to 4 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
Test the claim that the proportion of men who own cats is smaller than 90% at the .05 significance level.The null and alternative hypothesis would be: H0:μ=0.9H0:μ=0.9H1:μ>0.9H1:μ>0.9 H0:p=0.9H0:p=0.9H1:p>0.9H1:p>0.9 H0:p=0.9H0:p=0.9H1:p≠0.9H1:p≠0.9 H0:μ=0.9H0:μ=0.9H1:μ≠0.9H1:μ≠0.9 H0:μ=0.9H0:μ=0.9H1:μ<0.9H1:μ<0.9 H0:p=0.9H0:p=0.9H1:p<0.9H1:p<0.9 The test is: right-tailed two-tailed left-tailed Based on a sample of 65 people, 87% owned catsThe test statistic is: (to 2 decimals)The critical value is: (to 2 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis I dont understand this.
A well-known brokerage firm executive claimed that 90% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 100 people, 81% of them said they are confident of meeting their goals.Test the claim that the proportion of people who are confident is smaller than 90% at the 0.10 significance level.The null and alternative hypothesis would be: H0:p=0.9H0:p=0.9H1:p≠0.9H1:p≠0.9 H0:p≥0.9H0:p≥0.9H1:p<0.9H1:p<0.9 H0:μ≥0.9H0:μ≥0.9H1:μ<0.9H1:μ<0.9 H0:p≤0.9H0:p≤0.9H1:p>0.9H1:p>0.9 H0:μ=0.9H0:μ=0.9H1:μ≠0.9H1:μ≠0.9 H0:μ≤0.9H0:μ≤0.9H1:μ>0.9H1:μ>0.9 The test is: left-tailed two-tailed right-tailed The test statistic is: (to 3 decimals)The p-value is: (to 4 decimals)Based on this we:
A mobile game currently has 38% of the market share. The CEO of the company producing the game recently authorized an advertising campaign with the goal of increasing the game's market share. A random sample of 1041 mobile gamers found that 409 gamers played the game at least once a day. A .Setup the null and alternative hypotheses to see if there is a significant change in the market share of the game. B. Determine if you can use a normal model to approximate the distribution of sample proportions. C. Find the critical value for a level of significance of 0.03. D. Find the standardized test statistic. E. State the conclusion of the hypothesis test.
A local coach wants to see if training in light rooms or dark rooms impacts athlete performance. With the single team, the coach tries training in different conditions and takes measurements. She ran her analysis in Microsoft Excel and received the following output table. If the P value is less than 0.05, the null hypothesis may be rejected. Based on the data table, what decision should the coach make about the hypothesis? Coach should reject the null hypothesis. Coach should fail to reject the null hypothesis. Coach should accept the null hypothesis. Coach does not have enough information to decide.
Spam: A researcher reported that 71.8 % of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that greater than 69% of emails are spam? Use both a=0.01 and a= 0.05 levels of significance and the critical value method with the table. Part 1 of 5 State the appropriate null and alternate hypotheses. Ho: P- .69 H1: p> .69 This hypothesis test is a right-tailed V test. Part 2 of 5 Find the critical values. Round the answers to three decimal places. For a=0.01 , the critical value is 2.326 For a=0.05 , the critical value is 1.645 Part: 2/5 Part 3 of 5 Compute the test statistic. Do not round intermediate calculations. Round the answer to two decimal places.
A well-known brokerage firm executive claimed that 50% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 400 people, 51% of them said they are confident of meeting their goals.Test the claim that the proportion of people who are confident is larger than 50% at the 0.05 significance level.The null and alternative hypothesis would be: H0:μ=0.5H0:μ=0.5H1:μ≠0.5H1:μ≠0.5 H0:p≥0.5H0:p≥0.5H1:p<0.5H1:p<0.5 H0:μ≥0.5H0:μ≥0.5H1:μ<0.5H1:μ<0.5 H0:p=0.5H0:p=0.5H1:p≠0.5H1:p≠0.5 H0:μ≤0.5H0:μ≤0.5H1:μ>0.5H1:μ>0.5 H0:p≤0.5H0:p≤0.5H1:p>0.5H1:p>0.5 The test is: two-tailed left-tailed right-tailed The test statistic is: (to 3 decimals)The p-value is: (to 4 decimals)Based on this we: Fail to reject the null hypothesis Reject the null hypothesis
A well-known brokerage firm executive claimed that 50% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 800 people, 47% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is smaller than 50% at the 0.01 significance level. The null and alternative hypothesis would be: Ho:p > 0.5 Ho:p 0.5 Ho:p= 0.5 H1:p 0.5 Hị:µ + 0.5 H:µ > 0.5 Hị:p < 0.5 H1:p+0.5 The test is: right-tailed two-tailed left-tailed The test statistic is: (to 3 decimals) The p-value is: (to 4 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
A well-known brokerage firm executive claimed that 90% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 400 people, 84% of them said they are confident of meeting their goals. Test the claim that the proportion of people who are confident is smaller than 90% at the 0.10 significance level. The null and alternative hypothesis would be: Ho: p= 0.9 Hop≥ 0.9 Hopμ ≥ 0.9 H₁ μ = 0.9 Ho: p ≤ 0.9 Ho: ≤ 0.9 0 H₁: p0.9 H₁: p 0.9 H₁: >0.9 The test is: two-tailed left-tailed right-tailed o The test statistic is: The p-value is: Based on this we: Reject the null hypothesis O Fail to reject the null hypothesis (to 3 decimals) (to 4 decimals)
A new reading program may reduce the number of elementary school students who read below grade level. The company that developed this program supplied materials and teacher training for a large-scale test involving nearly 8,400 children in several different school districts. Statistical analysis of the results showed that the percentage of students who did not meet the grade-level goal was reduced from 14.7% to 14%. The hypothesis that the new reading program produced no improvement was rejected with a P-value of 0.026. Complete parts a) and b) below. a) Explain what the P-value means in this context. Choose the correct answer below. OA. There is a 97.4% chance of seeing a sample proportion of 14.7% (or more) of students failing the test by natural sampling variation if 14% is the true population value. B. There is only a 2.6% chance of seeing a sample proportion of 14.7% (or more) of students failing the test by natural sampling variation if 14% is the true population value. OC. There…

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