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 Ho:p = 0.7 Ho:µ = 0.7 Ho:p= 0.7 H1:µ < 0.7 H1:µ + 0.7 H1:p < 0.7 H1:p # 0.7 H1:µ > 0.7 H1:p > 0.7 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 Question Help: D Post to forum

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 Ho:p = 0.7 Ho:µ = 0.7 Ho:p= 0.7 H1:µ < 0.7 H1:µ + 0.7 H1:p < 0.7 H1:p # 0.7 H1:µ > 0.7 H1:p > 0.7 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 Question Help: D Post to forum

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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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.
Teenagers who do not sleep well or long enough may have a higher risk of raised blood pressure, which could lead to cardiovascular disease later in life. The odds of raised blood pressure increase 3.5 times for those who have trouble falling asleep at night or who wake up too early and 2.5 times for those with sleeping periods of fewer than 6.5 hours. A health psychologist is interested in studying teenagers who have diastolic blood pressure scores in the top 20%. Diastolic blood pressure scores for teenagers follow a normal distribution with μ = 72 and σ = 10. Use the Distributions tool to help answer the questions that follow. The minimum z-score necessary to be in the top 20% of the diastolic blood pressure distribution is z =0.842 . Use this z-score to determine X, the corresponding diastolic blood pressure score. X = 80.42 This score, X, is the 80th percentile of diastolic blood pressure scores among teenagers. The percentile rank of this score is 80%…
A well-known brokerage firm executive claimed that 20% 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, 21% of them said they are confident of meeting their goals.Test the claim that the proportion of people who are confident is larger than 20% at the 0.10 significance level.The null and alternative hypothesis would be:
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 clinical trial is planned to compare an experimental medication designed to lower blood pressure to a placebo. Before starting the trial, a pilot study is conducted involving eight participants. The objective of the study is to assess how systolic blood pressure changes over time untreated. Systolic blood pressures are measured at baseline and again 4 weeks later Is there a statistically significant difference in blood pressures over time? Run the test at a 5% level of significance. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show all your work to receive full credit. Partial credit is available. Baseline 4 weeks 135 136 150 142 143 135 160 158 152 155 143 140 133 128 146 129
Acne is a common skin disease that affects most adolescents and can continue into adulthood. A study compared the effectiveness of three acne treatments and a placebo, all in gel form, applied twice daily for 12 weeks. The study's 517 teenage volunteers were randomly assigned to one of the four treatments. Success was assessed as clear or almost clear skin at the end of the 12 week period. The results of the study can be seen in the table below. Using the appropriate statistical test, determine if there is significant evidence that the four treatments perform differently. If so, how do they compare.
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 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 psychologist is interested in people’s assessment of life happiness and how it relates to positive lifestyle changes. The researcher randomly assigns 42 individuals to one of two groups: Exercise Only, and Exercise and Healthy Eating. The researcher believes that Group 1 (n = 21; Exercise and Healthy Eating) will be significantly happier than Group 2 (n = 21; Exercise Only). Perform the correct test of significance, with alpha = .05. Be sure to write out your hypotheses and final decision.
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…