IQ scores: Scores on an IQ test are normally distributed. A sample of 25 IQ scores had standard deviation s= 8. The developer of the test claims that the population standard deviation is greater than o= 13. Do these data provide sufficient evidence to support this claim? Use the a=0.01 level of significance. Part: 0/ 5 Part 1 of 5 State the appropriate null and alternate hypotheses. Ho: (Choose one) ▼ H: (Choose one) V

IQ scores: Scores on an IQ test are normally distributed. A sample of 25 IQ scores had standard deviation s= 8. The developer of the test claims that the population standard deviation is greater than o= 13. Do these data provide sufficient evidence to support this claim? Use the a=0.01 level of significance. Part: 0/ 5 Part 1 of 5 State the appropriate null and alternate hypotheses. Ho: (Choose one) ▼ H: (Choose one) V

Name: IQ scores: Scores on an IQ test are normally distributed. A sample of
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: IQ scores: Scores on an IQ test are normally distributed. A sample of 25 IQ scores had standarddeviation s= 8. The developer of the test claims that the population standard deviation is greater thano= 13. Do these data provide sufficient evidence to

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

Claim: The standard deviation of pulse rates of adult males is more than 10 bpm. For a random sample of 175 adult males, the pulse rates have a standard deviation of 10.9 bpm. Complete parts (a) and (b) below. a. Express the original claim in symbolic form. bpm (Type an integer or a decimal. Do not round.) b. Identify the null and alternative hypotheses. Ho: bpm H1: bpm (Type integers or decimals. Do not round.)
A psychologist uses two tests to assess clients on the same psychological construct. Both tests have been used extensively and the scores for each are normally distributed. The population mean for Test 1 is 30 with a standard deviation (o) of 8. The population mean for Test 2 is 50 with a standard deviation (o) of 12. Q12: The score for a client on Test 1 is 40. What is the equivalent score on Test 2? Hint: Designate the raw score of 40 as X1 and the unknown raw score on Test 2 as X2. Find the z score for X1 and use it in a second z score equation to solve for X2.
Test Prep. A GRE prep company claims that average score improvement of their students is equal to 50 points. We randomly sample 40 students of their average improvement is 47 points with a standard deviation of 37 points. Evaluate the company's claim using a hypothesis test. Use a 5% level of significance. (a) Write the hypotheses for the test: OHo: H = 50 Hạ: p 47 %3D O H,: µ = 50 Ha: p > 50 %3D OHo: H = 47 Ha: p = 47 (b) Compute the p-value for the test: p = |(round to three sig figs) (c) What is the final conclusion? O There is not sufficient evidence to reject the company's claim. O There is sufficient evidence to support the company's claim. O There is not sufficient evidence to support the company's claim. O There is sufficient evidence to reject the company's claim. Question Help: M Message instructor D Post to forum Submit All Parts Jump to Answer
On a certain hearing ability test, the mean is 300 and the standard deviation is 20. Thebetter you can hear, the higher your score. Can people who clean their ears frequentlyhear better than others? You take a sample of 31 people who clean their earsfrequently. Their sample mean test score is 308. Do a hypothesis test with α = 0.01.H 0 : People who clean their ears regularly hear at the same level that everyone elsedoes: µ = 300. c. Calculate your test statistic and draw it on your null hypothesis curve. This is justyour sample statistic ( or ) converted to a Z-score on your null hypothesis samplingdistribution. Make sure you are using the appropriate SE formula! Use forproportions and for means. d. Treat your test statistic as a Z-score and look up the area beyond it in the Z-table.That is your p-value. It may be helpful to shade this on your curve. (Remember, ifyou have a < in your alternative hypothesis, you’re looking at the area below yourtest statistic. If you have a…
For the given scenario, determine the type of error that was made, if any. (Hint: Begin by determining the null and alternative hypotheses.) A radio station has accepted 2626 as the mean age of its listeners. One radio station executive claims that the mean age of its listeners is different from 2626. The radio station executive conducts a hypothesis test and rejects the null hypothesis. Assume that in reality, the mean age of its listeners is 2929. Was an error made? If so, what type?
Claim: The mean systolic blood pressure of all healthy adults is less than than 119 mm Hg. Sample data: For 283 healthy adults, the mean systolic blood pressure level is 118.53 mm Hg and the standard deviation is 15.06 mm Hg. The null and alternative hypotheses are Ho μ=119 and H₁: μ< 119. Find the value of the test statistic. The value of the test statistic is (Round to two decimal places as needed.)
Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.05 significance level. The sample consisted of 55 night students, with a sample mean GPA of 2.87 and a standard deviation of 0.04, and 59 day students, with a sample mean GPA of 2.30 and a standard deviation of 0.03. Use correct notation for all answers below. a.) The null and alternative hypothesis would be: b.) The p-value is: c.) State your conclusion. d.) Interpret your conclusion specific to this situation.
A random sample of 64 students was asked to respond on a scale from one (strongly disagree) to seven (strongly agree) to the proposition: "Cellphone helps raise our standard of living." The sample mean response was 4.60 and the sample standard deviation was 1.6. Find the minimum significance level at which we can show that the true mean is greater than 4. 1 =-T.INV(0.05,63) 2 =T.DIST(3,63) 3 =T.INV.2T(0.10,63) 4 =T.DIST.RT(2,63) 5 =T.DIST.RT(3,63)
A century ago, the average height of adult women in the United States was 63 inches. Researchers believe that the average might be greater today. A random sample of 40 adult women was selected from the population. The sample had mean 64.2 inches and standard deviation 2.9 inches. Assuming all conditions for inference are met, the researchers will perform an appropriate hypothesis test to investigate their belief. Which of the following is the correct test statistic for the hypothesis test? A t = 63-64.2 2.9 63-64.2 2.9 t 3= 63-64.2 29 40 64.2-63 t 3= 2.9 64.2-63 t = 2.9 US 1 DELL 么 & %24 %23 || E.
U21T. Construct a 99% confidence interval estimate for the mean u using the given sample information. (Give your answers correct to two decimal plac n = 20, x = 16.8, and s = 2.1 |3D 15.6 X to 18 You may need to use the appropriate table in Appendix B to answer this question. Need Help? Read It Watch It Talk to a Tutor
need help with G only
Type an integer or decimal, do not round.