Background: Henry Cavendish carried out a series of experiments in 1798 to determine the mean density ofthe earth, as an indirect means to calculate the gravitational constant. A measurement of 5.5 indicates theearth's density is 5.5 times that of water.Source: Cavendish, H. (1798). Experiments to determine the density of theearth. Philosophical Transactions of the Royal Society of London, 88 (Part II),469-527.Directions: Perform an appropriate significance test to determine whether ornot Henry Cavendish's measurements for the density of the earth weredifferent from the true value of 5.513.1. Click on the Data button shown below to display the data and copy andpaste it into a statistical software package. Click the Data button asecond time to hide the data.DataDensity5.55.575.425.615.435.374.885.625.635.075.295.345.265.445.565.555.345.35.365.795.755.3955.585.585.275.855.655.492. Use the statistical software package to compute a numerical summary.3. Perform the significance test.a. Choose the correct null and alternative hypothesis.OHo5.513H₂: μ> 5.513Ho: 5.513H₂ = 5.513OHo5.513Ha 5.513OHo5.513Ha 5.513b. Compute the test statistic. Round your answer to 4 decimal places.c. Compute the p-value. Round your answer to 4 decimal places.d. Interpret the results of the test.O The p-value provides strong evidence against the null hypothesis. The differencebetween Henry Cavendish's measurements for the density of the earth and the true valueare statistically significant.O The p-value provides little evidence against the null hypothesis. The difference betweenHenry Cavendish's measurements for the density of the earth and the true value are notstatistically significant.

Let μ be the mean density of the water. Given that, Population mean (μ) = 5.513 Sample size (n) = 29…

Answered: 2. Use the statistical software package…

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

K Use the IQ scores for the accompanying sample of individuals with low lead exposure to test the claim that they are from a population with a median IQ score of 100. Use a 0.01 significance level. Click the icon to view the data on IQ scores. F2 Find the null and alternative hypotheses. OA. Ho: Median = 100 H₁: Median #100 OC. Ho: Median = 100 H₁: Median 100 Find the test statistic. z=(Round to two decimal places as needed.) Find the P-value. F3 F4 F5 F6 W F7 DELL F8 Ō GET OB. Ho: Median # 100 H₁: Median = 100 OD. Ho: Median 100 H₁: Median = 100 F9 O F10 F11 :: F12 PrtScr 60 Submit quiz Insert Submit quiz 11:45 AM 4/9/2023 Delete
Find the P-value for the indicated hypothesis test with the given standardized test statistics z. Decide whether to reject H0 for the given level of significance α. Left-tailed test, z=-1.32, α=0.10 Right-tailed test, z=2.46, α=0.01 Two-tailed test, z=-1.68, α=0.05
Determine P-Value.
A sample mean, sample size, and population standard deviation are provided below. Use the one-mean z test to perform the required hypothesis test at the 10% signifigance level. Find the test statistics and P-Values. x bar-22 n=36 s=5 Ho: mean =24 Ha: mean < 24
Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below. H0:μ=8.2 seconds; Ha:μ<8.2 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.
represents the summary statistics for 44 blue catfish caught at this reservoir, and a 0.05 significance level to test the claim that the mean weight of the fish in the John H. Kerr Reservoir is greater than 40 pounds. n = 44; = 40.76 pounds; s = 5.12 pounds a) kdentify the null and alternative hypotheses? Ho: =V 40 H.:p>v40 b) What type of hypothesis test should you conduct (left-, right-, or two-tailed)? O left-tailed O right-tailed O two-tailed c) Identify the appropriate significance level. 0.05 d) Calculate your test statistic. (Round your final answer to two decimal places.) e) Calculate your p-value. (Round your final answer to four decimal places.) f) Do you reject the null hypothesis? O We reject the null hypothesis, since the p-value is less than the significance level. O We reject the null hypothesis, since the p-value is not less than the significance level. O We do not reject the null hypothesis, since the p-value is less than the significance level. O We do not reject the…
A sample mean, sample size, and population standard deviation are provided below. Use the one-mean z test to perform the required hypothesis test at the 1% signifigance level. Find the test statistic and p-value. Do we reject or fail to reject? x bar=17 n=36 standard deviation=6 Ho: mean=21 Ha: mean <21
A random sample of 10 independent healthy people showed the body temperatures given below (in degrees Fahrenheit). Test the hypothesis that the population mean is not 98.6 °F, using a significance level of 0.05. 98.3 98.7 99.0 96.6 98.9 98.9 97.3 99.3 98.3 97.1 Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: u= 98.6 О В. Но: и#98.6 Hg: µ = 98.6 Ос. Н: и- 98.6 Hạ: u 98.6 O E. Ho: µ> 98.6 Ha: u = 98.6 OF. Ho: u<98.6 Ha: u = 98.6 Check the conditions to see whether the test statistic will follow a t-distribution. The sample V random and the observations V independent. The distribution of the sample is Find the test statistic. t=D (Round to two decimal places as needed.) Find the p-value. p-value = (Round to three decimal places as needed.) Interpret the results of the test. Since the p-value is V the significance level, V Ho. There is V evidence to conclude that the population mean is not 98.6°F at a significance level of 0.05.
Assume a significance level of α = 0.05 and use the given information to complete parts (a) and (b) below. Original claim: Less than 44% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.2897. a. State conclusion about the null hypothesis. (Reject Ho or fail to reject Ho.) Choose the correct answer below. O A. Fail to reject H, because the P-value is less than or equal to a OB. Reject H, because the P-value greater than a. OC. Fail to reject Ho because the P-value is greater than a O D. Reject H, because the P-value is less than or equal to a. G
Please provide a clear and complete solution. Answer fast for I have 30 minutes left. Thank you very much.
Assume a significance level of α=0.05and use the given information to complete parts (a) and (b) below. Original claim: More than 51% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.0383. a. State a conclusion about the null hypothesis. (Reject H0or fail to reject H0.) Choose the correct answer below. A.Reject H0 because the P-value is less than or equal to α. B. Fail to reject H0 because the P-value is greater than α. C. Fail to reject H0 because the P-value is less than or equal to α. D. Reject H0 because the P-value is greater than α. b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion? A. There is sufficient evidence to support the claim that the percentage of adults that would erase all of their personal information online if they could is more than 51%. B. The percentage of…
If you use a 0.05 level of significance in a two-tail hypothesis test, what decision will you make if ZSTAT= -2.03?Question content area bottom Determine the decision rule. Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places to the right of the decimal point as needed.) State your conclusion. Choose the correct answer below.A.Since ZSTAT into the rejection region, do not reject HoB.Since ZSTAT into the rejection region, reject HoC.Since ZSTAT into the rejection region, do not reject HoD.Since ZSTAT into the rejection region, reject Ho

SEE MORE QUESTIONS

Recommended textbooks for you

MATLAB: An Introduction with Applications

Statistics

ISBN:

9781119256830

Author:

Amos Gilat

Publisher:

John Wiley & Sons Inc

Probability and Statistics for Engineering and th…

Statistics

ISBN:

9781305251809

Author:

Jay L. Devore

Publisher:

Cengage Learning

Statistics for The Behavioral Sciences (MindTap C…

Statistics

ISBN:

9781305504912

Author:

Frederick J Gravetter, Larry B. Wallnau

Publisher:

Cengage Learning

MATLAB: An Introduction with Applications

Statistics

ISBN:

9781119256830

Author:

Amos Gilat

Publisher:

John Wiley & Sons Inc

Probability and Statistics for Engineering and th…

Statistics

ISBN:

9781305251809

Author:

Jay L. Devore

Publisher:

Cengage Learning

Statistics for The Behavioral Sciences (MindTap C…

Statistics

ISBN:

9781305504912

Author:

Frederick J Gravetter, Larry B. Wallnau

Publisher:

Cengage Learning

Elementary Statistics: Picturing the World (7th E…

Statistics

ISBN:

9780134683416

Author:

Ron Larson, Betsy Farber

Publisher:

PEARSON

The Basic Practice of Statistics

Statistics

ISBN:

9781319042578

Author:

David S. Moore, William I. Notz, Michael A. Fligner

Publisher:

W. H. Freeman

Introduction to the Practice of Statistics

Statistics

ISBN:

9781319013387

Author:

David S. Moore, George P. McCabe, Bruce A. Craig

Publisher:

W. H. Freeman

SEE MORE TEXTBOOKS