b. Test the claim by constructing an appropriate confidence intervalThe 90% confidence interval is < (P₁-P₂) <(Round to three decimal places as needed.)What is the conclusion based on the confidence interval?Because the confidence interval limitsthereevidence to support the claim that the rate of left-handedness among males is less than that among femalesc. Based on the results, is the rate of left-handedness among males less than the rate of left-handedness among females?to be a significant difference between the rates of left-handedness among males and females. ThereOA. It is reasonable to conclude that the rate of left-handedness among males is less than the rate of left-handedness among femalesOB. It is reasonable to conclude that the rate of left-handedness among males is the same as the rate of left-handedness among females.OC. It is not reasonable to conclude that the rate of left-handedness among males is less than the rate of left-handedness among femalesOD. The assumptions required for the tests are not valid, so the results should not be used for inference. In a random sample of males, it was found that 25 write with their left hands and 217 do not. In a random sample of females, it was found that 57 write with their left hands and 456 do not. Use a0.05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) belowa. Test the claim using a hypothesis testConsider the first sample to be the sample of males and the second sample to be the sample of females. What are the null and alternative hypotheses for the hypothesis test?OA H₂: P2P₂H₁: P₁ P₂OD. Ho: P₁ P₂H₁: P₁ P₂Identify the test statistic.2-0(Round to two decimal places as needed.)Identify the P-valueP-value=(Round to three decimal places as needed.)What is the conclusion based on the hypothesis test?The P-value ismales is less than that among females.the significance level of 0.05, soOB. H₂: P1 SP₂H₁: P₂ P₂CELLEⒸE H₂: P₁ P₂H₁: P₁

There are two independent samples which are males and females. We have to test whether the rate of…

Answered: In a random sample of males, it was…

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 hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. When evaluated using the 5% level of significance, what is the decision and interpretation of the null hypothesis? Show all steps.
Find A. Hypothesis and identify claim B. Compute test value C. Find p value D. Make decision E. Summarize results
In a survey of 500 drivers from the South, 408 wear a seat belt. In a survey of 370 drivers from the Northeast, 288 wear a seat belt. At a = 0.07, can you support the claim that the proportion of drivers who wear seat belts is greater in the South than in the Northeast? Assume the random samples are independent. Complete parts (a) through (e). (a) Identify the claim and state Ho and Ha The claim is "the proportion of drivers who wear seat belts in the South is greater than in the Northeast." Let p, represent the population proportion for the South, and p2 represent the population proportion for the Northeast. State Hg and H Choose the correct answer below. O B. Ho: P1 = P2 OC. Ho: P, * P2 O A. Ho: P1 > P2 H: P, SP2 H: P, *P2 Ha: P1 = P2 O D. Ho: P1 P2 OF. Ho: P1 2 P2 H,: P, OC. z
Is Friday the 13th Unlucky? The table shows the number of hospital admissions for Friday the 6th and Friday the 13th. Assume that the two samples are randomly selected. At the 0.05 significance level, test the claim that there is no mean difference in the number of hospital admissions for Friday the 6th and Friday the 13th.(Be sure to subtract in the same direction). Friday 6th Friday 13th Difference 4 2 6 10 5 6 9 13 3 2 11 15 12 12 What are the correct hypotheses? (Select the correct symbols and values.)H0: Select an answer μ₁ μ(d) x̄₂ σ₁² x̄₁ s₁² p μ ? = < ≤ ≠ > ≥ Select an answer 0 μ₁ μ₂ x̄₁ p σ₁² s₁² x̄₂ μ H1: Select an answer x̄₂ μ₁ μ p μ₂ x̄₁ s₂² μ(d) σ₂² ? < ≥ ≠ = ≤ > Select an answer μ x̄₂ s₁² x̄₁ p 0 μ₂ σ₁² μ₁ Original Claim = Select an answer H₀ H₁ df = Based on the hypotheses, find the following: Test Statistic = (Round to three decimal places.)Critical value(s) = ±± (Round to three decimal places.)p-value = (Round…
The national percentage of automobile accident fatalities that are alcohol related is 39%. In a random sample of 96 automobile accident fatalities in the state of New York, 44 were alcohol related. Is this sufficient evidence to say that new York has a higher percentage of alcohol related automobile fatalities? a. Perform the appropriate test of hypothesis and paste the session window output below. b. What is the value of the test statistic? c. At a = .10, what is your decision?
Z?
6.5 Gender equality. The General Social Survey asked a random sample of 1,390 Americans the following question: “On the whole, do you think it should or should not be the government’s responsibility to promote equality between men and women?” 82% of the respondents said it “should be”. At a 95% confidence level, this sample has 2% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning.8 (a) We are 95% confident that between 80% and 84% of Americans in this sample think it’s the government’s responsibility to promote equality between men and women. (b) We are 95% confident that between 80% and 84% of all Americans think it’s the government’s respon- sibility to promote equality between men and women. (c) If we considered many random samples of 1,390 Americans, and we calculated 95% confidence intervals for each, 95% of these intervals would include the true population proportion of Americans who think…
The accompanying table contains the data for the amounts (in oz) in cans of a certain soda. The cans are labeled to indicate that the contents are 16 oz of soda. Use the sign test and a 0.01 significance level to test the claim that cans of this soda are filled so that the median amount is 16 oz. If the median is not 16 oz, are consumers being cheated? Click the icon to view the data. What are the null and alternative hypotheses? O A. Ho: Median volume ≤ 16 oz H₁: Median volume = 16 oz C. Ho: Median volume = 16 oz H₁: Median volume # 16 oz Find the test statistic. Test statistic (Round to three decimal places as needed.) = More Info 15.6 16.0 16.1 16.1 407 15.7 (…) 16.1 15.9 19:3 15.8 16.5 15.8 15.7 15.9 OB. Ho: Median volume ≤ 16 oz H₁: Median volume > 16 oz O D. Ho: Median volume = 16 oz H₁: Median volume < 16 oz Volume (in ounces) 15.7 15.5 15.7 15.7 15.8 420 15.9 103 16.3 15.7 15.7 16.1 16.3 15.8 15.7 15.6 15.5 15.6 15.5 20 16.0 400 15.8 19:0 15.6 15.6 15.5 16.4 15.9 D - X
A null hypothesis states that p=.7, the observed sample proportion is p=.653 with a sample size of n=90. Determine the z-score. -0.97 O 0.92 -1.48 O 1.35
The mean price of mid-sized cars in a region is $57,600. A test is conducted to see if the claim is true. Identify the null hypothesis. ? v? v Identify the alternative hypothesis. ? v:?v| ? v Identify the type I error. O You decide that the mean price of mid-sized cars is not $57,600, when the mean price of mid- sized cars actually is $57,600. O You decide that the mean price of mid-sized cars is $57,600, when the mean price of mid- sized cars actually is not $57,600. O You decide that the mean price of mid-sized cars is not $57,600, when the mean price of mid- sized cars actually is nor $57,600. O You decide that the mean price of mid-sized cars is $57,600, when the mean price of mid- sized cars actually is $57,600. Identify the type II error. O You decide that the mean price of mid-sized cars is not $57,600, when the mean price of mid- sized cars actually is $57,600. O You decide that the mean price of mid-sized cars is $57,600, when the mean price of mid- sized cars actually is not…
Is Friday the 13th Unlucky? The table shows the number of hospital admissions for Friday the 6th and Friday the 13th. Assume that the two samples are randomly selected. At the 0.05 significance level, test the claim that there is no mean difference in the number of hospital admissions for Friday the 6th and Friday the 13th.(Be sure to subtract in the same direction). Friday 6th Friday 13th Difference 4 2 6 10 5 6 9 13 3 2 11 15 12 12

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