3. At a certain college, it is estimated that fewer than 25% of the students have cars on campus. Does this seemto be a valid estimate if, in a random sample of 90 college students, 28 are found to have cars? Use a 0.05level of significance. IStep 1: State the null and alternative hypothesis in words. Step 3: Compute test statisticHO:H1:Step 2: Level of Significance and Critical Value/s and Step 4: Decision isRegionStep 5: State the conclusion.-2

We have given that Sample size n =90 Favorable cases x=28 Sample proportion p^=x/n =28/90 =0.3111

Answered: 3. At a certain college, it is…

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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One study claimed that 86% of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 238 college students and finds that 193 of them identify themselves as procrastinators. Does this evidence support the professor’s claim that fewer than 86% of college students are procrastinators? Use a 0.05 level of significance. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
A researcher wants to select a sample from a population with a mean (µ)= 60 and administer a treatment to individuals in the sample. He thinks the treatment will help improve scores. a. State the hypotheses for a one-tailed test. b. For a one-tailed test, would the critical region be located in the positive or right-hand tail of the distribution, or in the negative or left-hand tail of the distribution?
Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x=101.22, y=102.45, r=0.916, P-value=0.000, and y=−8.44+1.1x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 99? Use a significance level of 0.05.
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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.
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You claim that the proportion of students who favored ONLINE ASSESSMENT over FACE-TO-FACE ASSESSMENT is 65% at a 5% significance level. From the analysis, the Z-test is 1.16. The correct conclusion is we have a. sufficient evidence to reject the claim. O b. insufficient evidence to reject the claim. O c. insufficient evidence to support the claim. O sufficient evidence to support the claim.
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You wish to test the following claim (Ha) at a significance level of a = 0.002. Ho: P₁ P2 Ha P₁ P2 : You obtain a sample from the first population with 523 successes and 107 failures. You obtain a sample from the second population with 268 successes and 40 failures. test statistic = p-value [three decimal accuracy] [four decimal accuracy]
The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 317 homes in Oregon found that 166 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 5% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Ho: P Test Statistic = 0° p-value = H₁: p Based on the hypotheses, compute the following: #v oº 48 48 <-Need answer, please. <-Need answer, please OB 0° Based on the above we choose to Fail to reject the null hypothesis ✓ OB The correct summary would be: There is not enough evidence to support the claim that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.
I In 2000, 6.1% of people used marijuana. This year, a company wishes to use their employment drug screening to test a claim. They take a simple random sample of 2848 job applicants and find that 145 individuals fail the drug test for marijuana. They want to test the claim that the proportion of the population failing the test is lower than 6.1%. Use .10 for the significance level. Round to three decimal places where appropriate. Hypotheses: Ho:p=6.1%Ho:p=6.1% H1:p<6.1%H1:p<6.1% Test Statistic: z = Critical Value: z = p-value: Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis Conclusion About the Claim: There is sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.1% There is NOT sufficient evidence to support the claim that the proportion of the population failing the test is lower than 6.1% There is sufficient evidence to warrant rejection of the claim that the…

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