-City planners are concerned if the mean lifetime of the most commonlyplanted trees in urban settings is more than 30 years. A sample of 42 recently fallentrees yielded a mean age of 30.7 years with standard deviation of 3.2 years.(a).5% significance level.Using the above data, perform an appropriate hypothesis test at

Given that Hypothesized Population Mean (\mu)(μ) = 3030 Sample Standard Deviation (s)(s) =…

Answered: -City planners are concerned if the…

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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The following sample data are from a normal population: Sample size (n) = 25, Sample mean=14, Sample standard deviation=3.6 With 95% confidence, what is the margin of error for the estimation of the population mean?
Consider the time it takes, in minutes, for police officers to respond to a 911 call, where a life-threatening crime (an armed robbery, an assault, a shooting) is in progress. For a particular police jurisdiction, the average response time is 20 minutes with a standard deviation of 3 minutes. Assuming the data is symmetric and using the Empirical rule, what %% of the response times fall between 17 to 23 minutes.
You were told that the mean daily cost of a hotel room in downtown St. Louis is $120. You will be needing to rent a hotel room there in the near future and you want to find the best rate. You randomly select a sample of 18 hotel rooms and find that the average cost is $121.33 with a standard deviation of $18.046. At the α = 0.05 significance level, determine if there is enough evidence to reject the claim. What is the decision for this hypothesis test? You were told that the mean daily cost of a hotel room in downtown St. Louis is $120. You will be needing to rent a hotel room there in the near future and you want to find the best rate. You randomly select a sample of 18 hotel rooms and find that the average cost is $121.33 with a standard deviation of $18.046. At the α = 0.05 significance level, determine if there is enough evidence to reject the claim. What is the decision for this hypothesis test? Reject the null because the test statistic is in the critical region and the…
Can someone please help? Thank you.
Historically, evening long-distance calls from a particular city have averaged 15.1 minutes per call. In a random sample of 37 calls, the sample mean time was 13.3 minutes. Assume the standard deviation is known to be 5 minutes. Using a 0.05 level of significance, is there sufficient evidence to conclude that the average evening long-distance call has decreased?
A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is not equal to the mean of WaxCo. In a random sample of 12 floors of WaxWin and 20 of WaxCo. WaxWin had a mean lifetime of 27.6 with a standard deviation of 8.8 and WaxCo had a mean lifetime of 20.3 with a standard deviation of 7.6. Perform a hypothesis test using a significance level of 0.01 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: HA: P1 > µ2(claim) Ho: µ1 2 µ2 HA: µ1 < µ2(claim) Ho: µ1 = µ2 HA: H1 + H2(claim) Since the level of significance is 0.01 the critical value is 2.836 and -2.836 The test statistic is: (round to 3 places) The p-value is: (round to 3 places)
A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is less than the mean of WaxCo. In a random sample of 20 floors of WaxWin and 17 of WaxCo. WaxWin had a mean lifetime of 26.4 with a standard deviation of 8.6 and WaxCo had a mean lifetime of 27.2 with a standard deviation of 6.6. Perform a hypothesis test using a significance level of 0.05 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: Ο Ηο: μι < με HA₁₂(claim) Ο Ηρ: με 2 με HA: 1₂(claim) O Ho: ₁ = ₂ HAM12(claim) Since the level of significance is 0.05 the critical value is -1.69 (round to 3 places) The test statistic is: The p-value is: (round to 3 places)
From a random sample of 48 days in a recent year, U.S. gasoline prices had a mean of $2.34 and a standard deviation of $0.32. Construct a 95% confidence interval for the population mean
To evaluate if auditors might be helped in determining the chances of fraud if they carefully measure cash flow, samples of midlevel auditors from CPA firms were asked to indicate the chance of material fraud on a scale from 0 to 100 for a case. A random sample of 36 auditors used the cash-flow information. Their mean assessment was 35.51, and the sample standard deviation was 24.14. For an independent random sample of 36 auditors not using the cash-flow information, the sample mean and standard deviation were, respectively, 40.58 and 29.18. Assuming that the two population distributions are normal with equal variances, test against a two-sided alternative the null hypothesis that the population means are equal. (Use α = 0.05.) Click the icon to view a table of critical values for the Student's t-distribution. Let μx be the mean assessment for auditors who measured cash flow and μy be the mean assessment for auditors not using the cash-flow information. Determine the null and…

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