The proportion p of residents in a community who recycle has traditionally been 70%. A policy maker claims that the proportion is less than 70% now that oneof the recycling centers has been relocated. If 144 out of a random sample of 210 residents in the community said they recycle, is there enough evidence tosupport the policy maker's claim at the 0.05 level of significance?Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)(a) State the null hypothesis H. and the alternative hypothesis H₁.Ho :0H₁:0(b) Determine the type of test statistic to use.(Choose one) ▼(c) Find the value of the test statistic. (Round to three or more decimal places.)0(d) Find the critical value. (Round to three or more decimal places.)0(e) Is there enough evidence to support the policy maker's claim that theproportion of residents who recycle is less than 70%?ExplanationCheckLOнXALLXI5 1D=0XS 6OO2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | AccessibiW

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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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UBS Wealth Management UK periodically ask their customers to evaluate their financial consultants and services. The higher the customer rating, the better the perceived service. The maximum rating is 7. Independent samples of two financial consultants are given below. Consultant A has 10 years' experience, Consultant B has 1 years' experience. Use α = 0.05 and test to see whether the consultant with more experience has the higher population mean service rating. Consultant A Consultant B N1 = 16 N2 = 10 Mean1 = 6.82 Mean2 = 6.25 S1= 0.64 S2 = 0.75 State the null and alternative hypothesis, Compute the value of the test statistic. What is the p-value? What is your conclusion?
A poll found that 59% of a sample of 1020 teens in a certain large country go online several times a day. (Treat this sample as a simple random sample.) Complete parts a through e below. Find the margin of error for this poll if we want 90% confidence in our estimate of the percent of teens in the country who go online several times a day. Explain what the margin of error means. If instead we need to be 95% confident, will the margin of error be larger or smaller? Explain. Find that margin of error.
What do the results mean? What type of error could you be making and what would that mean? Redo as a one-tail test, expecting a decrease in the number of headaches. Pay attention to what changes and what doesn’t and why.
please do the following parts, this is one question! hypotheses, assumptions, and please check the assumptions (listed as a problem in the image)
According to a report on consumer fraud and identity theft, 22% of all complaints for a year were for identity theft. In that year, Wisconsin had 787 complaints of identity theft out of 3498 consumer complaints. Does this data provide enough evidence to show that Wisconsin had a higher proportion of identity theft than 22%? Test at the 1% level. State the hypotheses. Ho: p? v Ha: p|? v Calculate the test statistic. Round to four decimal places. Calculate the standardized test statistic. Round to three decimal places. Z = Find the p-value. Round to four decimal places. p-value = State your decision. O Since the p-value is less than .01, reject Họ. O Since the p-value is greater than .01, reject Ho. O Since the p-value is less than .01, fail to reject Ho. O Since the p-value is greater than .01, fail to reject Ho. Interpret the results. O At the 1% level of significance, there is enough evidence to show that the proportion of complaints due to identity theft in Wisconsin is not equal to…
One urban affairs sociologist claims that the proportion, p, of adult residents of a particular city who have been victimized by a criminal is at least 55%. A random sample of 160 adult residents of this city were questioned, and it was found that 71 of them had been victimized by a criminal. Based on these data, can we reject the sociologist's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. H. :0 H :0 (b) Determine the type of test statistic to use. (Choose one) ▼ O=0 OSO O20 (c) Find the value of the test statistic. (Round to three or more decimal places.) O
When a researcher changes their data analyses to get the p-value they desire, that is known as a. HARKing b. p-hacking c. multiple comparisons d. a Type II error
According to a report on consumer fraud and identity theft, 20% of all complaints for a year were for identity theft. In that year, Rhode Island had 717 complaints of identity theft out of 3498 consumer complaints. Does this data provide enough evidence to show that Rhode Island had a higher proportion of identity theft than 20%? Test at the 5% level. State the hypotheses. Họ: p ? Ha: P Calculate the test statistic. Round to four decimal places. Calculate the standardized test statistic. Round to three decimal places. z = Find the p-value. Round to four decimal places. p-value = State your decision. O Since the p-value is less than .05, reject Ho. O Since the p-value is greater than .05, fail to reject Hg. O Since the p-value is less than .05, fail to reject Hg. O Since the p-value is greater than .05, reject Ho. Interpret the results. O At the 5% level of significance, there is enough evidence to show that the proportion of complaints due to identity theft in Rhode Island is more than…
Several studies showed that after eating a low-fat cereal for two meals a day, subjects had lost some weight.A cereal companyfinanced this research. Identify what is wrong. Choose the correct answer below. A. Since the research is composed of voluntary response samples, there may be key data points missing. B. The data used in the studies is not reliable because it was not measured by the administrator. C. It is questionable that the sponsor is a cereal company because this sponsor can be greatly affected by the conclusion. D. It is not possible to take accurate measurements.
According to the USDA World Agricultural Supply and Demand Estimates report found that Americans consumed an average of 279 eggs per person in 2019. The farmers at Eggcellent Acres believes the number of eggs consumed by Americans has increased over the last few years. Based on the Minitab printout below, test to determine if the average number of eggs consumed by Americans has increased. Test at the 0.10 level of significance. (Show all six steps.) One-Sample T: Eggs Consumed Descriptive Statistics N Mean StRex SE Mean 78 285.00 22.59 2.56 i population mean of Sample Test Null hypothesis Ho: M = 279 Alternative hypothesis Hi: u > 279 T-Value P-Value 2.35 0.011 95% Lower Bound for y 280.74
The average spent nationally on Christmas gifts in 2020 was reported to be $650 per person. Below is a random sample of gift givers from Iowa and what they spent in 2020. Is there sufficient evidence to conclude that Iowans spend less than the national average? Estimate the average amount spent by Iowans on Christmas gifts in 2020 with a 90%-CI. 700 450 400 525 550 550 600 575 650 625 500

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