06:23A myopenmath.comYou are conducting a study to see if the proportion ofvoters who prefer the Democratic candidate issignificantly smaller than 73% at a level of significanceof a = 0.01. According to your sample, 40 out of 59potential voters prefer the Democratic candidate.a. For this study, we should useSelect an answerb. The null and alternative hypotheses would be:Ho: ? OSelect an answer >(pleaseenter a decimal)H1: ? 0Select an answer ©(Pleaseenter a decimal)c. The test statistic ? O(pleaseshow your answer to 3 decimal places.)d. The p-value(Please show your%3Danswer to 4 decimal places.)e. The p-value is ? 0f. Based on this, we should Select an answer thenull hypothesis.g. Thus, the final conclusion is that ...O The data suggest the populationproportion is not significantly smaller than73% at a = 0.01, so there is sufficientevidence to conclude that the proportionof voters who prefer the Democraticcandidate is equal to 73%.The data suggest the populatonproportion is significantly smaller than73% at a = 0.01, so there is sufficientevidence to conclude that the proportionof voters who prefer the Democraticcandidate is smaller than 73%The data suggest the populationproportion is not significantly smaller than73% at a = 0.01, so there is not sufficientevidence to conclude that the proportion%3Dof voters who prefer the Democraticcandidate is smaller than 73%.h. Interpret the p-value in the context of the study.O If the sample proportion of voters whoprefer the Democratic candidate is 67%and if another 59 voters are surveyed thenthere would be a 18.4% chance ofconcluding that fewer than 73% of allvoters surveyed prefer the Democraticcandidate.There is a 73% chance of a Type I error. prefer the Democratic candidate is 67%and if another 59 voters are surveyed thenthere would be a 18.4% chance ofconcluding that fewer than 73% of allvoters surveyed prefer the Democraticcandidate.There is a 73% chance of a Type I error.O If the population proportion of voters whoprefer the Democratic candidate is 73%and if another 59 voters are surveyed thenthere would be a 18.4% chance fewer than67% of the 59 voters surveyed prefer theDemocratic candidate.There is a 18.4% chance that fewer than73% of all voters prefer the Democraticcandidate.i. Interpret the level of significance in the contextof the study.There is a 1% chance that the proportionof voters who prefer the Democraticcandidate is smaller than 73%.If the population proportion of voters whoprefer the Democratic candidate is 73%and if another 59 voters are surveyed thenthere would be a 1% chance that we wouldend up falsely concluding that theproportion of voters who prefer theDemocratic candidate is smaller than 73%O If the proportion of voters who prefer theDemocratic candidate is smaller than 73%and if another 59 voters are surveyed thenthere would be a 1% chance that we wouldend up falsely concluding that theproportion of voters who prefer theDemocratic candidate is equal to 73%.There is a 1% chance that the earth is flatand we never actually sent a man to themoon.Hint:HelpHelpful Videos: CalculationsInterpretations [+][+] Setup [+]Question Help: O Message instructorSubmit Question

Since you have posted a question with multiple sub-parts, we will solve first three sub- parts for…

Answered: You are conducting a study to see if…

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 worldwide market share for a web browser was 20.1% in a recent month. Suppose that a sample of 200 random students at a certain university finds that 50 use the browser. At the 0.01 level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.1%? Determine the null and alternative hypotheses. A.H0: P≥0.201; H1: P<0.201 B.H0: P=0.201; H1: P≠0.201 C. H0: P≠0.201; H1: P=0.201 D. H0: P≤0.201; H1: P>0.201 ZStat=__ (Round to two decimal places) p-value=____ __A__ the null hypothesis. There is __B__ evidence to conclude that the market share at teh university is __C__ the worldwide market share of 20.1%. A: Reject or do not reject B: insufficient or sufficient C: at least, less than, equal to, at most, not equal to, greater than
In a random sample, 12 of 14 industrial accidents weredue to unsafe working conditions. Use the 0.01 level ofsignificance to test the null hypothesis θ = 0.40 againstthe alternative hypothesis θ Z 0.40.
A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test H0: p = 0.01 versus Ha: p > 0.01, where p is the actual proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than 1% defective circuits.) In this context, define Type I and Type II errors. (Select all that apply.) a. A Type I error is obtaining convincing evidence that more than 1% of a shipment is defective when in fact (at most) 1% of the shipment is defective. b. A Type I error is not…
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Method A: M1= 54, SS1= 400, n1= 15 Method B: M2=40, SS2= 440, n2=15 Using alpha=0.05 should the null hypothesis by rejected for the above data?
Find the Type II error given that the null hypothesis, H0, is: there are no more than 15% of structures in the county were built without permits.And, the alternative hypothesis, Ha, is: a building inspector claims that more than 15% of structures in the county were built without permits. Select the correct answer below: The building inspector concludes that more than 15% of the structures in the county were built without permits when, in fact, no more than 15% of the structures were built without permits. The building inspector concludes that more than 15% of the structures in the county were built without permits when, in fact, more than 15% of the structures were built without permits. The building inspector cannot conclude that more than 15% of the structures in the county were built without permits when, in fact, no more than 15% of the structures were built without permits. The building inspector cannot conclude that more than 15% of the structures in the county were…
In a local community past data shows that 17% of the residents own a skateboard, and the city is considering building more green lanes to accommodate these skate boarders as a recent survey showed that 63 out of 270 people supported the concept of more green lanes on the city's streets. a) State the null and alternative hypothesis. b) Find a Z and P-value, and state your conclusion using alpha = 5% c) Re-do part (b) using the Chi-square test, also stating your conclusion using alpha = 5%
A nationwide study of undergraduate students reported that the mean number of drinks consumed per week during the spring semester is 7.96. The mean number of drinks consumed per week at USC is 7.64 (s.d.=2.55, N=412 Health services is concerned that USC students are consuming significantly more alcohol per week than the national average. Using an alpha level of .05, Is there sufficient evidence to be concerned? Be sure to select the correct critical value for the alternative hypothesis, and then use this evidence to make your conclusion
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12) The personnel director for a small firm thinks that employees may be more likely to use sick days on Monday and Friday resulting in three-day weeks. a) What is the appropriate null and alternate hypothesis to determine if sick day absences depend on day of the week? b) The personnel director obtains sick day data for the past year and summarizes the number of absences by day of the week in the table below. Use JMP software to test the hypothesis of part a) c) Using alpha = 0.05, draw a conclusion for the hypothesis test and state the conclusion in the context of the problem. Day of Week Absences Monday 126 Tuesday 74 Wednesday 91 Thursday 82 Friday 115
About half of the police officers in Baltimore, Maryland, have completed a special course in community policing. As Captain of your division, you want to know if the course increased calls from the minority community for assistance from your precinct. Below is the results of before and after training in proportions for calls from the minority community in your precinct. Be sure and state whether you accept or reject the null hypothesis. Alpha is 0.05 for setting your Z critical value. After Training Before training Ps1 = 0.47 Ps2 = 0.43 N1 = 157 N2 = 113
A law enforcement agent believes that at least 84% of the drivers stopped for speeding on Saturday nights are under the influence of alcohol. A sample of 84 drivers who were stopped for speeding on a Saturday night was taken. Seventy-five percent of the drivers in the sample were under the influence of alcohol. (a) State the null and alternative hypotheses. (Enter != for ≠ as needed.) H0: Ha: (b) Compute the test statistic. (c) Using the p-value approach, test the hypotheses at the 0.05 level of significance. Find the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. Do not reject H0. There is insufficient evidence to conclude that less than 84% of the drivers stopped for speeding on Saturday nights are under the influence of alcohol.Do not reject H0. There is sufficient evidence to conclude that less than 84% of the drivers stopped for speeding on Saturday nights are under the influence of alcohol. Reject H0. There…

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