A manufacturer of light bulbs claims that on the average 2% of the bulbs manufactred by his firm are defective. A random sample of 400 bulbs contained 13 defective bulbs. On the basis of this sample, can you support the manufacture's claim at 5% LOS?
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A: A study, which randomly surveyed 3,700 households and drew on this information from the IRS, found…
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A: Given , n= 113 , x = 58 Here One prop Z test is applicable
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A: Given,sample size(n)=153sample mean(x¯)=10,586standard deviation(σ)=2509α=0.05
Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
A: In hypothesis testing, we start by stating the null hypothesis (H0) and the alternative hypothesis…
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A: GivenMean(x)=63.4standard deviation(s)=15.1sample size(n)=25α=0.025
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A: 1. For the vaccinated group: Number of vaccinated individuals studied = 35 million = 35,000,000.…
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A: Solution-:Probability of murders by a firearm p== 72.2% = 0.722, n=400We want to find,(a) If 400…
Q: The National Academy of Science reported that 36% of research in mathematics is published by US…
A: Given p=36%, n=278, X=85 Level of significance ɑ=0.02
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A: The provided information are: Number of products that have been returned (x)=1 Total number of…
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A: The treatment degrees of freedom is, dftreatment=k-1=3-1=2 The treatment degrees of freedom is 2.…
Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
A: The objective of this question is to determine whether there is sufficient evidence to support the…
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A: Population proportion, p = 0.31Sample size, n = 1199Number of successes, x = 350Significance level,
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A: Given Information : In a study that compared average time taken by a men and women to do a certain…
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A: sample size(n)=16Mean()=45.3standard deviation(s)=7.9significance level()=0.05
Q: The National Academy of Science reported that 31% of research in mathematics is published by US…
A: Given : Hypothesized proportion : p0=0.31 Claim : Proportion is different from 0.31 . Sample size :…
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A: If we want to compare two proportion then the Z test statistic is, Z=p1⏞-p2⏜P(1-P)(1n1+1n2) Here n1…
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A: It is given thatPopulation mean, μ=10941Sample mean, x¯=11425Population standard deviation,…
Q: Are the conditions for inference met
A:
Q: with children under the age of 18 were selected at random, and 501 of those 1163 adults reported…
A: We have given that p = 45% = 0.45 n = 1163 x = 501 p̂ = x/n = 501/1163 = 0.4308 Significance level…
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A: For the year A, sample proportion: sample size: For the year B, sample proportion: sample size:
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A:
Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
A:
Q: The U.S. Energy Information Administration claimed that U.S. residential customers used an average…
A: Given that Sample size n = 184 Sample mean = 11121 Population SD = 1856 Level of significance = 0.10
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A: a). Given, n1=39X1=21n2=41X2=23α=1-0.90=0.10 Now, p1=X1n1=2139=0.5385p2=X2n2=2341=0.5609 As per z…
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A: Step 1: Determine the population proportion (p0).In the given scenario:The population proportion is…
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A: Given,n=1025x=707sample proportion(p^)=xnp^=7071025=0.6898
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A: Ans : Null hypothesis H 0: \ µ = 720Another hypothesis Ha: \µ \≠ 720Sample means (X bar) =…
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Q: Answer 囲 Tables Keypad Keyboard Shortcuts We reject the null hypothesis and conclude that there is…
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A: Given, n = 20000 p = 11000 = 0.001 This is the case of binomial distribution: The probability…
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A: The given values are:Population mean (μ) = 10,476 kWhSample mean (x̄) = 10,767 kWhPopulation…
Q: The National Academy of Science reported that 37% of research in mathematics is published by US…
A: we have to find test statistics..
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A:
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A: Introduction: Denote the proportion of purchases for a particular gift item made by men as p1.…
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A: Formula :
Q: The National Academy of Science reported that 31 % of research in mathematics is published by US…
A: p=0.31 n=188 x=74
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A: From given information, Grand mean=25020=12.5 Number of groups = K = 3.
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- According to the Washington Post, 45% of all Americans are born with brown eyes. Consider a random sample of 80 adults from New York city. We found 32 with brown eyes.(a) Is there suffcient evidence at the 0.01 level of signicance to indicate that the proportion of brown- eyed adults in New York differs from the proportion of Americans who are born with brown eyes? (b) Is there suffcient evidence at the 0.05 level of signicance to indicate that the proportion of brown-eyed adults in New York is less than the proportion of Americans who are born with brown eyes?The National Academy of Science reported that 33% of research in mathematics is published by US authors. The mathematics chairperson of a prestigious university wishes to test the claim that this percentage is no longer 33%. He has no indication of whether the percentage has increased or decreased since that time. He surveys a simple random sample of 280 recent articles published by reputable mathematics research journals and finds that 108 of these articles have US authors. Does this evidence support the mathematics chairperson’s claim that the percentage is no longer 33%? Use a 0.01 level of significance. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: p = 0.33 Ha: p___ 0.33A pharmaceutical company proposed a vaccine that would combat paralytic polio. Two independent random samples were collected from the population. One random sample of 201,229 was given the placebo, and among them 100,910 contracted paralytic polio. The other random sample of 200,745 was given the vaccine, and among them 50,033 people ended up contracting paralytic polio. Let the true proportion of cases of paralytic polio given placebo be p1p1 and the true proportion of cases of paralytic polio given vaccine be p2. Calculate the upper bound of a 95% confidence interval for p1−p2. Round your answer to three decimal places.
- an automobile engineer claims that 1in 10 automobile accidents are due to driver fatigue . if a random sample of 5 accidents observed , then the probabilty that 3 accidentsare due to drive fatigue isThe National Academy of Science reported that 41% of research in mathematics is published by US authors. The mathematics chairperson of a prestigious university wishes to test the claim that this percentage is no longer 41%. He has no indication of whether the percentage has increased or decreased since that time. He surveys a simple random sample of 186 recent articles published by reputable mathematics research journals and finds that 92 of these articles have US authors. Does this evidence support the mathematics chairperson’s claim that the percentage is no longer 41%? Use a 0.05 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. circle the answer below. H0 p=0.41 ha: p⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯0.41 A.<B.≠C.> Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places Step 3 of 3: Draw a conclusion and interpret the decision. A. We reject the null hypothesis and conclude that there is insufficient evidence at a…The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,476 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 153 of their customers and calculates that these customers used an average of 10,767 kWh of electricity last year. Assuming that the population standard deviation is 2478 kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
- The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 115 of their customers and calculates that these customers used an average of 11,425 kWh of electricity last year. Assuming that the population standard deviation is 3217 kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 35 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 35 minutes for the cable technician to arrive. The customer asks a simple random sample of 4 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 39.l minutes with a standard deviation of 2.6 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.10 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.The Substance Abuse and Mental Health Services Administration (SAMHSA, 2017) estimates that 10.9% of the population of the United States age 18–24 had an episode of depression in the previous 12 months. Suppose a researcher takes a random sample of 225 United States adults aged 18–24. Determine the value of the sample proportion, p, such that 5% of samples of 225 United States adults age 18–24 are greater than p. You may find software or a z-table useful. Give your answer precise to three decimal places. p = II
- An educator believes that the proportion of finance majors seeking to earn an MBA degree equals the proportion of management majors seeking to earn an MBA degree in the population. A random sample of 100 finance majors is taken and 24 indicate that they will seek to earn an MBA degree. Likewise, a random sample of 400 management majors is taken and 65 indicate that they will seek to earn an MBA degree. Using this information, test the educator’s claim at the 0.05 levelA large company that produces a "tat-burner pill claims an average Ioss of 20 pounds in the first month. A consumer advocacy group believes that this claim is actually just "hype" intended to sell more of the compound. The advocacy group Would like to obtain statistical evidence about this issue and takes a random sample of 100 consumers who responded that they had purchased the pill but didn't know what the survey was about. They find that these 100 people lost an average of 18 pounds. Let the standard deviation of the population be 7.5 pounds. Clearly state the hypotheses, obtain the test statistic and p-value, and state the decision and conclusion. Show all your work.One cable company claims that it has excellent customer service. In fact, the company advertises that a technician will arrive within 45 minutes after a service call is placed. One frustrated customer believes this is not accurate, claiming that it takes over 45 minutes for the cable technician to arrive. The customer asks a simple random sample of 25 other cable customers how long it has taken for the cable technician to arrive when they have called for one. The sample mean for this group is 47.9 minutes with a standard deviation of 7.6 minutes. Assume that the population distribution is approximately normal. Test the customer's claim at the 0.01 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho : μ = 45 Ha:μ 45