A study of​ 420,095 cell phone users resulted in 135 who developed cancer of the brain or nervous system. When comparing this sample group to another group of people who did not use cell​ phones, it was found that there is a probability of 0.512 of getting such sample results by chance. What do you​ conclude?

Q: In a survey of 1,580 U.S. adults, 226 said that they were very confident in the nutritional…

A:

Q: If a study subject is randomly selected and it is found that the subject entered a guilty plea, what…

A: Let us define some events A : a person is sent to prison. B : a person is not sent to prison. E : a…

Q: A recent survey showed that the probability of a student using an on-line motorcycle taxi is 0.6.…

A: From the provided information, Probability of a student using an on-line motorcycle taxi is 0.6. Let…

Q: A recent study indicated that 67% of U.S. adults consider air conditioning a necessity. If a random…

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Q: In one study, researchers at a clinic looked at more than 120,000 patients enrolled in 14…

A: Among the patients with heart problems, researchers found at least one risk factor in 84.8% of the…

Q: The following table compares the gender vs the level of education of the patients seen in some…

A:

Q: If n=6 and mean= .40 , what is the probability that X>1?

A: Given,n=6 , mean=0.40A random variable X~Binomial(n, p)

Q: Dr. x was interested in studying the hand-washing habits of York KINE students, so they sampled 100…

A: Given that  Sample size n =100 Population mean μ =125 SEM=3

Q: Based on historical data, your manager believes that 26% of the company's orders come from first -…

A:

Q: In a survey of U.S. adults with a sample size of 2094​, 370 said Franklin Roosevelt was the best…

A:

Q: A recent study of cardiovascular risk factors reported that 30% of adults met criteria for…

A: Solution: It is given that a recent study of cardiovascular risk factors reported that 30% of adults…

Q: The Cincinnati Enquirer reported that, in the United States, 60% of adults and 82% of youths ages 12…

A: Let X be a random variable following normal distribution with mean μ and variance σ2, i.e.,…

Q: A Food Marketing Institute found that 28% of households spend more than $125 a week on groceries.…

A: given data  p (spend more than $125) = p = 0.28 n = 235 P(p^<0.31) = ?

Q: Atlantic Airways flights were late (arrived or departed 15 minutes late or more). A random sample of…

A: We have given that  Number of trials n=140 Success of probability p =20.4%=0.204 We use normal…

Q: If a group of 10 women is selected: What is the probability that at least two women will develop…

A: Answer n = 10p = 10% = 10/100 = 0.101- p = 1-0.10 = 0.90

Q: In the Data, there is no difference between men and women on a certain test. However, you found a…

A: Type I error: Reject the null hypothesis even though it is true. Type II error: Do not reject the…

Q: Out of a survey of 400 customers, under the assumption that 0.5 is the correct proportion, what is…

A: From the provided information, Sample size (n) = 400 Proportion (p) = 0.5

Q: Twenty five slips of paper, numbered 1, 2, 3, . .., 25, are placed in a box. If Amy draws six of…

A: Given problem is :

Q: Based on historical data, your manager believes that 40% of the company's orders come from…

A:

Q: A Food Marketing Institute found that 26% of households spend more than $125 a week on groceries.…

A:

Q: A telemarketing firm conducts surveys with a 40% historical response rate. What is the probability…

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Q: What is the probability that in a random sample of 440 voters, less than 48.8% say they will vote…

A:

Q: In a certain town, 62% of the residents enjoy riding roller coasters. Alfred takes a sample of 75…

A: Given Data: Using binomial to normal approximation Mean=n*p=0.62*75=46.5 Standard deviation=4.2036…

Q: Cynthia is president of an international chemical distribution company, Cchem. Weekly gross sales…

A:

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.