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Questions and Answers

(1071 solutions)

The p-value helps a researcher determine:

Whether or not the results are statistically significant.This is correct. While the p-value helps you determine whether the results are statistically significant this does not tell you about the size or strength of the effect which is why it’s a good idea to report test statistics, correlation coeff...

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Due to a limited budget, the researcher obtained opinions from a random sample of only 200 U.S. adults. With this sample size, the researcher can be 95% confident that the obtained sample proportion will differ from the true proportion (p) by no more than (answers are rounded):

7.1%Good job! Remember that for a sample size of n, the sample proportion pˆp^ will differ from the “true” population proportion by no more than 1n√1n. In this case, this is about 7.1%....

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These next two questions refer to the following information:

A researcher would like to estimate p, the proportion of U.S. adults who support stricter gun control laws.

Question 6 of 7

Question 6

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If the researcher would like to be 95% sure that the obtained sample proportion would be within 5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?

400...

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How large a sample of facebook users is needed in order to estimate μ with a 95% confidence interval of length  5.86 friends ?

903Good job! We would like our confidence interval to be a 95% confidence interval (implying that z* = 2) and the confidence interval length should be 5.86, therefore the margin of error (m) = 5.86/ 2 = 2.93. The sample size we need in order to obtain this is: 903:n=(2×442.93)2n=2×442.932n=(882.93)2...

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Which of the following will provide a more informative (i.e., narrower) confidence interval than the one in problem 3?

Both using a sample of size 400 (instead of 225) and using a 90% level of confidence (instead of 95%) are correct.Good job! In general, we can obtain a more informative (narrower) confidence interval in one of two ways: compromising on the level of confidence (in other words, choosing a lower level ...

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We are 95% confident that the mean number of friends that facebook users have is:

between 332.13 and 343.87Good job! The 95% confidence interval for the mean, μ, is x¯±2⋅σn√=338±2⋅44225√=38±.2⋅4415≈338±5.87=(332.13, 334.87)x¯±2⋅σn=338±2⋅44225=38±.2⋅4415≈338±5.87=(332.13, 334.87) ....

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The next four questions refer to the following information:

Robin Dunbar, a leading anthropologist at Oxford University who studies social networks, has suggested that there is a magical number for social networks that humans can manage: any grouping larger than 150 people becomes overwhelming. 

With that in mind, a group of researchers at Facebook wanted to study more about how their average user manages their social networks. One study of 225 Facebook users found a mean of 338 friends. Facebook knows from prior studies of their users that the standard deviation is 44.

Question 2 of 7

Question 2

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Based on this information, what would be the point estimate for μ?

338\Good job! The point estimate for the population mean μ is the sample mean, x¯x¯. In this case, to estimate the mean number of facebook friends from among the population of all facebook users, we used the sample mean obtained from the sample, therefore x¯x¯ = 338....

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¯
 is said to be an unbiased estimator for μ in the long run except for in which of the following cases? Select the best answer.

Both Choices A and BGreat job! This is correct. The sample mean and the sample proportion are truly unbiased estimates only when the sample is random and the design is not flawed. If the sample is not random or the design is flawed then the sample mean or proportion could be biased....

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The sampling distribution of a statistic is (select the best answer):

Good job! This is simply the explanation of what the sampling distribution of a statistic is. None of the others are correct....

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Which one of the following is true regarding the standard deviation of the sampling distribution of the sample proportion, p-hat, of "yes" responses?

   The standard deviation of the sampling distribution will be 3 times larger with sample size 100.For a sample size of 100, the sampling distribution of p-hat is: p(1−p)100−−−−−√p(1−p)100 . For a sample of size 900, the sampling distribution of p-hat is: p(1−p)900−−−−−√p(1−...

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