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Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Week 5

Practice problems on confidence intervals and statistical inference techniques.

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    Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Hamilton Week 5 Homework 4. Why is a 99% confidence interval wider than a 95% confidence interval? Solution) The definition of a confidence interval is that it contains the true population mean. If I have a 95% confidence interval, that means I am 95% certain that the true population mean is in the interval. If I want to be even more certain, I have to widen the interval. If I can be less certain, I can narrow the interval. So the widest interval will be 99%, and the narrowest would be 90%. 12. A person claims to be able to predict the outcome of flipping a coin. This person is correct 16/25 times. Compute the 95% confidence interval on the proportion of times this person can predict coin flips correctly. What conclusion can you draw about this test of his ability to predict the future? Solution) WE HAVE GIVEN THAT n = 25 and p = 16/25 And we need to construct the 95% C.I. for the proportion of times this person can predict coins flips correctly as, ṕ ± 1.96 * √ ( ṕ q^/n) =.64± 1.96 * √ (.64*.36/25) = .64 ± .1882 So the 95% C.I. is, ( 0.4518, 0.8282) So We Are 95 Out Of 100 Attempts are confident that the values of the samples are lies b/w (.4518,.8282) 15. You take a sample of 22 from a population of test scores, and the mean of your sample is 60. (a) You know the standard deviation of the population is 10. What is the 99% confidence interval on the population mean? Solution)

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    We have given that n = 22, sample mean =60 and σ = 10 The 99 % C.I. for the population mean is, = sample mean ± 2.58 * σ / √ n = 60 ± 2.58 * 10 / √ 22 = 60 ± 5.501 So, (54.499, 65.501 ) (b) Now assume that you do not know the population standard deviation, but the standard deviation in your sample is 10. What is the 99% confidence interval on the mean now? Solution) Here we have given that n = 22, sample mean = 60 and S = 10 Here we assume that the population stranded deviation is unknown, and also n<30 so we here use the student t distribution as, = sample mean ± t α /2,v s / √ n - 1 Where v= n - 1 degree of freedom and v= 22 - 1 = 21 = 60± 2.831 * 10/ √ 21 = 60±6.178 So, (53.822, 66.178 ) 18. You were interested in how long the average psychology major at your college studies per night, so you asked 10 psychology majors to tell you the amount they study. They told you the following times: 2, 1.5, 3, 2, 3.5, 1, 0.5, 3, 2, 4. (a) Find the 95% confidence interval on the population mean. Sol) Here we have given that sample mean = 2.25 n= 10 and σ = 1.0548 We use a standard normal distribution and assume that population standard deviation is known so The 95 % C.I. for the population mean is, = sample mean ± 1.96 * σ / √ n = 2.25 ± 1.96* 1.05 48 / √ 10 = 2.25 ± .6538 = (1.596, 2.904 ) (b) Find the 90% confidence interval on the population mean. THE 90% C.I. IS. = sample mean ± 1.645 * σ / √ n = 2.25 ± 1.645 * 1.0548/ √ 10 = 2.25 ± .549 So, (1.701, 2.799) 100. What is meant by the term “90% confident” when constructing a confidence interval for a mean?

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