Practice concerns for Home work 12 - confidence periods and speculation testing. • Read sections 10. installment payments on your 3 and 10. several of the textual content.

• Solve the practice problems under.

• Open up the Homework Assignment 12 and resolve the problems.

Observe that Quiz 12 covers confidence intervals. Therefore , do challenges 1, a couple of, 3, 4, 5, 6a, 7a ahead of the next questions.

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1 . (10 marks) A sample of size and = 75 produced the sample suggest of Back button population standard deviation σ = several, compute a 95% self-confidence interval intended for the population imply µ.

installment payments on your (10 marks) Assuming the people standard change σ = 3, how large should an example be to estimate the people mean µ with a margin of error not exceeding beyond 0. a few? 3. (10 marks) We all observed 28 successes in 70 self-employed Bernoulli trial offers. Compute a 90% assurance interval pertaining to the population proportion p.

5. (10 marks) The businesses manager of a large production herb would like to estimate the indicate amount of time a worker takes to assemble a fresh electronic element. Assume that the conventional deviation of the assembly period is a few. 6 a few minutes.

a) After observing a hundred and twenty workers putting together similar gadgets, the director noticed that their very own average time was 16. a couple of minutes. Build a 92% confidence period for the mean set up time.

b) How many workers needs to be involved in this study so as to have the imply assembly period estimated up to ±15 just a few seconds with 92% confidence?

five. (10 marks) Suppose a consumer advocacy group would like to execute a review to find the portion p of consumers who bought the newest technology of an Mp3 music player were satisfied with their purchase.

a) How big a sample d should they decide to try estimate s with 2% margin of error and 90% confidence?

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b) The advocacy group took a random test of one thousand consumers who have recently bought this Music player and found that 400 had been happy with their very own purchase. Look for a 95% assurance interval to get p.

six. (10 marks) In order to guarantee efficient usage of a server, it is necessary to estimate the imply number of contingency users. In respect to information, the sample mean and sample regular deviation of number of concurrent users at 100 arbitrarily selected times is thirty seven. 7 and 9. 2, respectively. a) Construct a 90% self-confidence interval to get the suggest number of concurrent users. b) Do these kinds of data present significant data, at 1% significance level, that the suggest number of contingency users is usually greater than thirty-five?

7. (10 marks) To assess the precision of a laboratory scale, a standard weight that may be known to consider 1 gram is consistently weighed 4x. The ensuing measurements (in grams) will be: 0. 96, 1 . 02, 1 . 01, 0. 98. Assume that the weighings by the scale when the true pounds is 1 gram are typically distributed with mean µ.

a) Make use of these info to calculate a 95% confidence interval for µ. b) Do these data give proof at five per cent significance level that the range is not really accurate? Solution this query by doing an appropriate check of speculation. 8. (10 marks) In their advertisements, a fresh diet program want to claim that their program brings about a mean weight

loss (µ) of more than twelve pounds in two weeks. To ascertain if this is a legitimate claim, the manufacturers of the diet

should test the null hypothesis H0: µ = 10 up against the alternative speculation: (A) H1: µ < 10

(B) H1: µ > 12

(C) H1: µ = 10

(D) H1: µ = 0

(E) None of the previously mentioned

9. (10 marks) Assume we would like to estimate the mean amount of money (µ) used on books by simply CS learners in a term. We have thefollowing data from 10 arbitrarily selected CS students: X = $249 and T = $30. Assume that the quantity spenton catalogs by CS students is normally distributed. To compute a 95% self confidence for µ, we will use the following crucial point: (A) z0. 025 = 1 ) 96

(B) z0. 05 = 1 ) 645

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