2) In the formula for the Confidence Interval for the Mean, if the Confidence Coefficient, z(α/2) = 1.65, what is the Confidence Level?

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1 Pg.431 1)The mean of the sampling distribution of means is equal to the mean of the population. T-F, and why or why not? True. If you were to take every possible sample from the population, and calculate each sample mean, the mean of those means would be equal to the mean of the entire population taken as a whole. The reason for this is that every element of the population would have been selected an equal number of times in all of the samples. The mean of the sample means is equivalent to summing the value of every selection made in all of the samples, and then dividing by the total number of selections. This would be the equivalent of taking the mean of the population directly. ) In the formula for the Confidence Interval for the Mean, if the Confidence Coefficient, z(α/) 1.65, what is the Confidence Level? 90% This is something that is probably shown in a chart in your textbook. If not, then you can determine the confidence level in this way: If z(α/) 1.65, then the tail area under the normal curve to the right of z 1.65 is 0.05 (from a table of the normal distribution). The total of the two tail areas (left and right) is then 0.05 * 0.10, and the confidence level is , or 90%. Pg Consider a population with µ 67.1 and σ5.76. (A) Calculate the z-score for x 65.7 from a sample of size 40. z x µ σ / n / (B) Could this z-score be used in calculating probabilities using Table 3 in Appendix B of the text? Why or why not? This is difficult to answer since I don t know what Table 3 in Appendix B looks like. If Table 3 is a table of the normal distribution, then yes, this z-score can be use to calculate the probability of obtaining a sample mean that is greater or less than 65.7.

2 .Given a level of confidence of 95% and a population standard deviation of 8, answer the following: (A) What other information is necessary to find the sample size (n)? The formula for calculating sample size is n z iσ α / E The information that is still needed is the Maximum Error of Estimate, E. (B) Find the Maximum Error of Estimate (E) if n 59. Rearranging the equation from part A gives: E z α / iσ n Substituting the given values gives: E 1.96( 8) A sample of 136 golfers showed that their average score on a particular golf course was 8.77 with a standard deviation of Answer each of the following (show all work and state the final answer to at least two decimal places.): (A) Find the 99% confidence interval of the mean score for all 136 golfers. For a 99% confidence interval, use z α/.576. The 99% confidence interval is then: x z α / s n < µ < x + z α / s n < µ < (.576) < µ < The 99% confidence interval is (81.77, 83.77)

3 (B) Find the 99% confidence interval of the mean score for all golfers if this is a sample of 90 golfers instead of a sample of 136. The 99% confidence interval is then: x z α / s n < µ < x + z α / s n < µ < < µ < The 99% confidence interval is (81.54, 83.00) (C) Which confidence interval is smaller and why? The confidence interval for the larger sample is smaller. The reason for this is that with a larger sample, we can be more certain that the sample mean accurately reflects the population mean. As a result, the interval needed to contain a given percentage of the possible sample means is smaller. 4. Assume that the population of heights of female college students is approximately normally distributed with mean µ of 67.6 inches and standard deviation σ of 5.96 inches. A random sample of 78 heights is obtained. Show all work. (A) Find the mean and standard error of the x distribution µ x x 67.6 σ x s n (B) Find P( x > 67.50) z x µ s / n / P( x > 67.50) P( z > )

4 5. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.8 inches and a standard deviation of 0.61 inches. Show all work. (A) What percentage of the grapefruits in this orchard is larger than 5.78 inches? z x µ σ P( x > 5.78) P( z > ) 0.56 Expressed as a percentage, the answer is 5.6% (B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample is greater than 5.78 inches? z x µ σ / n / P( x > 5.78) P( z > ) The probability is A researcher is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 99% confident that her estimate is correct. If the standard deviation is 4.11, how large a sample is needed to get the desired information to be accurate within 0.56 decibels? n z iσ α / E.576 i This should be rounded up to the next whole number. The required sample size is 358.

5 Pg Consider a normal population with µ 5 and σ 7.0. (A) Calculate the standard score for a value x of 4. z x µ σ (B) Calculate the standard score for a randomly selected sample of 45 with 4. z x µ σ / n / (C) Explain why the standard scores of 4 are different between A and B above. In part A the z-score is calculated for a single observation. In this instance the difference between the observation and the sample mean is divided by the standard deviation of the population (or of the sample, depending on which is known). In part B, the z-score is calculated for the sample mean. In this instance, as a result of the Central Limit Theorem, the difference between the sample mean and the hypothesized population mean is divided by the standard deviation of the sample means... Assume that the mean SAT score in Mathematics for 11th graders across the nation is 500, and that the standard deviation is 100 points. Find the probability that the mean SAT score for a randomly selected group of th graders is between 513 and 487. z / z /

6 P( 487 < x < 513) P( 1.59 < z < 1.59) P( z < 1.59) P( z < 1.59) Assume that a sample is drawn and z(α/) 1.96 and σ 30. Answer the following questions: (A)If the Maximum Error of Estimate is 0.0 for this sample, what would be the sample size? 1.96 i 30 n 0.0 8,643,600 (B)Given that the sample Size is 400 with this same z(α/) and σ, what would be the Maximum Error of Estimate? E z α / σn ( 1.96) (C)What happens to the Maximum Error of Estimate as the sample size gets smaller? The Maximum Error of Estimate gets larger as the sample size gets smaller. (D)What effect does the answer to C above have to the size of the confidence interval? The effect of the Maximum Error of Estimate getting larger is that the size of the confidence interval also gets larger.

7 4. By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 3.14 seconds. Answer each of the following (show all work): (A) How many measurements should be made in order to be 98% certain that the maximum error of estimation will not exceed 1.0 seconds? n n.36 * This should be rounded up to the next whole number. The required sample size is 54. (B) What sample size is required for a maximum error of 1.5 seconds? n.36 * This should be rounded up to the next whole number. The required sample size is A 99% confidence interval estimate for a population mean was computed to be (44.1, 6.9). Determine the mean of the sample, which was used to determine the interval estimate The mean of the sample is the midpoint of the confidence interval. Mean

8 6. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 190 was taken, and the mean amount spent was $ Assuming a standard deviation equal to $35.51, find the 98% confidence interval for µ, the mean for all such families For a 98% confidence interval, use z α/.36 The 98% confidence interval is: < µ < < µ < The 98% confidence interval is (8.0, 40.00) 7. A confidence interval estimate for the population mean is given to be (49.9, 58.07). If the standard deviation is and the sample size is 53, answer each of the following (show all work): (A) Determine the maximum error of the estimate, E. The maximum error of the estimate is equal to half the width of the confidence interval: E (B) Determine the confidence level used for the given confidence interval. z α / E n σ A value for z α/ 1.96 corresponds to a 95% confidence interval.