Exercises questions answers

EXERCISES

7.1 Define in your own words the following terms:

a. Central limit theorem

b. Standard error of the mean

c. Student’s t statistic

7.2 Calculate the standard error of the mean for the following sample sizes (μ = 100, σ = 10). Describe how the standard error of the mean changes as n in-creases.

a.     n = 4

b.     n = 9

c.      n = 16

d.     n = 25

e.      n = 36

7.3 The average fasting cholesterol level of an entire community in Michigan is μ = 200 (σ = 20). A sample (n = 25) is selected from this population. Based on the information provided, sketch the sampling distribution of μ.

7.4 The population mean (μ) blood levels of lead of children who live in a city is 11.93 with a standard deviation of 3. For a sample size of 9, what is the probability that a mean blood level will be:

a.     Between 8.93 and 14.93

b.     Below 7.53

c.      Above 16.43

7.5 Repeat Exercise 7.4 with a sample size of 36.

7.6 Based on the findings obtained from Exercises 7.4 and 7.5, what general statement can be made regarding the effect of sample size on the probabilities for the sample means?

7.7 The average height of male physicians employed by a Veterans Affairs med-ical center is 180.18 cm with a standard deviation of 4.75 cm. Find the proba-bility of obtaining a mean height of 184.93 cm or greater for a sample size of:

a.     5

b.     10

c.      20

7.8 A health researcher collected blood samples from a population of female medical students. The following cholesterol measurements were obtained: μ = 211, σ = 44. If we select any student at random, what is the probability that her cholesterol value (X) will be:

a.     P(150 < X < 250)

b.     P(X < 140)

c.      P(X >300)

What do you need to assume in order to solve this problem?

7.9 Using the data from Exercise 7.8, for a sample of 25 female students, calcu-late the standard error of the mean, draw the sampling distribution about μ, and find:

a.     P(200 <  < 220)

b.     P( < 196)

c.      P( > 224)

7.10 The following questions pertain to the central limit theorem:

a.     Describe the three main consequences of the central limit theorem for the relationship between a sampling distribution and a parent population.

b.     What conditions must be met for the central limit theorem to apply?

c.      Why is the central limit theorem so important to statistical inference?

7.11 Here are some questions about sampling distributions in comparison to the parent populations from which samples are selected:

a.     Describe the difference between the distribution of the observed sample values from a population and the distribution of means calculated from samples of size n.

b.     What is the difference between the population standard deviation and the standard error of the mean?

c.      When would you use the standard error of the mean?

d.     When would you use the population standard deviation?

7.12 The following questions relate to comparisons between the standard normal distribution and the t distribution:

a.     What is the difference between the standard normal distribution (used to determine Z scores) and the t distribution?

b.     When are the values for t and Z almost identical?

c.      Assume that a distribution of data is normally distributed. For a sample size n = 7, by using a sample mean, which distribution would you employ (t or Z) to make an inference about a population?

7.13 Based on a sample of six cases, the mean incubation period for a gastrointestinal disease is 26.0 days with a standard deviation of 2.83 days. The pop-ulation standard deviation (σ) is unknown, but μ = 28.0 days. Assume the data are normally distributed and normalize the sample mean. What is the probability that a sample mean would fall below 24 days based on this normalized statistic t where the actual standard deviation is unknown and the sample estimate must be used.

7.14 Assume that we have normally distributed data. From the standard normal table, find the probability area bounded by ±1 standard deviation units about a population mean and by ±1 standard errors about the mean for any distribu-tion of sample means of a fixed size. How do the areas compare?

Answer:

7.2 Since the population distribution is normal, the sample mean also has a nor-mal distribution. Its mean is also 100 but the standard deviation is 10/ √n, where n is the sample size. As n increases, the standard error of the mean decreases at a rate of 1/ √n.

a. In this case, n = 4 and so √n = 2 or the standard deviation is 10/2 = 5.

b. In this case, n = 9 and so √n = 3 or the standard deviation is 10/3 = 3.33.

c. In this case, n = 16 and so √n = 4 or the standard deviation is 10/4 = 2.50.

d. In this case, n = 25 and so √n = 5 or the standard deviation is 10/5 = 2.0.

e. In this case, n = 36 and so √n = 6 or the standard deviation is 10/6 = 1.67.

7.4 The population is normal with mean 11.93 and standard deviation 3 . So the standard error of the mean is 3/ √n. Since n = 9, the standard error of the mean is 3/3 = 1.0.

a. To find the probability that the sample mean is between 8.93 and 14.93, we first normalize it. The sample mean has a mean of 11.93 and a standard deviation of 1. So Z = (W – 11.93)/1 and P(8.93 < W < 14.93) = P(–3 < Z < 3) = 2P(0 < Z < 3) = 2(0.4987) = 0.9974.

b. To find the probability that the sample mean is below 7.53, we normalize first. Z = (W – 11.93) and P(W < 7.53) = P(Z < –4.4) = P(Z > 4.4) = 0.5 – P(0 < Z < 4.4) < 0.5 – 0.04990 = 0.0001.

c. To find the probability that the sample mean is above 16.43, we normalize first. Z = (W – 11.93) and P(W > 16.43) = P(Z > 1.5) = 0.5 – P(0 < Z < 1.5) < 0.5 – 0.4332 = 0.0668.

7.5 We repeat the calculations in 7.4 but with a sample size of 36. . So the stan-dard error of the mean is 3/√n. Since n = 36, the standard error of the mean is 3/6 = 0.5.

a. To find the probability that the sample mean is between 8.93 and 14.93, we first normalize it. The sample mean has a mean of 11.93 and a standard deviation of

1. So Z = (W – 11.93)/0.5 = 2(W – 11.93) and P(8.93 < W < 14.93) = P(–6 < Z < 6) = 2P(0 < Z < 6) > 2(0.4990) = 0.9980.

b. To find the probability that the sample mean is below 7.53, we normalize first. Z = 2(W – 11.93) and P(W < 7.53) = P(Z < –8.8) = P(Z > 8.8) = 0.5 – P(0 < Z < 8.8) < 0.5 – 0.04990 = 0.0001.

c. To find the probability that the sample mean is above 16.43, we normalize first. Z = 2(W – 11.93) and P(W > 16.43) = P(Z > 3.0) = 0.5 – P(0 < Z < 3.0) < 0.5 – 0.4987 = 0.0013.

7.7 X is normal with mean 180.18 cm and standard deviation 4.75 cm. Find the probability that the sample mean is greater than 184.93 cm when

a. The sample size n = 5. The mean for the sampling distribution of the sample average is 180.18 and it has a standard error of 4.75/√5 = 4.75/2.24 = 2.12. P(X > 184.93) = P(Z > 4.75/2.12) = P(Z > 2.24) = 0.5 – P(0 < Z < 2.24) = 0.5 – 0.4875 = 0.0125.

b. The sample size is 10, the mean for the sampling distribution of the sample average is 180.18, and it has a standard error of 4.75/√10 = 4.75/3.16 = 1.50. P(X > 184.93) = P(Z > 1.50) = 0.5 – P(0 < Z < 1.50) = 0.5 – 0.4332 = 0.0668.

c. The sample size is 20, the mean for the sampling distribution of the sample average is 180.18 and it has a standard error of 4.75/√20 = 4.75/4.47 = 1.06. P(X > 184.93) = P(Z > 1.06) = 0.5 – P(0 < Z < 1.06) = 0.5 – 0.3554 = 0.1446.

7.11 a. The observed data have a variance that is the same from one observation to the next; the sample average has a different distribution with a variance that is smaller by a factor of 1/n. It has the same mean, and if the samples do not have a normal distribution the sample mean will by the central limit theorem have a distri-bution that is closer to the normal than the population distribution.

b. The population standard deviation is the square root of the population vari-ance. The standard error of the mean is the standard deviation for the sampling dis-tribution of the sample average. For random samples, it differs from the population standard deviation by a factor of 1/√n.

c. The standard error of the mean is used to create a standard normal or a t sta-tistic for testing a hypothesis about a population mean based on a random sample. It is also used to construct confidence intervals for means when arandom sample is available.

d. The population standard deviation should be used to characterize the popula-tion distribution. It is used when you want to make statements about probabilities associated with individual outcomes such as the probability that a randomly select-ed patient will have a measurement between the values A and B.

7.13 The normalized statistic has Student’s t distribution with 5 degrees of freedom. The normalized statistic t = (X – 28)/(2.83/√6), where X is the sample mean. We ignore the fact that for our particular sample X = 26. We are only interested in the proportion of such estimates that would fall below 24 (our particular one did not since 26 > 24). We take 24 for X since the probability that the sample mean falls below 24 is the same with unknown variance as the probability that t < (24–28)/(2.83/√6) = –4/1.155 = –3.46. We look up t with 5 degrees of freedom and find that P(t < –3.46) = P(t > 3.46) < 1 – 0.99 = 0.01 since P(t > 3.365)1 – P(t < 3.365) = 1 – 0.99 = 0.01 for t with 5 degrees of freedom. We use the one-tailed probability.