estimation

Question 1

For a normal distribution, 68% of the population fall within 1sd of the mean, and 95% of the population fall within 2sd of the mean.

However, for a sample distribution, 68% fall within ________ of the mean, and 95% fall within __________ of the mean?

Answer 1

1 SE or sd/sqrt(n)

2 SE or 2sd/sqrt(n)

Question 2

If a sampling distribution has a standard error of 2.71 and a point estimate of 40, what is the estimated 95% confidence interval of the distribution?

Answer 2

xbar - 2sd/sqrt(n) < Mbt < xbar + 2sd/sqrt(n)
40 - 2(2.71) < Mbt < 40 + 2(2.71)
40 - 5.42 < Mbt < 40 + 5.42
34.58 < Mbt < 45.42

(34.58, 45.42) is known as the 95% confidence interval

Question 3

Using the z-table, what are the z-score values that bound 95% of the data (also known as the critical values)?

Answer 3

• 1.96, 1.96

Question 4

Using the exact z-scores, If a sampling distribution has a standard error of 2.71 and a point estimate of 40, what is the exact 95% confidence interval of the distribution?

Answer 4

40 - 1.96(2.71) < Mbt < 40 + 1.96(2.71)

34.69, 45.31

Question 5

What is the standard error of the mean that we would use to compare the sample mean .13 with sd = .107, compared with the means of other samples of the same size?

Answer 5

.13/sqrt(20) = .107/4.4721 = .024

Question 6

If a sampling distribution has a standard error of .024 and a point estimate of .13, what is the estimated 95% confidence interval of the distribution?

Answer 6

xbar - 2sd/sqrt(n) < Mbt < xbar + 2sd/sqrt(n)
.13 - 2(.024) < Mbt < .13 + 2(.024)
.083 < Mbt < .177

Question 7

What is the main problem with point estimates of population parameters?

Answer 7

They do not account for sampling error

Question 8

As the sample size increases, the range of the confidence interval __________ ?

Answer 8

decreases

Question 9

As the population standard deviation increases, the range of the confidence interval ___________ ?

Answer 9

increases

Question 10

A population is normally distributed with standard deviation 2.8. Compute the 95% confidence interval for the mean, based on the following random sample (n=6):

• 8, 9 ,12, 13, 14, 16

(Hint: first find the sample mean)

Answer 10

• the sample mean is (8 + 9 + 12 + 13 + 14 + 16) / 6 =

Question 11

A chemistry teacher wants to improve exam scores by incorporating more interactive in-class activities. The mean exam score for all her previous classes is 68% with standard deviation 10%. After trying out the interactive in-class activities for her current class, this class got a score of 75%. There were 25 students in the class. She decides to calculate a 95% confidence interval for what the average exam score would be if she continued this method for all classes. What is the standard error of the mean?

Answer 11

10/sgrt(25) = 10/5 = 2

Question 12

A chemistry teacher wants to improve exam scores by incorporating more interactive in-class activities. The mean exam score for all her previous classes is 68% with standard deviation 10%. After trying out the interactive in-class activities for her current class, this class got a score of 75%. There were 25 students in the class. She decides to calculate a 95% confidence interval for what the average exam score would be if she continued this method for all classes. What is the probability of obtaining this mean (75%) or greater?

Answer 12

• calculate the z-score
• 75 - 68/2 = 3.5
• consulting the z-chart, 3.5 = not on the table
• So, the answer is 0 since the value is not on the table

Question 13

A chemistry teacher wants to improve exam scores by incorporating more interactive in-class activities. The mean exam score for all her previous classes is 68% with standard deviation 10%. After trying out the interactive in-class activities for her current class, this class got a score of 75%. There were 25 students in the class. She decides to calculate a 95% confidence interval for what the average exam score would be if she continued this method for all classes. What is the margin of error?

Answer 13

• multiply the critical value by the SE
• Using the z-table, the z-score for 0.9750 (everything but the .025 of the upper or lower tail) is 1.96?
• 1.96 * 2 = 3.92

Question 14

A chemistry teacher wants to improve exam scores by incorporating more interactive in-class activities. The mean exam score for all her previous classes is 68% with standard deviation 10%. After trying out the interactive in-class activities for her current class, this class got a score of 75%. There were 25 students in the class. She decides to calculate a 95% confidence interval for what the average exam score would be if she continued this method for all classes. What is the confidence interval?

Answer 14

xbar - margin of error < Mbt < xbar + margin of error
75 - 3.92 < Mbt < 75 + 3.92
71.08 < Mbt < 78.92

Question 15

A medical doctor wants to reduce blood pressure in his patients by using meditation. He finds that the mean systolic blood pressure for the population of is 180 with sd of 18. With new method he obtains a new mean of 175. What is the probability of obtaining this 175 mean or lower if the SE is 6?

Answer 15

175-180 / 6 = -5/6 = -.83

consulting the z-table, this equals .2033

Question 16

A medical doctor wants to reduce blood pressure in his patients by using meditation. He finds that the mean systolic blood pressure for the population of is 180 with sd of 18. With new method he obtains a new mean of 175. The SE is 6. The critical value for 99% is 2.576. What is the margin of error?

Answer 16

• multiply the critical value by the SE

- 6 * 2.576 = 15.456

Question 17

A medical doctor wants to reduce blood pressure in his patients by using meditation. He finds that the mean systolic blood pressure for the population of is 180 with sd of 18. With new method he obtains a new mean of 175. The SE is 6. The critical value for 99% is 2.576. What is the confidence interval?

Answer 17

xbar - margin of error < Mbt < xbar + margin of error
175 - 15.456 < Mbt < 175 + 15.456
159.54 < Mbt < 187