WEEK 4- DISTRIBUTION OF SAMPLES:

Question 1

Sampling distribution

Answer 1

Sampling distribution is the distribution of a summary statistic.
Medical research often involves acquiring data from a sample of individuals and using the information gathered from the sample to make inferences about a broader group of individuals.

Question 2

Steps for sampling distribution:

Answer 2

Take samples of various sample sizes.
1. Repeated independent samples of the same size from the
same “population”.
2. Sample mean calculated for each sample.
3. Graph the results (sample means) on a Histogram.
This histogram is called the “sampling distribution”.

Question 3

Which of the following statements is INCORRECT regarding the mean of a sampling distribution?

It is the mean of the statistic for all of the samples in the distribution.
It depends on the sample size.
It is the same as the population parameter.

Answer 3

It depends on the sample size.

Question 4

“CENTRAL LIMIT THEOREM”

Answer 4

“For a large sample size, the distribution of the sample mean is normally distributed, even when the population distribution from which the sample has been drawn is decidedly non‐normal, with mean equal to the true mean of the sampled population and standard deviation equal to the standard error of the sample mean.”

Question 5

Implication of central limit theory

Answer 5

Solution: Take one sample.
 Calculate descriptive statistics (e.g. mean, standard deviation).
 Use the sample statistics (e.g. mean) of this sample to make
inferences about the population.
 i.e. We use the sample statistic (e.g. mean) to guess/infer what
the population statistic is.

Question 6

What is the “Standard Error”?

Answer 6

It is a measure of precision of the sample mean from a single sample in estimating the population mean. The smaller the SE, the more precisely the population mean is being estimated.

 Dispersion (spread) of the sampling distribution.
 Measures how precise the population mean is
estimated by the sample mean.
 Technically: Standard Deviation of the distribution of
the sample mean.

Question 7

As sample size increases, SE …..

Answer 7

decreases

Question 8

when z-score and T score is used?

Answer 8

Z‐score
 Used when the Population (true) Standard Deviation is known
T‐score
 Used when the Population (true) Standard Deviati

Question 9

Suppose that the blood cholesterol level of all men aged 20 to 30 is bell shaped with mean 186 mg/dl and an unknown standard deviation.
In a simple random sample of 100 men from this population the sample standard deviation is 41 mg/dl.

What is the value for the spread for the sampling distribution?

Answer 9

4.1

Question 10

T distribution

Answer 10

The t distribution is similar to the normal distribution and is appropriate for continuous data; both are symmetric and bell shaped.
Tails are a bit longer for the t distribution compared to the normal distribution.
The shape of the t distribution depends on the sample size. For large samples, the t distribution is more like the normal distribution
As the sample size (n) increases the t‐distribution approaches the normal distribution..

Question 11

Degrees of freedom (df)

Answer 11

The t-distribution is associated with calculation of degrees of freedom (df).
It denotes the number of independent pieces of information available to estimate another piece of information.
df = Number of pieces of information that was used to estimate
another

Question 12

For a large sample size, the distribution of the sample mean is …… distributed, even when the population distribution from which the sample has been drawn is decidedly non-normal, with ……. equal to the true mean of the sampled population and standard deviation equal to the standard ……. of the sample mean.

Answer 12

normally, mean, error

Question 13

According to the normal distribution probability law:

Answer 13

Approximately 68% of the sample means are expected to lie within one standard error of the true mean, that is, within +1SE and -1SE.
Approximately 95% of the sample means are expected to lie within 1.96 or approximately two standard errors of the true mean, that is, within -2SE and +2SE.
Approximately 99.7% of sample means are expected to lie within 2.97 or approximately three standard errors of the true mean, that is, within -3SE and +3SE.

Question 14

Which of the following statements is INCORRECT regarding the t-distribution?

It is appropriate for continuous data
A calculation of degrees of freedom is associated with it
When the sample size is small, it is more like the normal distribution
In general, it is less “peaked” than the normal distribution

Answer 14

When the sample size is small, it is more like the normal distribution

Question 15

Area under the sampling distribution: True SD Unknown

Answer 15

When population SD is unknown the sampling distribution follows a t‐distribution, with df = n – 1, where n = sample size.
As the sample size increases the distribution of t-score approaches the standard normal distribution.
1- Calculate SE = SD sample /square root of n
2- Calculate df= n-1
3- calculate the t- score = sample mean -true mean/ SE

Question 16

1. Question
   Suppose that the blood cholesterol level of all men aged 20 to 30 is bell shaped with mean 186 mg/dl and an unknown standard deviation.
   In a simple random sample of 100 men from this population the sample standard deviation is 41 mg/dl.

What is the probability that the sample mean takes a value between 183 and 189 mg/dl?

Answer 16

Between 50% and 80%

Question 17

Area under the sampling distribution: True SD known

Answer 17

Then irrespective of the sample size the area (other than reference ranges) under the sampling distribution for the sample mean can be calculated using the normal probability table.
This can be done by transforming the sample mean to a Z-score, where the Z-score is calculated by subtracting the true mean from the sample mean and dividing this difference by the standard error of the sample mean.

Question 18

1. Question
   Let us consider that pre‐op creatinine level is heavily right skewed in the population.

If you draw a large sample from this population what should be the shape of this variable (Y) in the sample?
 Right skewed
 Negatively sewed
 Can not be determined
 Normally distributed

Answer 18

The shape of pre‐op creatinine level in the sample would be right skewed. This would occur because a sample is representative of the population, thus it will have the same distribution as the population.

Question 19

2. Question
   Let us consider that pre‐op creatinine level is heavily right skewed in the population.

For large sample what will be the shape of the sampling distribution for the sample mean?

Answer 19

Normally distributed

Question 20

The age group to which Anne belongs has mean height 1.6 metre and standard deviation 0.1 metre. The age group to which Devi belongs has mean height 1.2 metre and standard deviation 0.08 metre. Anne is 1.7 metre tall. Devi is 1.36 metre tall. Which is the taller for their age?

Answer 20

Devi

Question 21

Juan makes a measurement in a chemistry laboratory and records the result in his lab report. The standard deviation of the students’ lab measurement is 10 milligrams. Juan repeats the measurement 4 times and records the mean of his 4 measurements.

What is the standard deviation of Juan’s mean result?
 25
 5
 10
 15

Answer 21

5
Standard deviation of Juan’s mean/average result is the standard error of the sample mean. The formula is: SE = SD/√n = 10/√4 = 5.

Question 22

5. Question
   Juan makes a measurement in a chemistry laboratory and records the result in his lab report. The variance of the students’ lab measurements is 100 milligrams. Juan repeats the measurements 4 times and records the sample mean. How many times must Juan repeat the measurement to reduce the standard deviation of the sample mean to 2?

8
16
25
32

Answer 22

25

SD/√n = 10/√25 = 10/2 = 2, Juan needs to repeat the measurement 25 times.

Question 23

Consider that the weight of a tumor in bladder cancer patients in the population follows the normal distribution with a mean 50g and standard deviation 5g.

If a bladder cancer patient is selected randomly what is the probability that the tumor is less than 45g?
 15.87%
 16%
 34.13%
 50%

Answer 23

15.87%

Question 24

Consider that the weight of a tumor in bladder cancer patients in the population follows the normal distribution with a mean 50g and standard deviation 5g.

If 4 of these patients are selected at random, calculate the probability that the average weight of the 4 tumors (assume each patient has only one tumor) will be greater than 55g?

0. 5
1. 4772
2. 0228
3. 345

Answer 24

we use Z-score because SD is known but this time we will use sampling distribution for sample mean to calculate the Z-score. The formula is: Z-score = (Sample mean – True mean)/SE, where SE = SD/√n = 5/√4=5/2=2.5. Z-score = (55-50)/2.5 = 2.0. The area above mean of 55 is the same as the area above Z-score of 2.0. The area b/w 0 and 2.0 is 0.4772. Hence the tail area is: 0.5 – 0.4772 = 0.0228.

Question 25

Consider the weight of adult Australians in a large sample follows a normal distribution with mean 78kg and standard deviation 10kg.
What is the median weight for the adult Australians?

78kg
Can not be determined
Mean and median have different values
None of the above

Answer 25

78 kg

Question 26

Consider the weight of adult Australians in a large sample follows a normal distribution with mean 78kg and standard deviation 10kg.
Find the probability that a randomly selected adult Australian will have weight between 58 and 98 kg.

68%
95%
99.7%
50%

Answer 26

95%

Question 27

Consider the weight of adult Australians in a large sample follows a normal distribution with mean 78kg and standard deviation 10kg.
Find the limits that include 99.7% of adult Australians.

50 to 100 kg
48 to 108 kg
25 to 50 kg
2 to 10 kg

Answer 27

48 to 108 kg