Welcome to the Unit 8
Seminar
Prof. Charles Whiffen

Sampling Distributions
• Sampling Distributions : A sampling distribution is a distribution of statistics obtained by selecting all the possible samples of a specific size from a population.
• Distribution of Sample Means : A sampling distribution of the mean gives all the values the mean can take, along with the probability of getting each value if sampling is random from the null-hypothesis population.
• Distribution of Sample Proportions: The distribution
ˆ p all possible samples of a given size.

Sampling Error
•
Sampling Error : The discrepancy between the statistic obtained from the sample and the parameter for the population from which the sample was obtained. For example, the mean ( ) calculated from a sample will not always equal the population mean (

).

• Central Limit Theorem : For any population with mean
 and standard deviation

, the
¯ distribution of sample means for sample size n will have a mean of
 and a standard deviation of

/
 n , and will approach a normal distribution as n approaches infinity ( n >30 is the general rule).
*
See Page 217

Distribution of Sample Means Example
• Consider the following data as a Population
2, 4, 6, 8
• The population mean is 5
• The population standard deviation is 2.236
• Now we are going to take ALL possible samples of n = 2 from this population.
• We will calculate the mean for each sample

Sampling Distribution of Means for
Samples of n = 2
Pick 1
8
8
6
6
8
8
6
6
4
4
2
4
4
2
2
2
Pick 2
2
4
6
8
6
8
2
4
6
8
8
2
4
2
4
6
Mean
5
6
6
7
4
5
5
6
5
3
4
2
3
4
7
8
80
2
0
0
2
18
8
8
2
2
8
0
2
8
18
2
0
Mean 2 Variance Standard
Deviation
36
49
25
36
25
36
16
25
4
9
16
25
9
16
49
64
440
0.000
1.414
2.828
4.243
1.414
0.000
1.414
2.828
2.828
1.414
0.00
1.414
4.243
2.828
1.414
0.00

Central Limit Theorem Applied
•
¯ x
= 80/16 = 5, which equals the population mean. So we have shown that the mean of the means is equal to mu or the population mean.
• S x
= √ 
X 2 – ( 
X) 2 /N / N
= √440 – (80) 2 /16 / 16 (notice we divide by N since this is a population).
= √40/16
= √2.5
= 1.58
• Now, we will calculate what the Central Limit Theorem tells us the standard deviation will be. It is
σ x
= σ/ √n
= 2.236 / √2
= 2.236 / 1.14142
= 1.58

The distribution of sample proportions is the distribution
all possible samples of a given size.
The larger the sample size, the more closely this distribution approximates a normal distribution.
In all cases, the mean of the distribution of sample proportions equals the population proportion.
the best estimate for the population proportion, p .

The margin of error for the 95% confidence interval is where s is the standard deviation of the sample.
N Note: Our text uses 2 but use 1.96 for 95% confidence instead.
We find the 95% confidence interval by adding and subtracting the margin of error from the sample mean. That is, the 95% confidence interval ranges from ( x – margin of error) to (x + margin of error)
We can write this confidence interval more formally as
– E < x or more briefly as ± E

95% Confidence Interval

Constructing a Confidence Interval
• A study finds that the average time spent by eighth-graders watching television is 6.7 hours per week, with a margin of error of 0.4 hour (for 95% confidence). Construct and interpret the 95% confidence interval
• The best estimate of the population mean is the sample mean,
¯ x = 6.7 hours.
• We find the confidence interval by adding and subtracting the margin of error from the sample mean, so the interval extends from 6.7 – 0.4 = 6.3 hours to 6.7 + 0.4 = 7.1 hours.

Interpreting the Confidence Interval
Figure 8.10

Interpreting the Confidence Interval
Figure 8.10 This figure illustrates the idea behind confidence intervals. The central vertical line represents the true population mean, μ. Each of the 20 horizontal lines represents the 95% confidence interval for a particular sample, with the sample mean marked by the dot in the center of the confidence interval. With a 95% confidence interval, we expect that 95% of all samples will give a confidence interval that contains the population mean, as is the case in this figure, for 19 of the 20 confidence intervals do indeed contain the population mean.
We expect that the population mean will not be within the confidence interval in 5% of the cases; here, 1 of the 20 confidence intervals
(the sixth from the top ) does not contain the population mean.

• Solve the margin of error formula [E =1.96s/√n] for n .
n



1 .
96 s
E


2 N Note: Our text uses 2 but use 1.96 for 95% confidence instead.
• You want to study housing costs in the country by sampling recent house sales in various (representative) regions. Your goal is to provide a 95% confidence interval estimate of the housing cost. Previous studies suggest that the population standard deviation is about $7,200. What sample size
(at a minimum) should be used to ensure that the sample mean is within
• a. $500 of the true population mean?
n



1 .
96

E


2


1 .
96

7 , 200
500
2

28 .
2
2 
796 .
6

• Considers the probability that the result of a study could have come about if the experimental procedure had no effect
• If this probability is low, scenario of no effect is rejected and the theory behind the experimental procedure is supported

 State the claim about the population mean
 Determine desired confidence level
 Select a random sample from the population
 Calculate the confidence interval for the desired level of confidence.
 If the claim is contained within the interval, the claim is reasonable; if it is not within the interval, the claim is not reasonable, at the given level of confidence.
 See Testing a Claim document in Doc Sharing

Results from an IQ test administered to 45 randomly selected high school seniors showed a mean IQ of 100 with a standard deviation of 16.
1.
Find the point estimate for the mean IQ for all high school seniors.
2.
Find a 95% CI (confidence interval) for the mean
IQ for all high school seniors.
3.
Find the MOE (margin of error).

N Note: StatCrunch will use 1.96 instead of 2 for
95% confidence level.

(1)100; (2) 95.3 < µ < 104.7; (3) MOE = 4.7

A study of 105 randomly selected restaurants in
Springfield found that 67 have a kids’ menu.
1.
Find the point estimate for the proportion of all restaurants which have a kids’ menu.
2.
Find a 90% CI (confidence interval) for the proportion of all restaurants in Springfield which have a kids’ menu.
3.
Find the MOE (margin of error).

(1)0.638; (2) 0.561 < p < 0.715; (3) MOE = 0.077

Review of Unit 8 Work
By Tuesday at Midnight you must complete:
• Initial post to one discussion question
• Two responses to other student posts to discussion questions
• Live Binder
• MSL HW
• MSL Quiz
30