estimating the value of a parameter using confidence intervals, Study notes of Statistics

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CHAPTER 9: ESTIMATING

THE VALUE OF A

PARAMETER USING

CONFIDENCE INTERVALS

1 By: Wandi Ding

DEVIATION KNOWN

1. Point estimate: is the value of a statistic that estimates the value of a parameter. For example, the sample mean is a point estimate of the population mean,. Suppose: we want to estimate the average weight for all students in MTSU for this semester, we could take a random sample of 100 students and find the average weight of these students, say, 130 pounds, this kind of estimate is called a point estimate. Often, there is another question to be asked, it is, how good is a point estimate? There is no way of knowing how close a particular point estimate is to the population mean if the population is large. For this reason, statisticians prefer another type of estimate, called interval estimate.
2. Interval estimate: an interval for unknown parameter is an interval or a range of values used to estimate the parameter with the specific confidence level of estimate. It is also called confidence interval. (^) 2

x

DEVIATION KNOWN

Ex: for 95% confidence interval for a population mean:

α=1-95%=5%; α/2=0.05/2=0.025, (Based on the standard normal table)

4

0. 025 1.^96

2

z   z 

α/2 (^) α/

-1.96 0 1.

DEVIATION KNOWN

Confidence interval estimates for the population mean can be written in this form, too. point estimate margin of error

=E is margin of error, also called maximum error of estimate. There are three factors which affect the margin of error.

1. level of confidence. Note: α=1-level of confidence.
2. sample size, n.
3. standard deviation of the population.

Note: The value of is called critical value of the distribution

and the next slide shows common critical values used lot in confidence intervals. (^) 5

 ( ) (^2) n

z ^ 

 2 z

DEVIATION KNOWN

Comments for confidence interval:

Lower bound:

Upper bound:

7

( ) ( ) (^2 2) n

x z n

x  z       ^ 

( ) (^2) n

x  z ^ 

( ) (^2) n

x  z ^ 

DEVIATION KNOWN

Ex: a survey of 30 adults found that the mean age of a person’s primary vehicle is 5.6 years. Assuming the standard deviation of the population is 0.8 year, find the 99% confidence interval of the population mean and interpret it.

Ans:

based on the 30 adults sample; n=30; is the standard deviation of population. confidence level is 99% and α=1-99%=1%,

8

( ) ( ) (^2 2) n

x z n

x  z       ^ 

x  5. 6

1. 005 2.^575 2

2. 01 2

z   z  z 

DEVIATION KNOWN

5. Calculating the necessary sample size:

(^22) 10

2

2

2

)

(

E

z n

E

z n

E n z

n

E z

a

a

a

a

















DEVIATION KNOWN

Ex: the college president asks the statistics teacher to estimate the average age of the students at their college. How large a sample is necessary? The statistics teacher would like to be 99% confident that the estimate should be accurate within 1 year. (the standard deviation of the age is known to be 3 years.)

Ans:

11

α=1-99%=1%=0.01, α/2=0.01/2=0.005, Za/2 =2.575, E=

59. 67 so,round up to 60

) 1

2. 575 * 3 (

)

( 2

2 2





 E

z n

a^ 

The sample size at least is 60 students.

DEVIATION UNKNOWN

Student’s t-distribution:

Suppose that a simple random sample of size n is taken from a population. If the population from which the sample is drawn follows a normal distribution. The distribution of

follows a student’s t-distribution with n-

Degree of freedom, where is the sample mean and s is the sample standard deviation.

Note: the t-statistic represents the number of sample standard errors is from the population mean,. 13

n

s t  x ^ 

x

x 

DEVIATION UNKNOWN

Properties of the t-distribution:

1. The t-distribution is different for different degree of freedom.

2. The t-distribution is centered at 0 and is symmetric about 0.

3. The area under the curve is 1. The area under the curve to the right of 0 equals to the area under the curve to the left of 0, which equals ½=0.5.

4. As t increases without bound, the graph approaches, but never equals, zero. As t decreases without bound, the graph approaches, but never equals, zero.

5. The area is the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution, because we are using s as an estimate of , thereby introducing further variability into the t-statistic. (^) 14

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DEVIATION UNKNOWN

Constructing a (1-α)100% confidence interval for unknown:*

16

upper bound: 2 ( ) n

s x  t 

lower bound : 2 ( ) n

s x  t 

2 (^ ) 2 ( ) n

s x t n

s x  t      

DEVIATION UNKNOWN

Ex: the below data represents a sample of the number of home fires started by candles for the past several years, find the 99% confidence interval for the mean number of home fires started by candles each year. 5460, 5900, 6090, 6310, 7160, 8440, 9930

Answer: step 1: find the sample mean and standard deviation based on the sample

Step 2: find the tα/

sample size n=7, so degree of freedom is n-1=7-1= tα/2 =3. 17

x  7041. 4 s  1610. 3

DEVIATION UNKNOWN

Summary:

1. If population standard deviation is known, and

a) sample size n ≥30, using z-interval. b) sample size n <30, but the sample is drawn from a normal population, using z-interval. If the sample not from normal population, using non-parametric method.

2. If population deviation unknown, and

a) sample size n≥30, using t-interval. b) sample size n<30, if sample is from normal population, then using t-interval, if not from normal population, using non-parametric method. 19

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