303
Chapter Statistics 7 Inference for Means
Inference for Means
In chapter 6 we made the following assumptions: Population distribution is normal Mean is unknown Simple Random Sample Population standard deviation is known If unknown Use sample standard deviation (s)
Because of the adjustment for estimating with s the sampling distribution of the sample mean is no longer normally distributed.
Z, the standard normal distribution is replaced by t-distribution.
Only one N(0,1)
Many different t-distributions
For each different sample size there is a different t-distribution
t-Distribution
We define one sample t-statistics
tn -1
x - = s n
Degrees of freedom The way we distinguish between various t-distributions is by finding the degrees of freedom (df) that correspond to the sample size . Degrees of freedom is sample size minus one, df = n-1
t-Distribution
Shape of each t-distribution is very similar to the Z-distribution, but is slightly flatter. The larger the sample size, the closer the t-distribution is to the Zdistribution. Example:
Spread is more than Z
Symmetric about 0
t-Distribution
tn -1 x - = s n
has the t distribution with n 1 degrees of freedom.
A table of t distribution critical values can be found in Table D (the last page of the book). NOTE: These values are areas to the right (Not areas to the left as in the Z-table).
Using t-Table
Note that these values are areas to the right for any `t*' it gives P(t>t*)=probability In Table D, the degrees of freedom are listed in the left column. The probabilities are on top (these probabilities are inside for the Z-table) The individual t-values are inside the table. Make sure to get acquainted with this table and how it differs from the Z-table.
EXAMPLE of USING t-TABLE
Let n=25 and probability P=0.05 then find corresponding t* from t-table.
From df column in t-Table
df=n-1=24 and P=0.05 So that t*=1.761
From row P of t-Table
Value from inside the t-Table
EXAMPLE of using t-Table
If C=99% and n=21 then find corresponding t-value Step1: C=99% then =1% , =0.01. Step2 : For ttable the probability is /2=0.01/2=0.005 is the probability. Step 3: For t-table df=n-1=21-1=20 Step4: t-value=2.845
EXAMPLE of using t-Table
If n=45 and P=0.05 then find corresponding t-value. Step 1: For t-table df=n-1=45-1=44 , where 44 is not listed as a df in the t-table. In this case you round it down to 40, which is in the t-table. And use P=0.05 Step4: t-value=1.684 RULE: Always round down if df is not in the table ,i.e., if the df is
not in the table then pick the largest df that is in the table but smaller than df you want since this will give slightly wider CIs and we will be slightly less likely to reject H0.
Inference for Means
The steps in producing confidence intervals and hypothesis tests are the same as we have seen previously except When we change to s Change z* to t* s is calculated from the data using the formula:
1 n ( xi - x ) 2 s= n - 1 i =1
Finding CI for with Unknown
The confidence interval is ( x m)
Calculated from the data.
x t
*
s n
Calculated from the data.
Sample size
t* is found in table D at the back of the book. It must correspond to the appropriate df = n 1. It is easiest to find the confidence level at the bottom of the table and go up to the correct df.
Confidence Interval Example
An economist wants to determine the average amount a family of four in the United States spends on housing annually. He randomly selects 85 families of size four and finds that the average amount they spent on housing the previous year is $6219 with a standard deviation of $1978. Estimate the mean with 99% confidence. INFORMATION GIVEN: n=85, x = $6,219 and s = $1,978 df = n 1 = 85 1 = 84. t* is found in table D. We first go to the 99% confidence level at the bottom. Then we go up to 80 df (always round down). Thus, t* = 2.639.
Confidence Interval for with Unknown
Confidence Interval Example
x t
*
s n
1,978 = 6,219 566.18 = 6,219 2.639 85 = (5652.82, 6785.18)
t* is found in table D. We first go to the 99% confidence level at the bottom. Then we go up to 80 df (always round down). Thus, t* = 2.639.
This is a 99% confidence interval for the true average amount a family of four in the United States spends on housing annually.
Hypothesis Test for with Unknown
The steps for a hypothesis test are 1. State the null and alternative hypothesis. 1. State the level of significance (i.e., = 0.05). 1. Calculate the test statistic
x - 0 t= s n
1. Find the P-value: For one side or two sided hypothesis we use:
P-value
For a two-sided test: HA:
P - value = Pr (T t or T - t ) = 2Pr (T t )
For a one-sided test HA: >
P - value =Pr (T t )
For a one-sided test HA: <
P - value =Pr (T t )
Because of the limited number of t-values given in Table D, it is more common to find a range for the P-value, rather than the exact value (as will be seen in the example). Computers can be used to obtain exact values.
Hypothesis Test for with Unknown
5. Reject or fail to reject H0 based on the P-value.
0
If the P-value is less than or equal to , reject H .
Note: Last two steps are exactly the same as for the case where is known.
P-value from t-table
We put bounds on the p-value. P-value depend on the HA Select the row for given df (=n-1) Find t-left and t-right with corresponding probabilities p-left and p-right respectively such that t-left < t < t-right so that we get p-left > p-value > p-right.
EXAMPLE
If a two-sided then, P-value=2*P(t>|t0|) Let df=21 and t-statistic value t=2.05
HA :
0 ,
From t-table, 1.721<2.05<2.080 Corresponding P-values 0.05>P(t > 2.05)>0.025 Multiplying by 2 now we get 0.1>2* P(t>2.05) >0.05 so that 0.05<p-value<0.10
EXAMPLE
When H : > ,a one-sided test. Suppose n=52 and t=1.91 find the range for p-value. Here df=n-1=51 and we select 1.676 <1.91 < 2.009 Converting to p-value 0.05>P(T>1.91)>0.025 0.025<p-value<0.05
A 0
T.V. Example
Suppose that the data collected from our class survey is a random sample from the entire university (which it obviously is not). We wish to see if there is evidence that the average amount of television watched for students here is more than 7 hours per week T.V. Example SOLUTION: Information given n = 38, x = 8.05 , s
3 3 2 20 5 20 6 2 1 9 1 4 10 5 10 10 3 10 3 3 5 4 30 10 10 10 3 0 21 9
= 7.46
5 1 30 4 6 10 15 3
df = n - 1 = 38 - 1 = 37
EXAMPLE
1. State the null hypothesis:
H0 : = 7 Ha : > 7
or H 0 : 7
2. State the alternative hypothesis:
3. State the level of significance Assume = 0.05
Hypothesis Test for with Unknown
4. Calculate the test statistic.
x - 0 t= s n
5. Find the P-value.
8.05 - 7 1.05 = = = 0.87 7.46 1.21 38
P - value = Pr ( T t ) = Pr ( T 0.87 )
Remember the table gives probabilities to the right so we do not use the technique of subtracting from 1.
= between 0.15 and 0.20
Use df = 30 (rounding down)
Hypothesis Test for with Unknown
6. Do we reject or fail to reject H0 based on the p-value? p-value = between 0.15 and 0.20 is greater than = 0.05.
0
we 7. Conclusion fail to reject H "There is not significant statistical evidence that the average amount of television watched is more than 7 hours per week at the 0.05 level of significance."
More than one sample
To this point we have only looked at tests for single samples . For two samples: Matched pairs design Comparison of Means Pooled Estimators
Matched Pairs
When each individual can be given both treatments, we can reduce the two samples to a single sample using a matched pair design. Examples
Students are each given a pre-test and a post-test to determine the amount of material learned in a given time interval (sample unit is student) To examine the effect of a new drug, a large group of identical twins is identified. One twin is given a treatment and the other a placebo. A ophthalmologist is examining the importance of the dominant eye in reading. A large group of subjects is asked to read a passage with dominant eye covered and again with the non-dominant eye covered. NOTE: It can be seen in each of these examples that something pairs the two responses.
Analysis of Matched Pairs
Reduce the data from two samples to one sample Then use one-sample techniques Step1: The data is reduced from two samples to one by subtracting one of the responses from the other.
We could subtract each pre-test score from each post-test score. We could subtract each placebo response from each treatment response. We could subtract the time taken to read the passage with the nondominant eye from the time taken to read the passage with the dominant eye.
Example: Keyboards
To compare two brands of computer keyboards
Keyboard 1 (standard keyboard)
Keyboard 2 (specially designed so that the keys need very little pressure to make them respond)
Claim of manufacturer of keyboard 2 Typing can be done faster using keyboard 2
Example: Keyboards
SAMPLE
A simple random sample of 30 teachers was selected from a of population high-school teachers attending a national conference. Each teacher typed the same page of text once using keyboard 1 and once using keyboard 2. For each teacher the order in which the keyboards were used was determined by the toss of a coin. For each teacher the variable measured was the time (in seconds) to correctly type the page of text" (from Graybill, Iyer and Burdick, Applied Statistics, 1998).
DATA
Subject 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Keyboard 1 Keyboard 2 348 350 435 442 369 356 357 360 376 373 412 405 396 376 317 314 366 366 340 337 347 352 315 303 349 338 330 328 335 322 345 351 374 361 374 370 380 375 319 318 387 382 313 317 303 310 404 393 355 362 364 364 348 355 361 368 301 291 348 323
STEP 1 Reduction to one sample Take the difference Sample size n=30
DIFFERENCE=Keyboard2-keyboard1
Difference = K2 - K1 2 7 -13 3 -3 -7 -20 -3 0 -3 5 -12 -11 -2 -13 6 -13 -4 -5 -1 -5 4 7 -11 7 0 7 7 -10 -25
df = n - 1 = 30 - 1 = 29
xdiff = -3.53
sdiff = 8.56
HYPOTHESIS TESTING
1. State the null hypothesis:
H0 : = 0
or H 0 : 0
2. State the alternative hypothesis:
Ha : < 0
(from carefully reading)
3. State the level of significance Assume = 0.05
HYPOTHESIS TESTING
4. Calculate the test statistic.
x - 0 - 3.53 - 0 - 3.53 t= = = = -2.26 s 8.56 1.56 n 30
P - value = Pr ( T t ) = Pr ( T -2.26 )
5. Find the P-value.
Remember the table gives probabilities to the right.
= Pr ( T 2.26 ) = between 0.01 and 0.02
Use df = 29
HYPOTHESIS TESTING
6. Do we reject or fail to reject H0 based on the P-value? P-value = between 0.01 and 0.02 is less than = 0.05.
0
7. Conclusion. reject H we "There is significant statistical evidence that the average amount of time needed to type the passage is lower for keyboard 2 than keyboard 1 at the 0.05 level of significance."
Confidence Interval for Matched Pairs
Step 1: Reduce the data to a single sample Step 2: Use the same formula as for a confidence interval for with unknown , namely,
x t
Mean of Differences
*
Standard Deviation of Differences Sample Size
s n
Matched Pairs Confidence Interval
Example: Golf Balls
"In the manufacture of golf balls two procedures are used. Method I utilizes a liquid center Method II a solid center. To compare the distance obtained using both types of balls, 12 golfers are allowed to drive a ball of each type, and the length of the drive (in yards) is measured." (from Milton, McTeer, and Corbet, Introduction to Statistics, 1997) The manufacturer wants to estimate the mean difference with 90% confidence.
Example: Golf Balls
Information given: Sample size: n = 12.
Golfer 1 2 3 4 5 6 7 8 9 10 11 12 liquid
Make one sample
xdiff = 9.52
sdiff = 3.12
df = n 1 = 12 1 = 11
solid difference (liquid - solid) 180 172.7 7.3 215.8 202.5 13.3 140.6 128.1 12.5 182.7 173.9 8.8 193.8 180.7 13.1 100.2 88.7 11.5 195.2 188.9 6.3 117.6 108.8 8.8 199 186.5 12.5 179.5 175.9 3.6 122.3 112.7 9.6 106.7 99.8 6.9
Golf Balls Confidence Interval
x t
*
s n
3.12 = 9.52 1.796 = 9.52 1.62 12 = (7.90,11.14)
t* is found in table D. We first go to the 90% confidence level at the bottom. Then we go up to 11 df. Thus, t* = 1.796.
This is a 90% confidence interval for the true average difference for the distance traveled for the two types of golf balls.
Comparison of Means
Suppose you have two populations: Population1
Population Standard Deviation Sample Mean Sample Size
Population 2
2 (KNOWN)
1 (KNOWN)
x1
n1
x2
n2
The two-sample z-statistic is then
z=
( x1 - x2 ) - ( 1 - 2 )
12
n1 +
2 2
n2
NOTE: It is very rare that both population standard deviations are known. We will examine the situation in which they are not known
Comparison of Means
When we are interested in comparing two population means and we are estimating the population standard deviations 1 and 2 with s1 and s2, the two-sample tstatistic is then
( x1 - x2 ) - ( 1 - 2 ) t=
2 s12 s2 + n1 n2
1
2
with degrees of freedom equal to the smaller of n -1 and n -1 (or an appropriate estimate using computer software).
Comparison of Means: Hypothesis Testing
The null hypothesis can be any of the following:
H 0 : 1 = 2
or H 0 : 1 2
or H 0 : 1 2
The alternative hypothesis can be any of the following (depending on the question being asked):
H a : 1 2
or H a : 1 < 2
or H a : 1 > 2
The other steps are the same as those used for the tests we have looked at previously.
Example: Tomatoes
There has been some discussion among amateur gardeners about the virtues of black plastic versus newspapers as weed inhibitors for growing tomatoes. To compare the two, several rows of tomatoes are planted. Black plastic is used around nine randomly selected plants and newspaper around the remaining ten. All plants start at virtually the same height and receive the same care. The response of interest is the height in feet after a month's growth. (from Milton, McTeer, and Corbet, Introduction to Statistics, 1997). Perform a test to see if there is any difference between the average heights with significance level 0.10.
Example: Tomatoes
Information given: Sample sizes: n1 = 9, n2 = 10.
black plastic 1.8 1.29 1.13 2.92 2.2 1.25 2.61 1.6 2.06 newspaper 2.57 1.59 1.78 1.37 1.22 1.34 1.43 1.06 1.44 1.12
x1 = 1.87
s1 = 0.63
x2 = 1.49
s2 = 0.43
df = n1 - 1 = 9 - 1 = 8 because n1 is smaller than n2
Example: Tomatoes
1. State the null hypothesi...