Empirical Rule for X

The Central Limit Theorem (CLT) is stated as follows:

Given a large random sample of size n from a population with mean  and standard deviation , then tThe sample mean X is approximately normally distributed with mean and standard deviation given by

 = 

X

 =  / n

X

Note: 1. n >30 is usually large enough for the CLT to apply.

If the population from which we sample is normal thenX is exactly normally distributed with mean and standard deviation as above for any sample size.

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Empirical Rule for X

Consider a sample of size n from a population with mean  and standard deviation . Suppose X is normal ( or approximately normal), with  =  and  = /n

X X

(This would be the case if the population is normal or if the sample size is large).

Find the probability that X will be within (a) 2 of  (b) 3 of  .

X X

(a))

P( X will be within 2 of  )

X

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(b) P( X will be within 3 of  )

X

In general the statement “X will be within k of  “ means that X lies between

X

 - k  and  +k 

X X

If X is normal ( or approximately normal), then

P( X will be within k of  ) = P(-k < Z <k)

X

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Z Confidence Interval

Suppose we are given the following:

Normal Population: Scores on a standardized test.

Population Mean :  (unknown)

Population S.D.:  =1.5

To estimate  we will take a srs of size n =25 and use X as our estimator. Recall that since the population is normal,

X is normally distributed with  =  and  =  /n = 1.5/5 =.3

X X

We would like to be able to express this estimate in the form X  E or

(X – E, X + E ). Here E is some error which determines the accuracy of our estimate. Let’s take E = 2  for now .

X

Thus we have

For any given sample this interval may or may not contain the true mean  . It would be useful to know what the probability is that this interval covers  .

If the interval covers the true mean  then  is somewhere in the interval above so thatX is in fact within 2  ( =0.6) of  .

X

Thus P [ (X - 2  , X + 2  ) covers ]

X X

= P (X is within 2  of  )

X

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To make the probability above a nice number, .95, we should replace 2 by 1.96.

Thus we can say

“ For 95% of all samples of size n =25, the interval (X - 1.96  , X + 1.96  )

X X

will cover the true value of  .”

Or,

“ For 95% of all samples of size n =25, X will be within 1.96 of the true

X

population mean .”

The 95% value is called the LEVEL OF CONFIDENCE. This tells us the probability the interval will cover .

The 1.96 = .588 is called the margin of error. This tells us how accurate X is

X

(i.e. how closeX will be to  for 95% of all samples).

The interval (X - 1.96  , X + 1.96  ) is called a 95%

X X

Z-CONFIDENCE INTERVAL.

The simulation below will illustrate how confidence intervals work.

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MTB > random 25 c1-c40;

SUBC> norm 10 1.5.

MTB > zint 95 1.5 c1-c40.

[ The first two command lines select 40 random samples each of size n =25 from a normal distribution with  =10 and  = 1.5. The third command line forms the 95%

Z-CONFIDENCE INTERVAL for each sample]

Confidence Intervals (The assumed sigma = 1.5)

Variable N Mean StDev SE Mean 95.0% CI

C1 25 10.459 1.661 0.300 ( 9.871, 11.047)

C2 25 9.826 1.486 0.300 ( 9.238, 10.414)

C3 25 10.388 1.600 0.300 ( 9.800, 10.976)

C4 25 9.741 1.297 0.300 ( 9.153, 10.329)

C5 25 10.441 1.766 0.300 ( 9.853, 11.029)

C6 25 10.331 1.637 0.300 ( 9.743, 10.919)

C7 25 8.941 1.264 0.300 ( 8.353, 9.529)

C8 25 10.205 1.627 0.300 ( 9.617, 10.793)

C9 25 10.163 1.560 0.300 ( 9.575, 10.751)

C10 25 10.009 1.619 0.300 ( 9.421, 10.597)

C11 25 10.455 1.787 0.300 ( 9.867, 11.043)

C12 25 10.365 1.220 0.300 ( 9.777, 10.953)

C13 25 10.626 1.475 0.300 ( 10.038, 11.214)

C14 25 10.090 1.677 0.300 ( 9.502, 10.678)

C15 25 10.339 1.103 0.300 ( 9.751, 10.927)

C16 25 10.208 1.480 0.300 ( 9.620, 10.796)

C17 25 10.356 1.508 0.300 ( 9.768, 10.944)

C18 25 9.943 1.388 0.300 ( 9.355, 10.531)

C19 25 10.015 1.318 0.300 ( 9.427, 10.603)

C20 25 9.924 1.473 0.300 ( 9.336, 10.512)

C21 25 10.037 1.271 0.300 ( 9.449, 10.625)

C22 25 9.490 1.345 0.300 ( 8.902, 10.078)

C23 25 9.972 1.484 0.300 ( 9.384, 10.560)

C24 25 10.330 1.644 0.300 ( 9.742, 10.918)

C25 25 9.635 1.609 0.300 ( 9.047, 10.223)

C26 25 9.292 1.558 0.300 ( 8.704, 9.880)

C27 25 10.053 1.072 0.300 ( 9.465, 10.641)

C28 25 9.484 1.726 0.300 ( 8.896, 10.072)

C29 25 10.666 1.402 0.300 ( 10.078, 11.254)

C30 25 9.896 1.640 0.300 ( 9.308, 10.484)

C31 25 9.942 1.583 0.300 ( 9.354, 10.530)

C32 25 10.100 1.657 0.300 ( 9.512, 10.688)

C33 25 9.483 1.496 0.300 ( 8.895, 10.071)

C34 25 9.691 1.623 0.300 ( 9.103, 10.279)

C35 25 10.390 1.369 0.300 ( 9.802, 10.978)

C36 25 10.569 1.178 0.300 ( 9.981, 11.157)

C37 25 9.813 1.326 0.300 ( 9.225, 10.401)

C38 25 9.905 1.489 0.300 ( 9.317, 10.493)

C39 25 10.442 1.405 0.300 ( 9.854, 11.030)

C40 25 9.945 1.919 0.300 ( 9.357, 10.533)

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QUESTIONS

(a) In theory, how many of the above intervals would you expect to cover the true population mean  (=10)?

(b) In fact how many actually do?

Suppose you selected 40 samples of size n =25 from a real population ( where typically the population mean and standard deviation are unknown).

(a)Could you form a 95% Z- confidence interval for each sample? Explain.

(b) If weyou knew  and formed forty 95% Z-confidence intervals, then we would expect 38 intervalshowmany of the intervals would you expect to cover the population . ? Could you tell which intervals will cover  ? Explain.

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Note: (i) 100(1-)% Z-confidence interval of  is given by

X  Z/2 ; where  =  /n

X X

(ii) For 95% Z –confidence interval ,  = .05. hence 95% Z-confidence interval of  is

X  1.96  ; where  =  /n

X X

(iii) 99% Z-confidence interval of  is

X  2.57658 ; where  =  /n

X X

(iv) 90% Z-Confidence Interval of  is

X  1.64549 ; where  =  /n

X X

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Sample Size for Desired margin of Error

Problem: Suppose you wish to estimate a population mean  with a specified margin of error m and level of confidence 100(1-)% . What sample size should be used?

Solution: We know that m =

Now we solve this equation for n.

m2 =

nm2 =

n =

n = [z/2 /m]2

Note: In practice, we will round the sample size to the next whole number.

Example 6.5; Page 425: Tim Kelley of Example 6.2 has decided that he wants his estimate of his monthly weight accurate to within 2 or 3 pounds with 95% confidence. How many measurements must he take to achieve these margins of error?

For this example it is known that

(i)For 95% confidence and margin of error of 2 pounds we have

n ===8.6(always round up to the next whole number)

(ii) For 95% confidence and margin of error of 3 pounds we have

n = = =3.8

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Problem 6.16; page 432: You are planning a survey of starting salaries for recent liberal arts major graduates from your college. From a pilot study you estimate that the standard deviation is about $8000. What sample size do you need to have a margin of error equal to $ 500 with 95% confidence?

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STATISTICAL INFERENCE

Let us begin with a review of some basic definitions.

POPULATION: The set of all measurements or objects of interest in a particular study.

If the entire population were available for analysis we would know everything about it. However, in practice one cannot know the entire population because it is either too expensive, or simply impossible or impractical to examine each member. Thus a sample from the population is used to obtain information about the population.

SAMPLE: A subset of the population.

The sample picked should be “representative” of the population from which it comes and should avoid any bias which might skew our view of the population. One way to achieve this is to use a SIMPLE RANDOM SAMPLE (srs) i.e. a sample chosen in such a way that each member of the population has an equal chance of being chosen.

INFERENTIAL STATISTICS: deals with procedures which use the sample to draw conclusions about the population (from which it was drawn). The procedures of interest to us are CONFIDENCE INTERVALS and HYPOTHESIS TESTS.

In particular we will be interested in drawing conclusions about certain characteristics of the population. Such characteristics are known as POPULATION PARAMETRS. Examples of such characteristics are a POPULATION MEAN (denoted by the Greek letter  ) and a POPULATION PROPORTION ( denoted by the letter p).

EXAMPLE: Consider the population of weights ( in kg) of all newborn babies in Canada for a particular year. In this case, the POPULATION MEAN  is the average weight of all newborns in the population. An investigator may want to use a simple random sample of these weights to determine if there is sufficient evidence to answer questions like:

Is  > 3.2 kg? or Is  < 3.2 kg? or Is   3.2 kg?

EXAMPLE: Consider the population of all lakes in Nova Scotia. A biologist may be interested in the following POPULATION PROPORTION:

p = the proportion of all lakes in Nova Scotia that are seriously affected by acid rain. She may want to use a simple random sample of lakes from this population to determine if there is sufficient evidence to answer questions like:

Is p>.7 ? or, Is p<.7? or, Is p.7 ?

When drawing conclusions about a population using information from a sample it is important to realize that one can NEVER be absolutely certain the conclusion is correct. This is because a sample, though it may be “representative” of the population, only contains part of all the information contained in the population.

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HYPOTHESIS TESTING

Example: A graduate student claims that over 70% of the lakes in Nova Scotia have been seriously affected by acid rain. To justify this claim she proposes the following `test`.

“ Choose a simple random sample of 15 lakes in Nova Scotia. If 11 or more of the sampled lakes are seriously affected by acid rain, the claim is justified.”

Formally, we set up this test as follows.

First notice that the population of interest to this graduate student is the set of all lakes in Nova Scotia. The parameter of interest in her investigation is

p=the true proportion of all lakes in Nova Scotia affected by the acid rain [p=the unknown population proportion].

NULL HYPOTHESIS ALTERNATIVE HYPOTHESIS

What we want to reject. Research Hypothesis.

The viewpoint opposite to Ha What we want to prove.

H0 : Ha:

TEST STATISTIC ( evidence from the sample used to make a decision)

X =

Distribution of X :

Now in conducting this test we should make use of the fact that large values of X would

be consistent with the ______hypothesis that p .7.

How large an X? Let’s pick some number c and decide that if Xc we conclude

that______. Thus if X<c we must conclude that______. The value c is

called a CRITICAL VALUE . In this example, the graduate student has decided to use c =11. Her method for making a decision can be described as follows.

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REJECTION OR CRITICAL REGION (rule for making a decision)

Now suppose she conducts her study and that she observes that X  11. Then she would claim to have shown that Ha : p > .7 is true. If you had to use her study to make a policy decision, the first question you should ask is

“ What is the probability that her claim is wrong? That is, what is the probability of getting X  11 when in fact H0 : p  .7 is true ?”

Let’s find out by doing the calculations below.

Suppose that H0 : p  .7 is true / Probability of a wrong decision
P(Reject H0 / H0 is true)
p = .5
p=.6
p=.7 / P(X  11 p =.5) = 1 – P (X  10)
=
=
P ( X  11 p =.6) = 1 – P ( X  10)
=
=
P(X  11 p = .7) = 1 – P(X  10)
=
=

The error of rejecting H0 when in fact H0 is true is called a TYPE 1 ERROR. Notice that in this example the largest probability of making a type 1 error is

______and that it occurs when the value of p is ______( that is on

the boundary between H0 and Ha). The largest probability of making a type 1 error is called the LEVEL OF SIGNIFICANCE or TYPE 1 ERROR RATE of the test and is denoted by the Greek letter .

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Conversely suppose that the graduate student observed X < 11 (i.e. X10), thus leading to the claim H0 : p  .7 is true. In this case you should ask

“ What is the probability that her claim is wrong? That is, what is the probability of getting X < 11 when in fact Ha : p > .7 is true?”

Let’s find out by doing the calculations below.

Suppose that
Ha: p>.7 is true / Probability of a wrong decision P (Accept H0 Ha true) / Probability of a correct decision P (Reject H0Ha true)
p=.8
p=.9 / P(X< 11p=.8)
=P(X 10)
=
P( X< 11 p =.9)
= P (X  10)
= / P(X11 p=.8)
= 1 – P( X  10)
=
=
P(X  11 p =.9)
=1- P(X  10)
=
=

The error of accepting H0 when in fact Ha is true is called a TYPE II ERROR. For a particular value of p say p1 in the alternative ( i.e. p1 >.7) the probability of making a type II error is called the TYPE II ERROR RATE evaluated at p = p1. This probability is denoted by (p1). Thus,

(p1) = P ( Accept H0  p = p1 in Ha)

Also for a particular value of p say p1 in the alternative (i.e. p1 > .7) we can calculate the probability of a correct decision ( see the last column of the table above). The probability of making a correct decision, that is, rejecting H0 when in fact Ha is true is called the POWER OF THE TEST AGAINST THE ALTERNATIVE p1 in Ha and is denoted K(p1). Thus

K(p1) = P (Reject H0  p = p1 in Ha)

Notice that K(p1) and (p1) are related by K(p1) = 1 -  (p1). If in fact Ha is true, power is a measure of a test’s ability to detect this. For example if in fact p were

actually .8(.9), this test will detect this with probability______(______).

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A good test that is one in whose results we can be confident of , will be one in which the probabilities of the type I and type II errors are small.

The ideas discussed above are refer to the ERROR STRUCTURE of a test. A summary is provided below.

ACTUAL SITUATION

DECISION / H0 is True / Ha is True ( H0 is false)
Accept H0 ( Do not reject H0) / Correct Decision / Type II Error
Reject H0 ( Accept Ha) / Type I Error / Correct Decision

QUESTION: For the student’s test above, state in words the consequence of

making a

(a)Type I Error:

(b)Type II Error:

ERROR RATES AND POWER OF A TEST

TYPE I ERROR

Reject H0 when H0 is true / P(Type I Error) = P (Reject H0 H0 true).
The largest possible probability of a type I error is denoted by  and is called the LEVEL OF SIGNIFICANCE or TYPE I ERROR RATE of the test. In calculating  = P ( reject H0  H0 true ) , use the value of p right on the boundary between H0 and Ha

TYPE II ERROR

Accept H0 when Ha is true / (p1) = P (Type II Error)
= P ( Accept H0 p = p1 in Ha)
POWER AGAINST the ALTERNATIVE p1 / K(p1) = P ( Reject H0  p = p1 in Ha )
= 1 -  (p1)
In the case that Ha is true, power is a measure of the sensitivity of the test i.e. the ability of the test to detect that Ha is true.

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Changing the Rejection Region

Question: If we use the same sample size, how can we modify this test in order to reduce the type I error rate  ?

Suppose we take c =14, so we reject H0 if X  14. What is  ?

In this case what will happen to the type II error rate (p1) and the power K(p1) ?

NOTE: Ideally, we would like  and (p) to be zero and K(p) to be 1; but for fixed n decreasing  causes (p) to increase and K(p) to decrease.

NOTE: The only way to decrease both  and (p) is to increase the sample size.

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The P-value

Consider the test: H0: p  .70, Ha: p > .7, n=30; Reject H0 if X  26.

Suppose we conduct the test and observe X to be x0 = 28. According to the rejection region we would reject H0 . We would in fact have rejected H0 even if our critical value had been 28. But with a critical value of 28, the type I error rate would be smaller.

The P-value is the smallest type I error rate at which one can reject H0 on the basis of the observed outcome x0 . It is obtained by replacing the critical value ‘c’ by x0 in the calculation of the type I error rate.

P-value = P (X  x0  H0 is true)

For example, consider the cases where x0 is 28 and x0 is 24.

Type I error rate  / P-value when x0 = 28 / P-value when x0 = 24
P(X26 p =.7)
= 1 – P (X  25  p = .7)
= 1 - .9698
=.0302 / P ( X  28  p =.7)
= 1 – P (X  27  p =.7)
= 1- .9979
=.0021 / P(X  24  p =.7)
= 1 – P ( X  23  p =.7)
= 1 - .8405
= .1595

Notice

If x0 is in the rejection region the p-value   .

If x0 is not in the rejection region then the p-value is >  .

Thus it is clear that we can conduct our test at  = .03 without using a rejection region. We just have to calculate the P-value and use the following rule.

If the P-value   then reject H0 .

If the P-value >  then do not reject H0.

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Summary: Hypothesis Testing

Concept / Left-Tailed Test / Right-Tailed Test
Hypotheses / H0: p  p0 , Ha : p < p0 / H0: p  p0 , Ha : p > p0
Critical Region / Reject H0 if X  c / Reject H0 if X  c
Type I Error Rate 
P(Reject H0H0 true) / P(Xc p =p0) / P(X c p = p0)
Type II Error Rate (p1)
P(Accept H0 p =p1 in H a) / P(X>c p = p1) / P(x<c p =p1)
Power K(p1)
P(Reject H0 p =p1 in Ha) / P(Xcp=p1)
or, 1-(p1) / P(Xc p= p1)
or, 1 -(p1)
P-Value / P(Xx0 p =p0) / P(Xx0 p =p0)

P-value Decision Rule

Reject H0 if the P-value  

Note: A similar theory also applies to a Two-tailed test, i.e., a test of

H0: p =p0, Ha: p  p0

While we will conduct such tests in our applications, we will not discuss the theory here.

An analogy of statistical hypotheses

In practice we use  = .01 or  = .05. Thus to reject H0 we need strong evidence.

In our judicial system, we use the phrase innocent until proven guilty beyond a reasonable doubt. We may define null and alternative hypotheses as follows:

H0: defendant is innocent

Ha: defendant is guilty.

To prove defendant is guilty we need strong evidence.

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CUMULATIVE BINOMIAL PROBABILITIES : P(Xx)

n x / 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 / x
15 0
1
2
3
4
5
6
7
8
9
10 11
12
13
14
15 / .2059 .0352 .0047 .0005 .0000 .0000 .0000 .0000 .0000
.5490 .1671 .0353 .0052 .0005 .0000 .0000 .0000 .0000
.8159 .3980 .1268 .0271 .0037 .0003 .0000 .0000 .0000
.9444 .6482 .2969 .0905 .0176 .0019 .0001 .0000 .0000
.9873 .8358 .5155 ..2173 .0592 .0093 .0007 .0000 .0000
.9978 .9389 .7216 .4032 .1509 .0338 .0037 .0001 .0000
.9997 .9819 .8689 .6098 .3036 .0950 .0152 .0008 .0000
1.000 .9958 .9500 .7869 .5000 .2131 .0500 .0042 .0000
1.000 .9992 .9848 .9050 .6964 .3902 .1311 .0181 .0003
1.000 .9999 .9963 .9662 .8491 .5968 .2784 .0611 .0022
1.000 1.000 .9993 .9907 .9408 .7827 .4845 .1642 .0127
1.000 1.000 .9999 .9981 .9824 .9095 .7031 .3518 .0556
1.000 1.000 1.000 .9997 .9963 .9729 .8732 .6020 .1841
1.000 1.000 1.000 1.000 .9995 .9948 .9647 .8329 .4510
1.000 1.000 1.000 1.000 1.000 .9995 .9953 .9648 .7941
1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 / 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
20 0
1
2
3
4
5
6
7
8
9
10 11
12
13
14
15
16
17
18
19
20 / .1216 .0115 .0008 .0000 .0000 .0000 .0000 .0000 .0000
.3917 .0692 .0076 .0005 .0000 .0000 .0000 .0000 .0000
.6769 .2061 .0355 .0036 .0002 .0000 .0000 .0000 .0000
.8670 .4114 .1071 .0160 .0013 .0000 .0000 .0000 .0000
.9568 .6296 .2375 .0510 .0059 .0003 .0000 .0000 .0000
.9887 .8042 .4164 .1256 .0207 .0016 .0000 .0000 .0000
.9976 .9133 .6080 .2500 .0577 .0065 .0003 .0000 .0000
.9996 .9679 .7723 .4159 .1316 .0210 .0013 .0000 .0000
.9999 .9900 .8867 .5956 .2517 .0565 .0051 .0001 .0000
1.000 .9974 .9520 .7553 .4119 .1275 .0171 .0006 .0000
1.000 .9994 .9829 .8725 .5881 .2447 .0480 .0026 .0000
1.000 .9999 .9949 .9435 .7483 .4044 .1133 .0100 .0001
1.000 1.000 .9987 .9790 .8684 .5841 .2277 .0321 .0004
1.000 1.000 .9997 .9935 .9423 .7500 .3920 .0867 .0024
1.000 1.000 1.000 .9984 .9793 .8744 .5836 .1958 .0113
1.000 1.000 1.000 .9997 .9941 .9490 .7625 .3704 .0432
1.000 1.000 1.000 1.000 .9987 .9840 .8929 .5886 .1330
1.000 1.000 1.000 1.000 .9998 .9964 .9645 .7939 .3231
1.000 1.000 1.000 1.000 1.000 .9995 .9924 .9308 .6083
1.000 1.000 1.000 1.000 1.000 1.000 .9992 .9885 .8784
1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 / 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
30 0
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 / .0424 .0012 .0000 .0000 .0000 .0000 .0000 .0000 .0000
.1837 .0105 .0003 .0000 .0000 .0000 .0000 .0000 .0000
.4114 .0442 .0021 .0000 .0000 .0000 .0000 .0000 .0000
.6474 .1227 .0093 .0003 .0000 .0000 .0000 .0000 .0000
.8245 .2552 .0302 .0015 .0000 .0000 .0000 .0000 .0000
.9628 .4275 .0766 .0057 .0002 .0000 .0000 .0000 .0000
.9742 .6070 .1595 .0172 .0007 .0000 .0000 .0000 .0000
.9922 .7608 .2814 .0435 .0026 .0000 .0000 .0000 .0000
.9980 .8713 .4315 .0940 .0081 .0002 .0000 .0000 .0000
.9995 .9389 .5888 .1763 .0214 .0009 .0000 .0000 .0000
.9999 .9744 .7304 .2915 .0494 .0029 .0000 .0000 .0000
1.000 .9905 .8407 .4311 .1002 .0083 .0002 .0000 .0000
1.000 .9969 .9155 .5785 .1808 .0212 .0006 .0000 .0000
1.000 .9991 .9599 .7145 .2923 .0481 .0021 .0000 .0000
1.000 .9998 .9831 .8246 .4278 .0971 .0064 .0001 .0000
1.000 .9999 .9936 .9029 .5722 .1754 .0169 .0002 .0000
1.000 1.000 .9979 .9519 .7077 .2855 .0401 .0009 .0000
1.000 1.000 .9994 .9788 .8192 .4215 .0845 .0031 .0000
1.000 1.000 ..9998 .9917 .8998 .5689 .1593 .0095 .0000
1.000 1.000 1.000 .9971 .9506 .7085 .2696 .0256 .0001
1.000 1.000 1.000 .9991 .9786 .8237 .4112 .0611 .0005
1.000 1.000 1.000 .9998 .9919 .9060 .5685 .1287 .0020
1.000 1.000 1.000 1.000 .9974 .9565 .7186 .2392 .0078
1.000 1.000 1.000 1.000 .9993 .9828 .8405 .3930 .0258
1.000 1.000 1.000 1.000 .9998 .9943 .9234 .5725 .0732
1.000 1.000 1.000 1.000 1.000 .9985 .9698 .7448 .1755
1.000 1.000 1.000 1.000 1.000 .9997 .9907 .8773 .3526
1.000 1.000 1.000 1.000 1.000 1.000 .9979 .9558 .5886
1.000 1.000 1.000 1.000 1.000 1.000 .9997 .9895 .8163 / 0
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z Test for a Population Mean
To test the hypothesis H0 :  = 0 based on an SRS of size n from a population with unknown mean  and known standard deviation  , compute the test statistic
z =
In terms of a standard normal random variable Z, the P-value for a test of H0 against
Ha : 0 is P(Z  z)
Ha : 0 is P(Z  z)
Ha :   0 is 2P(Z  z)
These P-values are exact if the population distribution is normal and are approximately correct for large n in other cases (Page 445-Text Book).

Confidence Intervals and Two-Sided Tests: A level  two-sided significance test rejects a hypothesis H0: =0 exactly when the value 0 falls outside a level 1- confidence interval for .

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To illustrate the test we consider the following problem.

Problem 6.45 (Page 455): The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. The mean score for U.S. college students is about 115, and the standard deviation is about 30. A teacher who suspects that older students have better attitudes toward school gives the SSHA to 20 students who are at least 30 years of age. Their mean score is x = 135.2.

(a)Assuming that  = 30 for the population of older students, carry out a test of

H0 :  = 115, Ha :  > 115.

Report the P-value of your test, and state your conclusion clearly.

(b)Your test in (a) required two important assumptions in addition to the assumption that the value of  is known. What are they? Which of these assumptions is most important to the validity of your conclusion in (a).

Solution: Given: n= , x = ,  =

Assume  = .05

(a) (i) Ha :  > 115; therefore this is right sided test.

(ii) The test statistic in this case is

z =

(iii) p-value = P (Z  3.01 )

(iv) Decision:

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(v) Concluding Sentence:

(b)Assumptions: (i)

(ii)

Example 6.16; Page 450: Bottles of a popular cola drink are supposed to contain 300 milliliters (ml) of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is normal with standard deviation  = 3 ml. A student who suspects that the bottler is underfilling measures the contents of six bottles. The results are

299.4297.7 301.0 298.9 300.2 297.0

Is this convincing evidence that the mean contents of cola bottles is less than the advertised 300 ml? The hypotheses are

H0 :  = 300, Ha :  < 300

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Problem 6.46; Page 456: The mean yield of corn in the United States is about 120 bushels per acre. A survey of 40 farmers this year gives a sample mean yield of = 123.8 bushels per acre. We want to know whether this is good evidence that the national mean this year is not 120 bushels per acre. Assume that the farmers surveyed are an srs from the population of all commercial growers and that the standard deviation of the yield in this population is  = 10 bushels per acre. Give the P-value for the test of

H0 :  =120, Ha :  120.

Are you convinced that the population mean is not 120 bushels per acre? Is your conclusion correct if the distribution of corn yields is some what non-normal? Why?

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Problem 6.87; Page 481: You have an srs of size n = 16 from a normal distribution with  = 1. You wish to test

H0 :  = 0

Ha :  > 0.

You decide to reject H0 if > 0 and the accept H0 otherwise.

(a)Find the probability of a Type I error , that is, the probability that your test rejects H0 when in fact  = 0.

(b)Find the probability of a Type II error when  = 0.2. This is the probability that your test accepts H0 when in fact  = 0.2.

(c)Find the probability of a Type II error when  = 0.6.

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Problem 6.84; Page 480: Example 6.16 discusses a test about the mean contents of cola bottles. The hypotheses are

H0 :  = 300

Ha :  < 300 .

The sample size is n =6, and the population is assumed to have a normal distribution with  = 3. A 5% significance test rejects H0 if z  -1.645, where the test statistic z is

z = .

Power calculations help us see how large a shortfall in the bottle contents the test

can be expected to detect.

(a)Find the power of this test against the alternative  = 298.

(b)Find the power of this test against the alternative  = 294.

found in (b) ? Explain why this result makes sense?

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The t-distribution

The t-distribution depends on a single parameter. This parameter is called its degrees of freedom (df). If sampling is done from a normal distribution whose mean is  and standard deviation , then

X - 

Z = 

 /n

follows standard normal distribution. Since,  in practice is mostly unknown; therefore, we can replace it by its estimate s. The random variable