Chapter 5 - YU, Chi Wai PDF

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MATH2411: Applied Statistics Dr. YU, Chi Wai Chapter 5: HYPOTHESIS TESTING 1 WHAT IS HYPOTHESIS TESTING? As its name indicates, it is about a test of hypothesis. To be more precise, we would first translate our question of interest into a hypothesis about an unknown parameter like or , and then test...

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MATH2411: Applied Statistics | Dr. YU, Chi Wai

Chapter 5: HYPOTHESIS TESTING HAT YPO THES ESIS STIN ING 1 WHA T IS HYP OTH ES IS TEST IN G? As its name indicates, it is about a test of hypothesis. To be more precise, we would first translate our question of interest into a hypothesis about an unknown parameter like 𝜇𝑋 or 𝜎𝑋2 , and then test it. Unlike what we did in Chapter 4 to use a sample of data to get a point-valued or an interval-valued estimate of the unknown parameter, we now would have a hypothesized value of the parameter which is assumed to be true first and then we use data to see if the assumption should be rejected or not be rejected.

2 STATIS TI CA PO TH ESE TISTI TICA CALL HY HYPO POTH THE SESS: 𝑯𝟎 A AND ND 𝑯𝟏

In hypothesis testing, we first need to study the following key terms:

 The null hypothesis 𝑯𝟎 : It is the hypothesis that is assumed to be true and then tested to be rejected or not to be rejected formally. It always contains “=” sign. (i.e. =, ≤, ≥).

 The alternative hypothesis 𝑯𝟏 : It is the hypothesis that typically represents the underlying research question of the investigator and is the complement of 𝐻0 , i.e. it contains the values of parameter we accept if we reject 𝐻0 . It never contains “=” sign, EXCEPT a simple test.

In this chapter, both of them are with respect to the parameter 𝜇𝑋 or 𝜎𝑋2 .  Test statistic: An estimator used for the parameter in a test.

Throughout our course, the test statistics used for testing the hypotheses for 2 , respectively. 𝜇𝑋 and 𝜎𝑋2 are 𝑋 and 𝑆𝑛−1 ~1~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

EXAMPLE XAMPLE •

An agronomist may want to decide on the basis of experiments whether or not a new fertilizer would produce a higher yield of soybeans than an old one whose mean is known to be 10. In this case the agronomist has to test 𝜇𝑋 > 10, where 𝜇𝑋 is the mean of the random variable of the yield of soybean by the new fertilizer, assuming a normal population. Then, we have

𝑯𝟎: 𝝁𝑿 = 𝟏𝟎 against 𝑯𝟏: 𝝁𝑿 > 𝟏𝟎.

•

A manufacturer of pharmaceutical products may decide on the basis of samples whether or not 90% of all patients given a new medication will recover from a certain disease. In this case we might say that the manufacturer has to decide whether or not the parameter p of a binomial population equals 0.90. We have

𝑯𝟎: 𝝁𝑿 = 𝒑 = 𝟎. 𝟗 against 𝑯𝟏: 𝝁𝑿 ≠ 𝟎. 𝟗.

OFF HYPOT OTHE HESI SISS TESTI STIN 3 TYPE O HE SI NG According to the form of the alternative hypothesis, we can have the following

Four types of tests: I)

SIM IMP PLE TES TEST T

II)

ONE-SID SIDE ED RIG IGH HT TE TESST

III)

ONE-SID SIDE ED LEFT TES TEST T

IV)

TWO-SID SIDE ED TES EST T

𝐻 : 𝜇 = 𝜇0 { 0 𝑋 𝐻1 : 𝜇𝑋 = 𝜇1 𝐻 :𝜇 { 0 𝑋

= 𝜇0 𝐻1 : 𝜇𝑋 > 𝜇0

𝐻 : 𝜇 = 𝜇0 { 0 𝑋 𝐻1 : 𝜇𝑋 < 𝜇0

𝐻 : 𝜇 = 𝜇0 { 0 𝑋 𝐻1 : 𝜇𝑋 ≠ 𝜇0

~2~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

 In this course, we will NOT study any test with an inequality null hypothesis.  Note that the hypotheses should be stated BEFORE looking at the data.  Unless we have enough information to do a simple test or a onesided test, in practice we would opt for the default which is a two-sided test.

4 MAI N CO NCE PT O THES IS TESTI NG AIN CON CEP OFF HYPO POTHES THESIS TIN The Basic Idea of doing hypothesis testing is a “CONTRADICTION” with the following three steps:

Step 1:

Determine 𝐻0 and 𝐻1 .

Step 2:

Under 𝐻0 , define a rare event --- the event which happens with a very small probability in one experiment of getting 𝑛 data. Step 3:

Collect data. If data contradicts 𝐻0 , then we can say that 𝐻0 is false and reject 𝐻0 , while if data do not contradict 𝐻0 , then we canNOT say that 𝐻0 is true and accept it, but we can say that we do NOT reject 𝐻0 .

~3~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

Example: We want to know whether or not a coin is fair. Consider a random experiment of flipping the coin, say 10 times.

Step 1: (𝐻0 ) The coin is fair, i.e. P({H})=P({T}) = 1/2, and (𝐻1 ) The coin is NOT fair.

Step 2: Under 𝐻0 , i.e. the coin is assumed to be fair, the probability of getting 10 tails in ONE experiment is (1/2)10 ≈ 0.00098. So, we can define the event of getting 10 tails to be the rare event under 𝐻0 .

Step 3: Perform the experiment to collect data, i.e. we now flip the coin 10 times. If we finally get 10 tails, then the collected data tell us that getting 10 tails is NOT a rare event, i.e. they contradict 𝐻0 . Therefore, we would say that we have evidence to suspect the reliability of 𝐻0 and thus reject 𝐻0 ( = accept 𝐻1 ).

In hypothesis testing, we only use the data to see if there is enough evidence to reject 𝑯𝟎 .

 If we have enough evidence to reject 𝐻0 , we can have great confidence that 𝐻0 is false and 𝐻1 is true.

 However, if we do not have enough evidence to reject 𝑯𝟎 , then it does not mean that we have great confidence in the truth of 𝐻0 . In this case, we should say "do not reject 𝑯𝟎 ", instead of "accept 𝑯𝟎 ".

~4~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

5 TEST EERR RROR ORSS AND ERRO ROR PR OBAB ABIL ILITI ITIES RR OR RP ROB AB IL ITI ES We would use a test statistic (a point-valued estimator) to formulate the so-called a statistical test statement (hereafter, a test statement) of hypothesis 𝑯𝟎 ------ a procedure/statement/condition based upon the observed values of the random variable of our interest that leads to the rejection or non-rejection of the hypothesis 𝐻0 . Note that there is no perfect test statement. Each test statement must lead to the following two kinds of errors.

If 𝐻0 is true If 𝐻0 is false

Not reject 𝐻0 No error TYPE II ERROR

Reject 𝐻0 TYPE I ERROR No error

TYP ERR ERR IS IN IN FFACT YPE E I ERR RROR OR: THE ERR RRO OR O OFF R RE EJE JECT CT CTIING H0 WH WHE EN IIT T IS ACT TR TRUE UE. TYP YPE E II ER ERR ROR: TH THE E ER ERR ROR OF N NO OT R RE EJE JECT CT CTIN IN ING G H0 W WHEN HEN IIT T IISS IIN N FFACT ACT FFA ALSE. Correspondingly, we have 

𝛼 = 𝑃(𝑇𝑦𝑝𝑒 𝐼 𝑒𝑟𝑟𝑜𝑟) = 𝑃(𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 𝑖𝑓 𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒)

It is the probability of making a wrong decision to reject 𝐻0 . 

𝛽 = 𝑃(𝑇𝑦𝑝𝑒 𝐼𝐼 𝑒𝑟𝑟𝑜𝑟) = 𝑃(𝑁𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 𝑖𝑓 𝐻0 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒)

It is the probability of making a wrong decision not to reject H0.

Ideally, we want to formulate a test statement such that these two error probabilities can be minimized. However, in general, we cannot control both error probabilities simultaneously (when the sample size is fixed). The following example illustrates this problem.

~5~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

EXAMPLE XAMPLE Suppose we knew that the light bulbs produced from a standard manufacturing process have life times distributed as normal with a standard deviation 𝜎𝑋 = 300 hours. However, we did not know the mean lifetime 𝜇𝑋 . For simplicity, assume that we were sure that the mean lifetime should be either 1200 or 1240.

𝐻0 : 𝜇𝑋 = 1200 𝐻1 : 𝜇𝑋 = 1240

Then we may set up the following simple test:

{

Suppose that we draw a sample of 100 light bulbs and measure their lifetimes. The sample mean 𝑋 is used to estimate the true population mean 𝜇𝑋 .

Since the hypothesized value in 𝑯𝟏 is larger than the hypothesized value in 𝑯𝟎 , intuitively, we can say that a large value of  𝒙 will lead to the rejection of 𝑯𝟎 , or we can set the test statement

Reject 𝑯𝟎 if  𝒙 > 𝒄.

We would later discuss how to determine the constant 𝑐, which is often called a critical value in practice. Now we simply use the above statement to get

𝛼 = 𝑃(𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 𝑖𝑓 𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) = 𝑃(𝑋 > 𝑐 𝑖𝑓 𝜇𝑋 = 1200)

𝛽 = 𝑃 (𝑁𝑜𝑡 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 𝑖𝑓 𝐻0 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒) = 𝑃(𝑋 ≤ 𝑐 𝑖𝑓 𝜇𝑋 = 1240)

and

Note that (Step 2) under 𝐻0 (i.e. when 𝜇𝑋 = 1200), the event {𝑋 > 𝑐 } occurs with a very small probability 𝛼 . Thus, {𝑋 > 𝑐 } is a rare event under 𝐻0 . So, in  > 𝒄 IF 𝑯𝟎 is TRUE. ONE experiment of getting 𝑛 data, we should NOT get 𝒙

(Step 3) In other words, getting 𝑥 > 𝑐 in one experiment would contradict 𝐻0 , and then we would reject 𝐻0 . ~6~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

The following picture illustrates why we cannot minimize these two error probabilities simultaneously. (More details would be discussed in lecture!)

Since there is a trade-off between the two types of error: making 𝜶 smaller will lead to a larger 𝜷, and vice versa. So, in designing a test statement, we can only control one of the errors, normally guarantee 𝜶 in a desired low value, and then find a test statement with 𝜷 as small as possible.

DETER TERMIN MIN MINATI ATI ATIO ON OF A CR CRITI ITI ITICA CA CALL VA VALLUE Recall that when we design a test of hypothesis, in general we cannot control the two error probabilities at the same time, and what we can do is to control the Type I error probability 𝜶 in a desired low level, (often use 0.01, 0.05 or 0.1), and then reduce the Type II error probability 𝜷 as much as we could. How to design a test statement with this restriction of 𝜶?

𝐻0 : 𝜇𝑋 = 1200 𝐻1 : 𝜇𝑋 = 1240

With reference to the previous example, we now want to test

at 𝜶 = 𝟎. 𝟎𝟓.

{

~7~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

Again, in this example, a large value of the sample mean will lead to the rejection of the null hypothesis 𝐻0 . So, we consider

Reject 𝑯𝟎 cifshould  𝒙 >be𝒄determined . Now, we can expect that the critical value by the fixed Type I error probability. From now, we would call 𝜶 a significance level.

After some technical steps (discussed in lecture), we have 𝑐 = 1249.35 and thus the complete test statement:

Reject 𝑯𝟎 at a significance level 0.05 if 𝒙  > 𝟏𝟐𝟒𝟗. 𝟑𝟓.

Suppose that if the observed value of the sample mean is  𝒙 = 𝟏𝟐𝟑𝟕, then we could conclude that we DO NOT HAVE ENOUGH EVIDENCE TO REJECT 𝑯𝟎 at a level 𝜶 = 0.05. ~8~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

6 POW OWER TEST STATE ATEMEN MENT ER OF A TES T ST ATE MEN T

A power of the test statement is defined as 1 − 𝛽, i.e. the probability of rejecting 𝑯𝟎 if 𝑯𝟎 is false. It is often used to assess the goodness of the test statement.

For the comparison of two different test statements, first we need both test statements to have a common 𝛼, and then the test statement is said to be better if it has a higher power. Question: How can we increase the power of a given test statement when the value 𝛼 remains unchanged? Answer:

QUES ESTION TION Find the power of the test statement

“Reject 𝑯𝟎 at a significance level 0.05 if 𝒙  > 𝟏𝟐𝟒𝟗. 𝟑𝟓.“

in the previous example.

~9~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

What would be the test statement and its power if the sample size is changed from 100 to 400?

~ 10 ~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

ULAT ATIO ION OFF A TTES EST TATEME TEMEN AB OUT 7In the FOfollowing, RMUL AT IO NO ES T SSTA TA TEME NT A BOU T 𝝁𝑿 (NOR ORMA MA MALL CA CASSE) we ONLY consider the formulation of the test statement for one-sided tests and two-sided test in the case that X follows a Normal distribution, i.e. 𝑋 ∼ 𝑁(𝜇𝑋 , 𝜎𝑋2 ). Recall that when 𝑋 ∼ 𝑁(𝜇𝑋 , 𝜎𝑋2 ), we have the result that

𝟐 Thus, when 𝝈𝑿

𝜎𝑋2 ) . 𝑋 ∼ 𝑁 (𝜇𝑋 , 𝑛

is known, we have the following results:

1. One-sided right test: Consider

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 { 𝑯 𝟏 : 𝝁𝑿 > 𝝁𝟎

.

Intuitively, we reject 𝐻0 if 𝑥 > 𝑐. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if 𝑥 (when

𝜎𝑋2

is KNOWN).

~ 11 ~

> 𝜇0 + 𝑧𝛼

𝜎𝑋

√𝑛

MATH2411: Applied Statistics | Dr. YU, Chi Wai

2. One-sided left test: Consider

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 { 𝑯 𝟏 : 𝝁𝑿 < 𝝁𝟎

.

Intuitively, we reject 𝐻0 if 𝑥 < 𝑐. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if 𝑥 (when

𝜎𝑋2

is KNOWN).

3. Two-sided test: Consider

< 𝜇0 − 𝑧𝛼

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 { 𝑯 𝟏 : 𝝁𝑿 ≠ 𝝁𝟎

𝜎𝑋

√𝑛

.

Intuitively, we reject 𝐻0 if 𝑥 < 𝑎 OR 𝑥 > 𝑏. Consequently, we would Reject 𝐻0 at a significance level 𝛼 if (when 𝜎𝑋2 is KNOWN).

|

~ 12 ~

| > 𝑧𝛼

−𝜇0 𝑥 𝜎𝑋 √𝑛

2

MATH2411: Applied Statistics | Dr. YU, Chi Wai 2 2 Previously in Chapter 4, when 𝜎𝑋2 is Unknown and we used 𝑠𝑛−1 to replace 𝜎𝑋 , we had a 𝑡 distribution to derive the formulas of the random and confidence interval for 𝜇𝑋 . Similarly, in the following we would also use the 𝑡 distribution to get the test statement of the hypothesis about 𝜇𝑋 when 𝜎𝑋2 is Unknown.

Recall that

If 𝑿 ∼ 𝑵(𝝁𝑿 , 𝝈𝟐𝑿), then

 − 𝝁𝑿 𝑿 𝟐 √𝑺𝒏−𝟏

𝒏

∼ 𝒕𝒏−𝟏

which means that the random variable on the left follows a 𝒕 distribution with 𝒏 − 𝟏 degrees of freedom. Thus, when 𝝈𝑿 𝟐

is UNknown, we have the following results:

1. One-sided right test: Consider

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 { 𝑯 𝟏 : 𝝁𝑿 > 𝝁𝟎

.

Intuitively, we reject 𝐻0 if 𝑥 > 𝑐. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if 𝑥 (when

𝜎𝑋2

is UNKNOWN).

~ 13 ~

> 𝜇0 + 𝑡𝑛−1,𝛼

𝑠𝑛−1 √𝑛

MATH2411: Applied Statistics | Dr. YU, Chi Wai

2. One-sided left test: Consider

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 { 𝑯 𝟏 : 𝝁𝑿 < 𝝁𝟎

.

Intuitively, we reject 𝐻0 if 𝑥 < 𝑐. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if 𝑥 (when

𝜎𝑋2

is UNKNOWN).

3. Two-sided test: Consider

< 𝜇0 − 𝑡𝑛−1,𝛼

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 { 𝑯 𝟏 : 𝝁𝑿 ≠ 𝝁𝟎

𝑠𝑛−1 √𝑛

.

Intuitively, we reject 𝐻0 if 𝑥 < 𝑎 OR 𝑥 > 𝑏. Consequently, we would Reject 𝐻0 at a significance level 𝛼 if (when 𝜎𝑋2 is UNKNOWN).

| 𝑠𝑛−1 | > 𝑡𝑛−1,𝛼

~ 14 ~

−𝜇0 𝑥 √𝑛

2

MATH2411: Applied Statistics | Dr. YU, Chi Wai

Remark that for the above three tests when 𝜎𝑋2 is UNknown, we often call the −𝝁 𝒙𝒔𝒏−𝟏𝟎 a t value. term √𝒏

In R (https://www.r-project.org/), we can use the function t.test to get the t value in the case of UNKNOWN 𝝈𝟐𝑿, when all collected data are given.

EXA XAMPLE MPLE Frequencies, in hertz (Hz), of 12 elephant calls: 14, 16, 17, 17, 24, 20, 32, 18, 29, 31, 15, 35

Assume that the population of possible elephant call frequencies (𝑋) is a normal distribution, Now a scientist is interested in the expected frequency 𝜇𝑋 of 𝑋. Do a two-sided test with 𝑯𝟎 : 𝝁𝑿 = 𝟏𝟎 at a 0.05 level of significance.

Note that R only provides us a t value only. When we do a two-sided test, we need to find the absolute value of the t value by ourselves.

How to find 𝒕𝒏−𝟏,

𝜶 𝟐

= 𝒕𝟏𝟏,

𝟎.𝟎𝟐𝟓

in R?

~ 15 ~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

QUESTION ESTION The average length of time for students to register for classes at a certain college has been 46 minutes. A new registration procedure using modern computing machines is being tried. If a random sample of 12 students had an average registration time of 42 minutes with a standard deviation of 11.9 minutes under the new system. Test the hypothesis that the population mean length of time under the new system is now less than 46. Use a 0.05 level of significance with the assumption that the data are from a normal distribution.

~ 16 ~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

𝟐

(NOR ORM MAL C CAS AS ASE E) 82 FisORthe MUL AT IO NO ES T SSTA TA TEME NT A BOU T 𝝈𝑿 ULAT ATIO ION OFF A TTES EST TATEME TEMEN AB OUT test statistic we would use to formulate the test statement about 𝜎2 . 𝑋 𝑆𝑛−1 2 2 Similar to our procedure of using 𝑆𝑛−1 to construct a random interval for 𝜎𝑋 and then get a confidence interval in Chapter 4, we would use the following theoretical result to get the test statement:

If 𝑿 ∼ 𝑵(𝝁𝑿 , 𝝈𝟐𝑿), then

(𝒏 − 𝟏)𝑺𝟐𝒏−𝟏 𝝈𝟐𝑿

𝟐 ∼ 𝝌𝒏−𝟏

which means that the random variable on the left follows a 𝝌𝟐 distribution with 𝒏 − 𝟏 degrees of freedom. Therefore, we can write down the following general result

1. One-sided right test: Consider

𝑯𝟎 : 𝝈𝑿𝟐 = 𝝈𝟎𝟐 { 𝑯𝟏 : 𝝈𝑿𝟐 > 𝝈𝟎𝟐

.

Intuitively, we reject 𝐻0 if 𝑠2𝑛−1 > 𝑐. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if

2 (𝑛−1)𝑠𝑛−1

~ 17 ~

𝜎20

2 > 𝜒𝑛−1,𝛼

.

MATH2411: Applied Statistics | Dr. YU, Chi Wai

2. One-sided left test: Consider

𝑯𝟎 : 𝝈𝑿𝟐 = 𝝈𝟐 𝟎 { 𝑯𝟏 : 𝝈𝑿𝟐 < 𝝈𝟎𝟐

.

Intuitively, we reject 𝐻0 if 𝑠2𝑛−1 < 𝑐. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if

3. Two-sided test: Consider

2 (𝑛−1)𝑠𝑛−1

𝜎20

2 < 𝜒𝑛−1,1−𝛼

𝑯𝟎 : 𝝈𝑿𝟐 = 𝝈𝟎𝟐 { 𝑯𝟏 : 𝝈𝑿𝟐 ≠ 𝝈𝟎𝟐

.

.

2 Intuitively, we reject 𝐻0 if 𝑠2𝑛−1 < 𝑎 OR 𝑠𝑛−1 > 𝑏. Consequently, we would

Reject 𝐻0 at a significance level 𝛼 if

2 (𝑛−1)𝑠𝑛−1

𝜎20

2 < 𝜒𝑛−1,1− 𝛼 2

OR

(𝑛−1)𝑠2𝑛−1

𝜎02

~ 18 ~

2 > 𝜒𝑛−1, 𝛼 . 2

MATH2411: Applied Statistics | Dr. YU, Chi Wai

QUESTION ESTION A manufacturer of car batteries claims that the life of his batteries is normally distributed with a standard deviation equal to 0.9 year. If a random sample of 10 of these batteries has a standard deviation of 1.2 years, do you think that 𝜎 > 0.9 year? Use a 0.05 level of significance to draw a conclusion.

How to find 𝝌𝟐𝒏−𝟏,𝜶 = 𝝌𝟗𝟐,𝟎.𝟎𝟓 in R?

~ 19 ~

MATH2411: Applied Statistics | Dr. YU, Chi Wai

9 TWO-SI ORMAL AL CASE) SIDE DED YPO THES ESIS STIN ING NFIDE DENCE TERVAL DE D HYP OTH ES IS TEST IN G VS CONFI DE NCE INTER VAL (NORM Recall that when we do a two-sided test for 𝜇𝑋 with unknown 𝜎𝑋2 , i.e. test

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎 {𝑯𝟏 : 𝝁𝑿 ≠ 𝝁𝟎

at a significance level 𝛼,

we would reject 𝐻0 at a significance level 𝛼 if

−𝜇 𝑥

| 𝑠𝑛−10 | > 𝑡 √𝑛

𝛼 . 𝑛−1,2

Or equivalently, we would NOT REJECT 𝑯𝟎 at a significance level 𝛼 if

𝑥 − 𝜇0 | ≤ 𝑡𝑛−1,𝛼 | 𝑠 𝑛−1 2 𝑛 √

𝝁𝟎 ∈  𝒙 ± 𝒕𝒏−𝟏,𝜶

This is exactly saying that

𝟐

~ 20 ~

𝒔𝒏−𝟏 √𝒏

.

MATH2411: Applied Statistics | Dr. YU, Chi Wai

Thus, we have the following approach to use C.I. to draw a conclusion for twosided testing of hypothesis:

Given that [a,b] is the 100(𝟏 − 𝜶)% C.I. for 𝜇𝑋 . Thus, to test

{

𝑯𝟎 : 𝝁𝑿 = 𝝁𝟎

𝑯 𝟏 : 𝝁𝑿 ≠ 𝝁𝟎

at a significance level 𝜶, it suffices to check if

𝝁𝟎 is inside OR outside [a, b].

Remark

 Inside  Not reject H0 at the level of significance 𝜶

 Outside  Reject H0 at the level of significance 𝜶

 Note tha...