[PDF] Using the TI-83/84 Plus Chapter 9: Hypothesis Testing - Two Samples

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Using the TI-83/84 Plus

Chapter 9: Hypothesis Testing - Two Samples

Here we see how to use the TI 83/84 to conduct hypothesis tests about mean dierences, dierences in means,

and dierences in proportions between two samples. The software will calculate the test statistic and the

P-value for the test statistic. It does not give you the critical value. For tests about means, you can either

input raw data via a list or simply enter the sample statistics. In all cases you will need to input a value from

the null hypothesis and whether the test is left, right, or two-tailed. All of these test functions can be found by pressing theSTATbutton and highlightingTESTS. Each topic has its own page or you can go directly to the videos.

Chapter 9.1 - Tests for Mean Dierences: Paired DataT-Test2Play Video.Chapter 9.2 - Tests for Two Means: Independent Data2-SampTTest3Play Video.Chapter 9.3 - Tests for Two Proportions2-PropZTest4Play Video.Entering Data Into Lists5Play Video.

2

Tests for Mean Dierences - Paired Data:T-TestPlay Video.This is just a T-Test where the mean dierence and standard deviation of dierences are used.

1.

P ressthe STATbutton and highlightTESTS.

2.

S crolldo wnto 2:T-Test:::

3.

Hig hlightDataorStats

Data: If you chooseDatayou need to create a list of sample dierences and reference this list. Then you have to test on this single list by choosing two-tailed (6=0), left-tailed ( < 0) or right-tailed ( > ). If your list of dierences is in standard form (not a frequency distribution), setFreq:to 1. Stats: If you choose,Stats, you have to enter the mean dierence from the null hypothesis (0), the sample mean (dierence) (x), the sample standard deviation for the dierences (Sx), the sample size (number of pairs) (n), and select two-tailed (6=0), left-tailed ( < 0) or right-tailed ( > 0). 4.

Hig hlightCalculateand hitENTER

5.

It giv esy outhe test statistic ( t) and theP-value for the test statistic (p) based on your choice of

two-tailed, left-tailed, or right-tailed test (so make sure you entered that correctly). Warning:TheP-value may be given in scientic notation.

For example2.143E6= 2.143106= 0.000002143.

Example:Here is the data for the cholesterol levels of 10 men diagnosed with high cholesterol. The rst row

gives the cholesterol levels before taking a certain medication (x). The second row gives the levels after one

year of regular medication (y). I put in a third row forxy, which will be the actual data we analyze and

our problem boils down to a one-population test. Here, the single population is that of dierences. Test the

claim that, on average, the medication lowers cholesterol for all men diagnosed with high cholesterol. Test

this claim at the 0.01 signicance level.Cholesterol Levels in mg/dLmeans 2s Before (x)237289257228303275262304244233263.2811.128.5 After (y)194240230186265222242281240212231.2864.029.4 d=xy434927423853202342132.0238.015.4

Here we are claiming thed>0 whered=yx.Claim:d>0H

o:d= 0H

1:d>0Getting There:STATInputOutput

Conclusion:TheP-value (5.131050:0000513) is less than(0.01). We reject the null hypothesis and the data supports the claim. 3

Tests for Two Means - Independent Data:2-SampTTestPlay Video.1.P ressthe STATbutton and highlightTESTS.

2.

S crolldo wnto 4:2-SampTTest:::

3.

Hig hlightDataorStats

Data: If you chooseDatathe calculator does all the work but you have to select two-tailed (1

6=2), left-tailed (1< 2) or right-tailed (1> 2) and then selectYesorNotoPooled(this

asks if you want to pool the variances -Nois the safer answer). If your data is a standard list, set Freq:to 1. If your data is entered as a frequency distribution, put in the list which contains the frequencies. Stats: If you choose,Stats, you have to enter the the sample means (x1 and x2), the sample standard deviations (Sx1 and Sx2), the sample sizes (n1 and n2), and select two-tailed (16=2), left-tailed (1< 2) or right-tailed (1> 2). SelectNoforPooled. 4.

Hig hlightCalculateand hitENTER

5.

It giv esy outhe test statistic ( t) and theP-value for the test statistic (p) based on your choice of

two-tailed, left-tailed, or right-tailed test (so make sure you entered that correctly). Warning:TheP-value may be given in scientic notation.

For example2.143E6= 2.143106= 0.000002143.

Demonstration Example:Here is the data for the cholesterol levels of men diagnosed with high cholesterol.

This time we assume the rst row comes from 10 men who don't use the drug (x1) and the second row comes

from 10dierentmen who took the drug for one year (x2). Test the claim that the mean cholesterol level

for all men who use the drug is less than the mean for those who do not use the drug. Use a 0.05 signicance

level.Cholesterol Levels in mg/dLmeans 2s No Drug (x1)237289257228303275262304244233263.2811.128.5 Drug (x2)194240230186265222242281240212231.2864.029.4 Here we are claiming that2< 1which means1> 2or12>0.Claim:12>0H o:12= 0H

1:12>0Getting There:STATInput (1)Input (2)Output

Conclusion:TheP-value (0.01184) is less than(0.05). We reject the null hypothesis and the data supports

the claim. 4 Tests for Two Proportions:2-PropZTestPlay Video.1.P ressthe STATbutton and highlightTESTS. 2.

S crolldo wnto 6:2-PropZTest:::

3. Y ouha veto en terthe n umberof succ essesin samples 1 and 2 ( x1andx2), the number of trials (or sample sizes) (n1andn2), then select two-tailed (p16=p2), left-tailed (p1 p2). 4.

Hig hlightCalculateand hitENTER

5.

It giv esy outhe test statistic ( z) and theP-value for the test statistic (p) based on your choice of

two-tailed, left-tailed, or right-tailed test (so make sure you entered that correctly). Warning:TheP-value may be given in scientic notation.

For example2.143E6= 2.143106= 0.000002143.

Example:A popular cold-remedy was tested for it's ecacy only this time the control group took a placebo

remedy. In a sample of 150 people who took the remedy upon getting a cold, 117 (78%) had no symptoms one

week later. In a sample of 120 people who took the placebo upon getting a cold, 90 (75%) had no symptoms

one week later. The table summarizes this information.# who are Symptom Freetotal #proportion groupafter one week (x)in group (n)^p=x=nRemedy1171500.78

Placebo901200.75

The Test:Test the claim that the proportion of all remedy users who are symptom-free after one week is

greater than the proportion for placebo users. Test this claim at the 0.05 signicance level. Letp1represent the proportion of all remedy-users who are symptom-free after one week andp2represent the proportion of all placebo-users who are symptom-free after one week.Claim:p1p2>0H o:p1p2= 0H

1:p1p2>0Getting There:STATInputOutput

Conclusion:TheP-value (0.2812) is greater than(0.05). We fail to reject the null hypothesis. There is

not enough data to support the claim that the proportion of all remedy users who are symptom-free after

one week is greater than the proportion for placebo users. 5 Entering Data into Lists:Play Video.1.P ressthe STATbutton. 2. Hig hlightthe EDIToption (using the arrows) and hitENTER. 3. C hoosea list (from L1, L2;:::, L6) using the arrows and enter the values by column. HitEnteror the down-arrow to move down the column. 4. Y oucan cleara list by highlighting the list name and hitting theClearbutton.quotesdbs_dbs10.pdfusesText_16

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