Using the TI 83/84

Summary Statistics

•Press STAT, ENTER (for EDIT)•If you do not see L1, scroll left. If you still don’t see it, press INSERT, 2nd, L1 (this is the 1 key)•If there are old data under L1:

– Press the up arrow, then CLEAR, ENTER– DON’T use DELETE, ENTER. This causes L1 to disappear

instead. Use INSERT, 2nd, L1 to get it back.

•Enter data values in L1 one at a time, pressing ENTER after each.

– If you make an error, use the up or down arrows to highlight the error, then enter the correct value. Use the arrows to get to the bottom of the list for the next value, if necessary.

– Be sure to press ENTER after the last data value.

Summary Statistics, Continued

• Press STAT, Right Arrow (for CALC), ENTER

• Press ENTER (for 1-Var Stats)

• Press ENTER again

• Read results– The Standard Deviation is labeled Sx

Normal Probablilities

• Find P(90 < x < 105) if x follows the normal model with mean 100 and standard deviation 15:

• P(90 < x < 105)

= normalcdf( 90 , 105 , 100 , 15)

= .378

x1 x2

Normal Quantiles

• We must find a so that P(x < a) = 2% when x has a normal distribution with a mean of 100 and a standard deviation of 15.

• With the TI 83/84: a = invNorm( .02, 100 , 15) = 69.2

x

Calculating r and Regression Coefficients

• The first time you do this:– Press 2nd, CATALOG (above 0)– Scroll down to DiagnosticOn– Press ENTER, ENTER– Read “Done”– Your calculator will remember this setting

even when turned off

Calculating r and Regression Coefficients Continued

• Press STAT, ENTER• If there are old values in L1:

– Highlight L1, press CLEAR, then ENTER

• If there are old values in L2:– Highlight L2, press CLEAR, then ENTER

• Enter predictor (x) values in L1• Enter response (y) values in L2

– Pairs must line up

• Press STAT, > (to CALC)• Scroll down to LinReg(ax+b), press ENTER,

ENTER• Read a, b, r and r2

Binomial Probabilities• To find binomial probability distribution values –

probabilities for a particular number of successes, say 3 successes in 5 trials with the chance of success = .9:– Let n = 5, p = .9, and x = 3 (Note the order: n, p, x)– Press 2nd, DISTR [VARS]– Scroll down to 0 and press ENTER, or just press 0– “binompdf(“ appears. Press 5, .9, 3 ) ENTER

(Note that the commas are required, and note the order: n, p, x)– .0729 appears– If the values for x = 2, 3, and 4 are all needed

• Press 2nd, DISTR [VARS]• Scroll down to 0 and press ENTER, or just press 0• “binompdf(“ appears. Press 5, .9, {2, 3, 4} ) ENTER• All 3 values appear

– In general, press 2nd, DISTR, 0, then enter n, p, x

Binomial Probabilities Continued

• To find the binomial probability for a range of successes, say 3 or fewer successes in 5 trials with the chance of success = .9 :– Let n = 5, p = .9, and x = 3– Press 2nd, DISTR [VARS]– Scroll down to A and press ENTER– “binomcdf(“ appears. Press 5, .9, 3 ) ENTER

(Note that the commas are required)• .0815 appears

– If the probability for fewer than 3 successes is needed: • Enter binomcdf(5, .9, 2 ); read .00856

– If the probability for more than 3 successes is needed: • Enter 1 - binomcdf(5, .9, 3 ); read .919

Confidence Intervals for Proportions• Press STAT and use the cursor

to highlight TESTS. • Scroll down to A: 1-PropZInt…

and press ENTER.• The TI remembers your previous

problem and shows the entries for it. You will overwrite these with the new entries. After x: enter the number of successes in the sample.

• After n: enter the sample size.• After C-Level: enter the

confidence level as a decimal fraction.

• When the cursor blinks on Calculate, press ENTER again.

11

Hypothesis Test for Proportions• Of 4276 households sampled, 4019 had telephones. Test, at the 1%

level, the claim that the percentage of households with telephones is now greater than 93%.

Claim: p > .93

H0: p < .93

(or p = .93)

HA: p > .93

Press STAT, TESTS, 1-PropZTest,

Enter

0

0

After p : enter .35

(the null hypothesis value)

After x : enter 4019

(number of successes in the sample

After n : enter 4276

(the sample size)

After prop highlight > p

(this is )

Highlight CALCULATE

Press ENTER

AH

.93 = .01

12

Hypothesis Test for Proportions, Continued• Of 4276 households sampled, 4019 had telephones. Test the claim

that the percentage of households with telephones is greater than the 93%.

H0: p < .93

(or p = .93)

HA: p > .93

Read

1-PropZTest

prop > .35

80.87315928

0

ˆ .9398971001

4276

z

p

p

n

Test Performed

HA

Sample Size

.93

2.54 z-score

.940Sample Estimate for p

.00560 p-Value

13

Example: For the racial profiling sample data, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped.

We have that:

nB= 200xB = 24nW = 1400xW = 147

And,H0: pB ≤ pW HA: pB > pW

xz

On the TI-83/84:

Press STAT, arrow to TESTS and scroll down to 2-PropZTest. Press Enter.

Make these entries:

Highlight Calculate and press ENTER

Hypothesis Test for Two Proportions

14

Hypothesis Test for Two Proportions, Continued

Your TI screen should look like this:

This is HA

This is the p- value.

This is p

If you press the down arrow twice, the TI will tell you the sample sizes, as a check.

15

n = 106

y = 98.20o

s = 0.62o

= 0.05

Confidence Interval for a Mean

16

Hypothesis Test for a Mean

0

1

: 1017

: 1017

: 1017

Claim

H

H

83/ 84 entries:

STAT TESTS 2

Stats Enter

TI

0: 1017: 1040x: 207xs: 20n

0: Enter

1

TI 83/ 84 entries (continued):

Highlight Calculate

Enter

Read

>1017 ( )

.4969...

.312787...

1040

207

20x

H

t

p

x

s

n

Press STAT and arrow over to TESTS.

Scroll down to 2-SampTTest and press ENTER.

Highlight STATS and press ENTERSet μ0: 0

Enter McGwire’s stats for sample 1

Enter Bonds’s stats for sample 2

Hypothesis Test for Two Means

McGwire

n1 70

x1 418.5

s1 45.5

Bonds

n2 73

x2 403.7

s2 30.6

Highlight HA as μ1: ≠ μ2

Your screen should look like this:

Select Pooled: No

Select Calculate

Press ENTER

Hypothesis Test for Two Means, Continued

Your screen should look like this:

HA

Test StatisticP-Value

Scrolling down reveals some to the statistics you entered earlier. The sample data support the

claim that there is a difference between the mean home run distances of Mark McGwire and Barry Bonds.

Because the P-value is less than .05, we reject The Null Hypothesis

Using the sample data given in the preceding example, construct a 95%confidence interval estimate of the difference between the mean home run distances of Mark McGwire and Barry Bonds. We need E so that

Confidence Interval for Two Means

(x1 – x2) – E < (µ1 – µ2) < (x1 – x2) + E

As before, Press STAT, go to TESTS, and now scroll down to 2-SampTInt and press ENTER.

The TI remembers the entries from the last time 2-SampTTest or 2-SampTInt was used. Since these are the same, we need only scroll down to C-Level to make sure it is set to .95 (NOT 95%!). Pooled is always No for our work.

Highlight CALCULATE and press ENTER.

Matched Pairs Test

TI-83/84 users may test paired sample hypotheses from raw data as follows:STAT EDIT Highlight L1 CLEAR ENTER

Enter first half of each pair in L2 Enter second half of each pair in L3

2nd QUIT 2nd L2 – 2nd L3 STO> 2nd L1

STATTESTS2: T-TestHighlight DataENTERProceed as for one-samplet - Test

Highlight L2 CLEAR ENTERHighlight L3 CLEARENTER

H0: d = 0

H1: d 0

Claim: there is a difference between the actual low temperatures and the low temperatures that were forecast five days earlier

That is, μd ≠ 0

Matched Pairs Test, Continued

Matched Pairs Confidence Interval

Follow the instructions given above for entering the 2 samples into L2 and L3, then storing the difference in L1.

We now need only compute a one-sample confidence interval using STATTESTSTIntervalENTERDataList: L1Freq: 1C-Level: .95CalculateENTER

Goodness-of-Fit Test

1. STAT, EDIT, CLEAR L1, L2 and L3

2. Enter observed counts in L2.

2

4. 2nd Quit

Store

(Obs-Exp)in L1

Exp

5. STAT, CALC, 1-Var Stats, ENTER.Write down value of Σx

6. 2nd, DISTR, 8:X2cdf, ENTER, Σx, 1000, k-1)

Read P-Value

3. Enter expected counts in L3.

Test for HomogeneityClaim: There is no difference in the distribution of the continent of origin for student and staff cars.

H0: There is no difference in the distribution of the continent of origin for student and staff cars.HA: There is a difference in the distribution of the continent of origin for student and staff cars.

Set α = .05