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Using the TI 83/84

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Summary Statistics Press STAT, ENTER (for EDIT) If you do not see L1, scroll left. If you still don’t see it, press INSERT, 2 nd, 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, 2 nd, 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.

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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

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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 x1x1 x2x2

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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

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Calculating r and Regression Coefficients The first time you do this: –Press 2 nd, CATALOG (above 0) –Scroll down to DiagnosticOn –Press ENTER, ENTER –Read “Done” –Your calculator will remember this setting even when turned off

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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 r 2

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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 2 nd, 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 2 nd, 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 2 nd, DISTR, 0, then enter n, p, x

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

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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.

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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 H 0 : p <.93 (or p =.93) H A : p >.93 Press STAT, TESTS, 1-PropZTest, Enter.93 =.01

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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%. H 0 : p <.93 (or p =.93) H A : p >.93 Test Performed HAHA Sample Size z -score.940 Sample Estimate for p p -Value

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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: n B = 200 x B = 24 n W = 1400 x W = 147 And, H 0 : p B ≤ p W H A : p B > p W x z 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

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14 Hypothesis Test for Two Proportions, Continued Your TI screen should look like this: This is H A 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.

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15 n = 106 y = o s = 0.62 o = 0.05 Confidence Interval for a Mean

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16 Hypothesis Test for a Mean

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Press STAT and arrow over to TESTS. Scroll down to 2-SampTTest and press ENTER. Highlight STATS and press ENTER Set μ 0 : 0 Enter McGwire’s stats for sample 1 Enter Bonds’s stats for sample 2 Hypothesis Test for Two Means McGwire n 1 70 x s Bonds n 2 73 x s Highlight H A as μ1: ≠ μ2 Your screen should look like this: Select Pooled: No Select Calculate Press ENTER

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Hypothesis Test for Two Means, Continued Your screen should look like this: H A Test Statistic P-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

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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 (x 1 – x 2 ) – E < (µ 1 – µ 2 ) < (x 1 – x 2 ) + 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.

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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 2 nd QUIT 2 nd L2 – 2 nd L3 STO> 2 nd L1 STAT TESTS 2: T-Test Highlight Data ENTER Proceed as for one-sample t - Test Highlight L2 CLEAR ENTER Highlight L3 CLEAR ENTER

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H 0 : d = 0 H 1 : d 0 Claim: there is a difference between the actual low temperatures and the low temperatures that were forecast five days earlier That is, μ d ≠ 0 Matched Pairs Test, Continued

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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 STAT TESTS TInterval ENTER Data List: L1 Freq: 1 C-Level:.95 Calculate ENTER

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Goodness-of-Fit Test 1. STAT, EDIT, CLEAR L1, L2 and L3 2. Enter observed counts in L2. 5. STAT, CALC, 1-Var Stats, ENTER. Write down value of Σx 6. 2 nd, DISTR, 8: X 2 cdf, ENTER, Σx, 1000, k-1) Read P -Value 3. Enter expected counts in L3.

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Test for Homogeneity Claim: There is no difference in the distribution of the continent of origin for student and staff cars. H 0 : There is no difference in the distribution of the continent of origin for student and staff cars. H A : There is a difference in the distribution of the continent of origin for student and staff cars. Set α =.05

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