John Wiley & Sons TI-83 User Manual download pdf (Page 68)

Solutions

1. Since we do not have the population standard deviations, we will use a T-based confidence

interval, 2-SampTInt, and select Stats. We enter the sample statistics and choose No for the

Pooled prompt since we do not know that the population standard deviations are equal.

The confidence interval is (3.82, 7.38).

2. This is paired data, so we need to examine the differences of the values. We enter the two

sets of measurements into L1 and L2, and take their difference as L3. Then we apply T-

Test to see if the population mean of the differences is 0. We choose Data and set µ

= 0,

use L3 as our list, and our alternative hypothesis is that µ øµ0. The p-value is 0.002,

which is less than our significance level of 5%. We reject the null hypothesis and have

evidence that the means of the two populations do differ.

3. We want to perform a significance test on two different population proportions so we use

2-PropZTest. We will test H

: p1 = p2 versus H

: p1 < p2. We have n

and n

= 143

but we do not have x

and x

, so we will need to compute them. x

= p

= 23%(89) =

20 and x

= p

= 37%(143) = 53. Note that we round x

and x

to the nearest integers;

otherwise we would receive an error message when using 2-PropZTest.

The p-value is 0.01, which is less than our significance level 5%. We reject H

and

conclude that p

< p

4. We will use 2-StampTTest since we do not know the population standard deviations. We

do not used pooled standard deviations since we do not have any reason to assume that the

populations standard deviations are equal. The p-value rounds to 0, so we reject the null

hypothesis and conclude that µ1 > µ2.

5. We want to compute a confidence interval for the difference between two different

population proportions, so we use 2-PropZInt. x

= 74%(200) = 148 and x

= 65%(195) =

127. The confidence interval is (.013, .165).

6. This is paired data, so we need to examine differences of the values. We enter the two sets

of measurements into L1 and L2, and take their difference as L3. Then we apply TInterval

to compute the confidence interval, selecting Data and using L3 as our list. The confidence

interval is (-64.02, 50.02).

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John Wiley & Sons TI-83 User Manual Page 68

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