section 7.2 tables 7.2-1 and 7.2-2 summarize hypothesis tests about one mean. (these tests assume...

Section 7.2

Tables 7.2-1 and 7.2-2 summarize hypothesis tests about one mean. (These tests assume that the sample is taken from a normal distribution, but the tests are robust when the sample is taken from a non-normal distribution with finite mean and variance.)

The relationship between hypothesis testing and confidence intervals can be described by saying that

rejecting the hypothesized value of a parameter with significance level is expected to correspond to that hypothesized value not being contained in the 100(1 – )% confidence interval,

not rejecting the hypothesized value of a parameter with significance level is expected to correspond to that hypothesized value being contained in the 100(1 – )% confidence interval.

Confidence intervals have the following properties:(1)

(2)

Hypothesis tests have the following properties:(1)

(2)

For fixed sample size n, the length of the confidence interval tends to be __________ as the confidence level 100(1 – )% is increased, and the length of the confidence interval tends to be __________ as the confidence level 100(1 – )% is decreased.

For fixed confidence level 100(1 – )%, the length of the confidence interval tends to be __________ as the sample size n is increased, and the length of the confidence interval tends to be __________ as the sample size n is decreased.

smaller

smaller

larger

larger

For fixed sample size n, as the significance level is decreased, a type II error probability is __________, and as the significance level is increased, a type II error probability is __________.

For fixed significance level , as the sample size n is increased, a type II error probability is __________, and as the sample size n is decreased, a type II error probability is __________.

increaseddecreased

increaseddecreased

1. It is assumed that the cereal weight in a “10-ounce box” is N(, 2). To test H0: = 10.1 versus H1: > 10.1 with a 0.05 significance level, a random sample of size n = 16 boxes is selected where it is observed that x = 10.4 and s = 0.4.

(a) Calculate the appropriate test statistic, describe the appropriate critical region, and write a summary of the results.

The test statistic is t =

The one-sided critical region with = 0.05 is

x – 10.1——— = s /n

t 1.753 .Since t = 3.000 > t0.05(15) = 1.753, we reject H0. We conclude

that the mean weight of cereal per box is greater than 10.1 ounces.

10.4 – 10.1————— = 0.4 /16

3.000

(b)

The p-value of the test is

P(T 3.000) is

slightly less than 0.005.

X – 10.1——— 3.000 ; = 10.1 = S /n

P

a t(15) random variable

Find the approximate p-value of this hypothesis test.

2.

(a)

Use the Analyze > Compare Means > One-Sample T Test options in SPSS to do Text Exercise 7.2-4.

The test statistic is t =

The two-sided critical region with = 0.05 is

x – 7.5——— = s /n

|t| 2.262 .

x – 7.5——— s /10

2.262 = t0.025(9)

– 2.262

(b) One-Sample Statistics

10 7.5500 .10274 .03249thickN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

1.539 9 .158 .05000 -.0235 .1235thickt df Sig. (2-tailed)

MeanDifference Lower Upper

95% ConfidenceInterval of the

Difference

Test Value = 7.5

From the SPSS output, we find t = 1.539 .

(c)

Since t = 1.539 < t0.025(9) = 2.262, we fail to reject H0. We conclude that the mean thickness for vending machine spearmint gum is not different from 7.5 hundredths of an inch.

From the SPSS output, we find that the limits of the 95% confidence interval for are 7.4765 and 7.6235. The hypothesized mean = 7.5 is contained in these limits, just as we would expect.

Note that to calculate the p-value of this test, we need to find

X – 7.5——— 1.539 ; = 7.5 S /n

P

3.

(a)

(b)

(c)

(d)

Use the Analyze > Compare Means > One-Sample T Test options in SPSS to do Text Exercise 7.2-6.

H0: = 3.4

H1: > 3.4

The test statistic is t =x – 3.4——— = s /n

x – 3.4——— s /9

The one-sided critical region with = 0.05 is t 1.860 .

1.860 = t0.05(8)

(e)

(f)


One-Sample Statistics

9 3.5556 .16667 .05556litersN Mean Std. Deviation

Std. ErrorMean

One-Sample Test

2.800 8 .023 .15556 .0274 .2837literst df Sig. (2-tailed)

MeanDifference Lower Upper


Difference

Test Value = 3.4

Since t = 2.800 > t0.05(8) = 1.860, we reject H0. We conclude that the mean FVC for the volleyball players is greater than 3.4 liters.

3.-continued

(g) The p-value of the test is

P(T 2.800) is

between 0.01 and 0.025, from Table VI in Appendix B.

X – 3.4——— 2.800 ; = 3.4 = S /n

P

From the SPSS output, we find the p-value of the test to be0.023 / 2 = 0.0115.


4.

(a)

Use the Analyze > Descriptive Statistics > Descriptives options in SPSS to do Text Exercise 7.2-8. We shall perform this hypothesis test about the mean under the assumption that the variance in weight of home-born babies is 5252.

The test statistic is z =

The one-sided critical region with = 0.01 is

x – 3315———— = 525 /n

z 2.326 .

x – 3315———— 525 /11

(b)

(c)

Descriptive Statistics

11 2657.00 3856.00 3385.9091 336.31606 113108.5

11

grams

Valid N (listwise)

N Minimum Maximum Mean Std. Deviation Variance

From the SPSS output, we find x = 3385.91 . x – 3315z = ———— = 525 /11

0.448

Since z = 0.448 < z0.01 = 2.326, we fail to reject H0. We conclude that the mean weight of home-born babies is not greater than 3315 grams.


P(Z 0.448) =

1 – (0.448) = 1 – 0.6736 = 0.3264 .

X – 3315———— 0.448 ; = 3315 = 525 /n

P

5. Note that a hypothesis test concerning the mean difference between two random variables measured on the same units is essentially a one-sample test about a mean when the data consist of differences between paired observations. Use the Analyze > Descriptive Statistics > Paired-Samples T Test options in SPSS to do Text Exercise 7.2-14.

Paired Samples Statistics

256.1765 17 18.70239 4.53600

251.4118 17 14.93343 3.62189

dist_a

dist_b

Pair1

Mean N Std. DeviationStd. Error

Mean

Paired Samples Test

4.76471 9.08659 2.20382 .09281 9.43660 2.162 16 .046dist_a - dist_bPair 1Mean Std. Deviation

Std. ErrorMean Lower Upper


Difference

Paired Differences

t df Sig. (2-tailed)

The test statistic is t =d – 0—— =sd /n

d – 0———sd /17

The one-sided critical region with = 0.05 is t 1.746 .


Since t = 2.162 > t0.05(16) = 1.746, we reject H0. We conclude that the mean distance is greater with brand A golf balls than with brand B.


P(T 2.162) is

between 0.01 and 0.025, from Table VI in Appendix B.

D – 0——— 2.162 ; D = 0 =SD /n

P

From the SPSS output, we find the p-value of the test to be0.046 / 2 = 0.023.


section 7.2 tables 7.2-1 and 7.2-2 summarize hypothesis tests about one mean. (these tests assume...

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