Consider the following computer output for an experiment to test effect of factor treatments and blocks. Use alpha = 0.05. Analysis of Variance Source of Variance df SS MS F P-value Factor ? 126 63 ? ? Block ? 54 18 ? ? Error 6 ? 2.8 Total 11 197 1) Find the degree of freedom for the factor and the block. (2 pts) 2) Find the number of levels of the factor and the number of blocks used in this experiment. (2 pts) 3) Estimate the P-value for the F-statistic of the factor. (2 pts) 4) Estimate the variance of the error term, sigma square hat. (2 pts) 5) Since the P-value for the F-statistic of the block is less than alpha = 0.05, there is no effect of the block in this experiment. True or False. (2 pts)

Table 2.2: Differences in runstitching times (standard − ergonomic).

1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42

.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43

-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82

A paired t-test is the standard procedure for testing this null hypothesis.

We use a paired t-test because each worker was measured twice, once for Paired t-test for

each workplace, so the observations on the two workplaces are dependent. paired data

Fast workers are probably fast for both workplaces, and slow workers are

slow for both. Thus what we do is compute the difference (standard − er-

gonomic) for each worker, and test the null hypothesis that the average of

these differences is zero using a one sample t-test on the differences.

Table 2.2 gives the differences between standard and ergonomic times.

Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-

ences in the sample. We assume that these differences are independent sam-

ples from a normal distribution with mean µ and variance σ

2

, both unknown.

Our null hypothesis is that the mean µ equals prespecified value µ0 = 0

(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the

workers to be faster in the ergonomic workplace.

The formula for a one sample t-test is

t =

¯d − µ0

s/√

n

,

where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test

sample size, and s is the sample standard deviation (of the differences)

s =

vuut

1

n − 1

Xn

i=1

(di − ¯d )

2 .

If our null hypothesis is correct and our assumptions are true, then the t-

statistic follows a t-distribution with n − 1 degrees of freedom.

The p-value for a test is the probability, assuming that the null hypothesis

is true, of observing a test statistic as extreme or more extreme than the one The p-value

we did observe. “Extreme” means away from the the null hypothesis towards

the alternative hypothesis. Our alternative here is that the true average is

larger than the null hypothesis value, so larger values of the test statistic are

extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of

freedom from the observed t-value to the right. (If the alternative had been

µ < µ0, then the p-value is the area under the curve to the left of our test

Step-by-step explanation: Through the laid down FORMULA ABOVE

this story was prepared by jim

mccullaugh in los angeles and

al stewart in new york.

new york the colossal success of "top gun" appears to have dramatically altered madison avenue's

perception of corporate sponsorship

of home video projects.

the paramount home video re- lease, featuring a 60- second diet

pepsi spot, was by no means the

first prerecorded video to carry an advertiser's plug, but it was the

first major theatrical release to do so. in the aftermath of the title's

success, paramount is widely ex- pected to ink a similar deal for

"crocodile dundee" or "star trek

iv" or both.

corporate america's new attraccds push british

recording sales

to 19% gain

london fueled by the compact

disk boom, the value of u.k. manu- facturer shipments of recordings

rose 19% in the 12 -month period ended march 1987, according to the

british phonographic industry.

total value of u.k. shipments for

the period was $755 million for all

configurations. (all value figures

are based on an exchange rate of

$1.68 to the pound sterling.)

cd sales totaled 10.7 million units

worth $121.3 million- compared

with 3.7 million units worth $39.5

million in the prior 12 months -and

accounted for two -thirds of the

overall gain in trade value. at present growth rates, says the bpi, cd

sales for calendar 1987 will reach 20

(continued on page 79)

step-by-step explanation: