PROBLEM 8.58

Since each person is used as their own control, a paired t-test should be used. hypotheses (one sided paired t-test) H0 : µd = 0 HA : µd > 0

PROBLEM 8.59

Under Ho, the t-distribution has df = 7 and tobserved = 0.263 The critical value at an alpah = 0.05 with df = 7, t = 1.86

Because 0.263 < 1.86 (tobs < t), we fail to reject the null hypothesis. That is, the mean decline in urinary salt excretion over a 1-week period is not significantly different from 0.

x <- 1.14
s <- 12.22
n <- 8 

t <- x/(s/sqrt(n))
t
## [1] 0.2638631
abs(qt(0.05,7))
## [1] 1.894579

PROBLEM 8.60

The 95% C.I. at an alpha level of .05 is (-9.076176, 11.356176).

se <- s/sqrt(n)
x + c(-1,1)*qt(0.975, 7) * se
## [1] -9.076176 11.356176

PROBLEM 8.67

The best test to use in this cause is the independent two-sample t-test with unequal variances.

PROBLEM 8.68

H0 : µ1 = µ2 HA : µ1 != µ2

From this, we obtain the test statistic t=1.08 and df=32.2 (so df= 32).

The value for a t test with df-2=30 at an alpha level of 0.05 is 2.04

Because the tobs = 1.08 < t = 2.04, we reject the null hypothesis. The two means are identical.

t30,.85 = 1.055, and t30,.9 = 1.310 and the p-value is between 0.2 < p < 0.3

p = 0.28

x1x2 <- (0.101-0.042)
mu1mu2<-0
s1 <- 0.253
s2 <- 0.106
n1 <-25
n2 <-25

t <- (x1x2-mu1mu2)/sqrt((s1^2/n1) + (s2^2/n2))
t
## [1] 1.075433
d <- ((s1^2/n1)+(s2^2/n2))^2/(((s1^2/n1)^2)/(n1-1) + ((s2^2/n2)^2)/(n2-1))
d
## [1] 32.17395
qt(0.975,30)
## [1] 2.042272
2*(0.1)
## [1] 0.2
2*(0.15)
## [1] 0.3
x <- pnorm(1.08, lower.tail = FALSE)
2*x
## [1] 0.2801422

PROBLEM 8.69

A 90% C.I. would be narrower than a 95% C.I. This is because we are less confident that our true mean would lie within our interval.

PROBLEM 8.70

We would use a two sample t test with equal variances.

Here we assume: -the distribution of balance scores is normal in each group with the same variance -Under H0, the means of the two groups are assumed to be the same -Under HA, the means of the two groups are assumed to be different

PROBLEM 8.71

sp^2 = 8.74 and t = 1.25

t actual is 2.0002

Since t = 1.25 < 2.0002, we fail to reject the null hypothesis at the 95% confidence level and conclude that there is no significant difference in mean balance score between the two groups.

Since t < t120,0.9 = 1.289 < t64,0.90, and 1.046 > t64,0.85, p > 2(1 ??? .9) = .2, and p < 2(1 ??? .85) = .3. Thus, .2 < p < .3.

p=0.22

s1 <- 3
n1 <- 36
s2 <- 2.8
n2 <- 30

spsquared <- ((s1^2)*(n1-1) + (s2^2)*(n2-1))/(n1+n2-2)
spsquared
## [1] 8.474375
x1 <- 3.4
x2 <- 2.5

t <- ((x1-x2)-0)/sqrt(spsquared*(1/n1 + 1/n2))
t
## [1] 1.250629
qt(0.975, 60)
## [1] 2.000298
x <- pnorm(1.25, lower.tail=FALSE)
2*x
## [1] 0.2112995

PROBLEM 8.76