16.3.60 one-way ANOVA test

Researchers analyzed IQ data on preterm children at age​ 7-8 years. The mothers of the children in the study had chosen whether to provide their infants with breast milk within 72 hours of delivery. The researchers used the following designations. Group​ I: mothers declined to provide breast​ milk; Group​ IIa: mothers had chosen but were unable to provide breast​ milk; and Group​ IIb: mothers had chosen and were able to provide breast milk. The summary statistics on IQ are at the right. At the 1​% significance​ level, do the data provide sufficient evidence to conclude that a difference exists in mean IQ at age​ 7-8 years for preterm children among the three​ groups? Critical values are given at the right.

Notation in one-way ANOVA:

  • k = number of populations

  • n = total number of observations

  • \(\bar x\) = mean of all n observations

  • \(n_j\) = size of sample from Population j

  • \(\bar{x_j}\) = mean of sample from Population j

  • \(s_j^2\) = variance of sample from Population j

  • \(T_j\) = sum of sample data from Population j

Defining formulas from sums of squares in one-way ANOVA:

  • SST = \(\sum (x_i - \bar x)^2\)

  • SSTR = \(\sum n_j(\bar{x_j} - \bar{x})^2\)

  • SSE = \(\sum (n_j-1)s_j^2\)

One-way ANOVA identity: SST = SSTR + SSE

Computing formulas from sums of squares in one-way ANOVA:

  • SST = \(\sum x_i^2 - (\sum x_i)^2/n\)

  • SSTR = \(\sum (T_j^2/n_j) - (\sum x_i)^2/n\)

  • SSE = SST - SSTR

The way they define \(\sum (T_j^2/n_j)\) is different from the one for x

Mean squares in one-way ANOVA:

  • MSTR = \(\frac{SSTR}{k-1}\)

  • MSE = \(\frac{SSE}{n-k}\)

  • SSE = SST - SSTR

Test statistic for one-way ANOVAA (independent samples, normal populations, and equal population standard deviations):

  • F = \(\frac{MSTR}{MSE}\)

with df = (k - 1, n - k)

Confidence interval for \(\mu_i - \mu_j\) in the Tukey multiple-comparison method (independent samples, normal populations, and equal population sstandard deviations):

  • \((\bar{x_i} - \bar{x_j}) \pm \frac{q_{\alpha}}{\sqrt{2}}.s\sqrt{\frac{1}{n_i} + \frac{1}{n_j}}\)

where s = \(\sqrt{MSE}\) and \(q_{\alpha}\) is obtained for a q-curve with parameters k and n - k

Test statistic for a Kruskal-Wallis test (independent samples, same-shape populations, all sample sizes 5 or greater):

  • \(K=\frac{SSTR}{SST/(n-1)}\) or

  • \(K=\frac{12}{n(n+1)}\sum_{j=1}^{k} \frac{R_j^2}{n_j} - 3(n+1)\)

where SSTR and SST are computed for the ranks of the data, and \(R_j\) denotes the sum of the ranks for the sample data from Population j. K has approximately a chi-square distribution with df = k -1

(a) Find the null and alternative hypotheses. Choose the correct answer below.

Since the question asks that “a difference exists in mean IQ at age​ 7-8 years for preterm children among the three​ groups,” the correct hypothesis is.

\(H_0: \mu_I = \mu_{IIa} = \mu_{IIb}\)

\(H_a:\) Not all the means are equal.

First we need to get the data from the question.

nj <- c(88, 17, 195)
x <- c(98.6, 97.3, 101.3)
s <- c(11.5, 20.6, 13.1)

Names of variables

n = total number of observations: n

We find SSTR, SSE, SST by using defining formula

n = sum(nj)
k = length(x)

Find SSTR using the formula SSTR = \(\sum n_j(\bar{x_j} - \bar{x})^2\)

SSTR = sum(nj*(x- sum(nj*x)/sum(nj))^2)
SSTR
## [1] 602.2155
round(SSTR, 2)
## [1] 602.22

Find SSE using the formula SSE = \(\sum (n_j-1)s_j^2\)

SSE = sum((nj-1)*s^2)
SSE
## [1] 51587.85
round(SSE, 2)
## [1] 51587.85

Find SST, we use formula SST = SSE + SSTR 

SST = SSE + SSTR
SST
## [1] 52190.07
round(SST, 2)
## [1] 52190.07

Find df treatment 

k = length(x)
k-1
## [1] 2

Find SS treatment 

SSTR
## [1] 602.2155

Find MS treatment 

MSTR = SSTR/(k-1)
MSTR
## [1] 301.1077

Find Error df 

n - k
## [1] 297

Find Error SS 

SSE
## [1] 51587.85

Find Error MS 

MSE = SSE / (n - k)
MSE
## [1] 173.6965

Find F-statistic treatment 

MSTR / MSE
## [1] 1.733528
round(MSTR / MSE, 2)
## [1] 1.73

Find df total

n - 1
## [1] 299

Find SS total

SST
## [1] 52190.07

Find the critical value.

Since \(\alpha = .01\)

alpha = 0.01
qf(1-alpha, k-1, n-k)
## [1] 4.67732

Round to decimal places

round(qf(1-alpha, k-1, n-k), 2)
## [1] 4.68

Since our test statistic = 1.73 < our critical value \(F_{\alpha}\)=4.68 and it is a right-tailed test, we do not have enough evidence to reject the hypothesis.

Hope that helps!