Question #298165

An investigator thinks that people under the age of forty have vocabularies that are different than those of

people over sixty years of age. The investigator administers a vocabulary test to a group of 31 younger

subjects and to a group of 31 older subjects. Higher scores reflect better performance. The mean score for

younger subjects was 14.0 and the standard deviation of younger subject's scores was 5.0. The mean score

for older subjects was 20.0 and the standard deviation of older subject's scores was 6.0. Does this

experiment provide evidence for the investigator's theory?


1
Expert's answer
2022-02-16T11:16:22-0500

Group 1 (Younger subjects)

n1=31xˉ1=14s1=5n_1=31\\\bar x_1=14\\s_1=5

Group 2(Older subjects)

n2=31xˉ2=20s2=6n_2=31\\\bar x_2=20\\s_2=6

To perform this test, we first check whether population variances for the two groups are equal.

We test,

H0:σ12=σ22vsH1:σ12σ22H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma^2_2

The test statistic is,

Fc=s22s12=3625=1.44F_c={s^2_2\over s_1^2}={36\over25}=1.44

The critical value is,

Fα2,n21,n11=F0.025,30,30=2.073944F_{{\alpha\over2},n_2-1,n_1-1}=F_{0.025,30,30}= 2.073944

The null hypothesis is rejected if, Fc>F0.025,30,30F_c\gt F_{0.025,30,30}

Since Fc=1.44<F0.025,30,30=2.073944F_c=1.44\lt F_{0.025,30,30}=2.073944, we fail to reject the null hypothesis and conclude that the population variances are equal.

We now perform hypothesis test on difference in means.

H0:μ1=μ2vsH1:μ1μ2H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\not=\mu_2

The test statistic is,

tc=(xˉ1xˉ2)sp2(1n1+1n2)t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}

where sp2sp^2 is the pooled sample variance given as,

sp2=(n11)s12+(n21)s22n1+n22=(30×25)+(30×36)60=183060=30.5sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(30\times25)+(30\times36)\over60}={1830\over60}=30.5

Therefore,

tc=(1420)30.5(131+131)=61.4028=4.28t_c={(14-20)\over \sqrt{30.5({1\over 31}+{1\over 31})}}={-6\over1.4028}=-4.28

tct_c is compared with the table value at α=0.05\alpha=0.05 with n1+n22=31+312=60n_1+n_2-2=31+31-2=60 degrees of freedom.

The table value is,

t0.052,60=t0.025,60=2.000298t_{{0.05\over2},60}=t_{0.025,60}= 2.000298

The null hypothesis is rejected if tc>t0.025,60.|t_c|\gt t_{0.025,60}.

Now, tc=4.28>t0.025,60=2.000298|t_c|=4.28\gt t_{0.025,60}=2.000298 therefore, we reject the null hypothesis and conclude that there is enough evidence to support the researcher's claim that people under the age of forty have vocabularies that are different than those of people over sixty years of age at 5% significance level.


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