Answer to Question #295106 in Statistics and Probability for hash cake

Question

1. A researcher thinks that people under the age of twenty have vocabularies that are different than those of people over sixty years of age. Thus, to test this hypothesis, the researcher administers a vocabulary test to a group of 25 younger samples and 30 older samples. The mean score of the younger samples was 20.0 and the standard deviation of their score was 5.0. On the other hand, the mean score of the older samples was 14.0 and the standard deviation of their scores was 6.0. Does this experiment provide evidence for the investigator’s theory?
2. The mean number of close friends for the population of people living in the city is 5.7. An investigator predicts that the mean number of close friends for introverts will be significantly less than the mean of the population. The mean number of close friends for a sample of 26 introverts is 6.5 with the standard deviation of 1.3. Are the given data support the investigator’s prediction?

Answer 1

$1)$

Group 1

$n_1=25\\\bar x_1=20\\s_1=5$

Group 2

$n_2=30\\\bar x_2=14\\s_2=6$

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

We test,

$H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma^2_2$

The test statistic is,

$F_c={s^2_2\over s_1^2}={36\over25}=1.44$

The critical value is,

$F_{{\alpha\over2},n_2-1,n_1-1}=F_{0.025,29,24}= 2.217443$

The null hypothesis is rejected if, $F_c\gt F_{0.025,29,24}$

Since $F_c=1.44\lt F_{0.025,29,24}=2.217443$, we fail to reject the null hypothesis and conclude that the population variances are equal. 

We now perform hypothesis test on difference in means.

$H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\not=\mu_2$

The test statistic is,

$t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}$

where $sp^2$ is the pooled sample variance given as,

$sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(24\times25)+(29\times36)\over53}={1644\over53}=31.02$

Therefore,

$t_c={(20-14)\over \sqrt{31.02({1\over 25}+{1\over 30})}}={6\over1.5082}=3.98$

$t_c$ is compared with the table value at $\alpha=0.05$ with $n_1+n_2-2=25+30-2=53$ degrees of freedom.

The table value is,

$t_{{0.05\over2},53}=t_{0.025,53}= 2.005746$

The null hypothesis is rejected if $t_c\gt t_{0.025,53}.$

Now, $t_c=3.98\gt t_{0.025,53}=2.005746$ 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 twenty have vocabularies that are different than those of people over sixty years of age at 5% significance level.

$2)$

$n=26\\\bar x=6,5\\s=1.3$

The hypotheses tested are,

$H_0:\mu=5.7\\vs\\H_1:\mu\gt5.7$

The test statistic is given as,

$t_c={\bar x-\mu\over{s\over\sqrt{n}}}={6.5-5.7\over{1.3\over\sqrt{26}}}={0.8\over 0.254951}= 3.137858$

The critical value is,

$t_{\alpha,n-1}=t_{0.05,25}=1.708141$

We reject the null hypothesis since $t_c=3.137858\gt t_{0.05,25}=1.708141$ and conclude that the mean number of close friends for introverts is significantly greater than the mean of the population at 5% significance level.

However, the provided data does not support the investigator's prediction.