Answer to Question #193162 in Statistics and Probability for Ahmad

Question

the head of the math department claims that the mean height of grade 7 students is 163 cm. the mean height of randomly selected grade 7 students is 161 cm. Using 0.05 significance level can it be concluded that the mean height of grade 7 students is shorter than 163 cm? Provide step by step solution briefly.

Answer 1

The following information has been provided:

Sample Mean 1 $\overline{X_1}$ $=161$

Sample Standard Deviation 1 $\left(s_1\right)$ $=1$

Sample Size $\left(n_1\right)\:$ $=3$

Sample Mean 2 $\:\overline{X_2}\:$ $=163$

Sample Standard Deviation 1 $\left(s_2\right)$ $=1$

Sample Size $\left(n_2\right)$ $=3$

Significance Level $\left(\alpha \right)\:\:\:$$=0.05$

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_o:\mu _1​=\mu \:_2\:$

$H_a:\mu \:_1​\not= \mu \:\:_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is $\alpha=0.05$

and the degrees of freedom are $df = 4$

In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

$df\:Total​\:=df\:1​\:+df\:2​\:=2+2=4\:$

Hence, it is found that the critical value for this two-tailed test is $\:\:t_c\:=2.776$ for $\alpha =0.05$ and $\:\:df=4$ .

The rejection region for this two-tailed test is $R=\left\{t:∣t∣>2.776\right\}.$

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t=\:\frac{\overline{X_1}-\overline{X_2}}{\sqrt{\frac{\left(n_1-1\right)s^2_1+\left(n_2-1\right)s^2_2}{n_1+n_2-2}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

$t=\:\frac{161-163}{\sqrt{\frac{\left(3-1\right)1^2+\left(3-1\right)1^2}{3+3-2}\left(\frac{1}{3}+\frac{1}{3}\right)}}=−2.449$

(4) Decision about the null hypothesis

Since it is observed that $∣t∣=2.449≤t c ​ =2.776$ it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is $p=0.0705$ , and since $p=0.0705$ , it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu _1$ is different than $\mu _2$ , at the $\alpha=0.05$ significance level.