Chapter 11, Problem 46CE

A goal of financial literacy for children is to learn how to manage money wisely. One question is: How much money do children have to manage? A recent study by Schnur Educational Research Associates randomly sampled 15 children between 8 and 10 years old and 18 children between 11 and 14 years old and recorded their monthly allowance. Is it reasonable to conclude that the mean allowance received by children between 11 and 14 years is more than the allowance received by children between 8 and 10 years? Use the .01 significance level. What is the p-value?

To determine

State whether it is reasonable to conclude that the mean allowance received by children between 11 and 14 years is more than the mean allowance received by children between 8 and 10 years.

Obtain the p-value.

Answer to Problem 46CE

Yes, there is enough evidence to conclude that it is reasonable to conclude that the mean allowance received by children between 11 and 14 years is more than the mean allowance received by children between 8 and 10 years.

The p-value is 0.007.

Explanation of Solution

It may be expected that the mean allowance received by children between 11 and 14 years is more than the mean allowance received by children between 8 and 10 years.

Therefore, the test hypotheses are given below:

Denote ${\mu }_1$ as the mean allowance received by children between 8 and 10 years and ${\mu }_2$ as the mean allowance received by children between 11 and 14 years.

Null hypothesis: $H_0:{\mu }_1\ge {\mu }_2$.

That is, the mean allowance received by children between 8 and 10 years is at least the mean allowance received by children between 11 and 14 years.

Alternative hypothesis: $H_a:{\mu }_1<{\mu }_2$

That is, the mean allowance received by children between 8 and 10 years is less than the mean allowance received by children between 11 and 14 years.

In this context, the level of significance is 0.01.

Necessary assumptions required for using the formula:

• The sampled populations are approximately normally distributed.
• The two samples are independent.
• The standard deviations for the two populations are equal.

In this context, the two populations are independent and distributed to normal. The population standard deviations are unknown.

Test statistic for the two-sample test of means-unknown$\sigma$:

$t=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{s_P^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Where $n_1\text{and }{n}_2$ are the sample sizes of two populations, $s_1^2\text{and}{s}_2^2$ are the sample variances, and $s_p^2$ is the pooled estimate of ${\sigma }^2$,

Here,

 $s_p^2=\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}$.

Excel procedure to find the mean and standard:

• Enter the data values in column H and column I.
• Obtain the sample mean 1 (${\overline{X}}_1$) in cell A1, enter the formula “=AVERAGE(H1:H15)”.
• Press “Enter”.
• Obtain the sample variance 1$\left(s_1^2\right)$ in cell B1, enter the formula “=VAR.S(H1:H15)”.
• Press “Enter”.
• Obtain the sample mean 2 (${\overline{X}}_1$) in cell A2, enter the formula “=AVERAGE(I1:I18)”.
• Press “Enter”.
• Obtain the sample variance 2 $\left(s_2^2\right)$ in cell B2, enter the formula “=VAR.S(I1:I18)”.
• Press “Enter”.

Output obtained using EXCEL is given below:

The pooled variance is obtained as given below:

Substitute $s_1^2$ as 10.029, $s_2^2$ as 19.18, $n_1$ as 15, and $n_2$ as 18.

$\begin{array}{c}{s}_p^2=\frac{\left(15-1\right)\left(10.029\right)+\left(18-1\right)\left(19.18\right)}{15+18-2}\\ =\frac{466.466}{31}\\ =15.047\end{array}$

The test statistic is given below:

Substitute ${\overline{X}}_1$ as 8.8, ${\overline{X}}_2$ as 12.33, $s_p^2$ as 15.047, $n_1$ as 15, and $n_2$ as 18.

$\begin{array}{c}t=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{s_P^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ =\frac{8.8-12.33}{\sqrt{15.047\left(\frac{1}{15}+\frac{1}{18}\right)}}\\ =\frac{-3.53}{1.356}\\ =-2.603\end{array}$

Rejection region:

In context, the level of the test, $\alpha$, is 0.01.

Here, the alternative is the left-tailed test. Hence, the rejection region will be $t<-t_{\alpha }$.

The critical value has to be obtained for $t_{0.01,\left(n_1+n_2-2\right)}=t_{0.01,31}$.

Critical value:

From the “Appendix B, Table B.5 Student’s t Distribution”, the critical value for 31 df for the level of significance 0.01 is 2.453.

Thus, the rejection region under the level of significance of 0.01 is $t<-2.453$.

Decision rule:

• If $t<-2.453$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the test statistic, t falls in the rejection region.

Therefore, by the decision rule, reject the null hypothesis.

Therefore, there is evidence to conclude that the mean allowance received by children between 8 and 10 years is less than the mean allowance received by children between 11 and 14 years.

p-value:

Step-by-step procedure to obtain the p-value using EXCEL:

• In a cell A1, enter the formula “=T.DIST.RT(-2.603,31)”.
• Press “Enter”.

Output obtained using EXCEL is given below:

Thus, the p-value is 0.007.