Answer to Question #280274 in Statistics and Probability for myli

Question

the mean of marriage age in Asia region IS at 25.2 years. A random of sampling is taken and as the new result, the of ages of 35 couples had a mean of 28.7 years and a standard deviation of 4.6

years. According to this new sampling, test at the 0.05 level of significance if the mean age of married couples is different from 25.2 years.

Answer 1

Solution:

Null hypothesis:

$H_{0}: \mu=25.2$ years

That is, the mean of marriage age is not different from 25.2 years.

Alternative hypothesis:

$H_{1}: \mu \neq 25.2$ years

That is, mean of marriage age is different from 25.2 years at the time of their first birth.

Find the critical value for two tailed test with n=35 and $\alpha=0.05$ .

The degree of freedom is obtained below:

$\begin{aligned} \mathrm{df} &=n-1 \\ &=35-1 \\ &=34 \end{aligned}$

Thus, from the "Appendix Table F, The t-distribution", the value in the $34^{\text {th }}$ row corresponds with the fifth column gives the required value of $t_{0.05}$ .

Hence, the critical value for the two-tailed test with $\alpha=0.05$ is Critical value of $t=\pm 2.032.$

Compute the test value by using MINITAB.

The formula is given below:

$t=\dfrac{\bar{X}-\mu}{\dfrac{s}{\sqrt{n}}}$

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 1-Sample t.

Step 2: In Summarized data, enter the sample size 35 and mean 28.7.

Step 3: In Standard deviation, enter a value $\mathbf{2 5 . 2}$ .

Step 4: In Perform hypothesis test, enter the test mean 25.2.

Step 5: Check Options, enter Confidence level as 95.0.

Step 6: Choose not equal in alternative.

Step 7: Click OK in all dialog boxes.

MINITAB output:

One-Sample T

Test of $\mu =25.2 \ vs\ \mu \ne25.2$

$\begin{array}{rrrrrrrr}\mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & 958 \text { CI } & \text { T } & \text { p } \\ 35 & 28.700 & 4.600 & 0.778 & (27.120, & 30.280) & 4.50 & 0.000\end{array}$

From the above MINITAB output, the test value is 4.50.

From the MINITAB output, the test value is 4.5, which falls in the critical region, as shown in Figure (1).

Thus, there is an evidence to reject the null hypothesis $\left(H_{0}\right)$ at 5% level of significance.

There is enough evidence to infer that mean of marriage age is different from 25.2 years at 5% level of significance.