Answer to Question #347637 in Statistics and Probability for aaa

Question

Write the six steps in hypothesis testing in each of the following studies. Take note, the p-value is already given.

The average family size in the Philippines was reported as 4.25. A random sample of families in a particular street resulted in the following family sizes:

5, 6, 7, 4, 3, 8, 2 and 9

\At 0.10, is the average family size more than the national average?

(p-value = 0.0961)

Answer 1

Sample mean 

$$\bar{x}=\dfrac{1}{8}(5+6+7+4+3+8+2+9)=5.5$$

Sample variance

$$s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}=6$$ $$s=\sqrt{6}\approx2.45$$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le4.25$

$H_1:\mu>4.25$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=7$ and the critical value for a right-tailed test is $t_c =1.4149236.$

The rejection region for this right-tailed test is $R = \{t:t>1.414923\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{5.5-4.25}{2.45/\sqrt{8}}=1.443075$$

Since it is observed that $t=1.443075>1.414923=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=7$ degrees of freedom, $t=1.443075$ is $p=0.0961,$ and since $p= 0.096107<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 4.25, at the $\alpha = 0.10$ significance level.