Question

A simple random sample of 15 people from a certain population has mean age of 25 with a standard deviation of 10. Can we conclude that the mean age of the population is younger than 25? Let alpha = 0.05

• Formulate the hypothesis: the null hypothesis and the alternative hypothesis.
• Set the significance level for a. type of test critical value of z/t
• Compute the z-statistic or t-statistic
• Decision: compare the computed value of z or t with the Critical value.
• Conclusion

can anyone help me with this? please :(

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge25$

$H_a:\mu<25$

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

2. Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=14$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.76131.$

The rejection region for this left-tailed test is $R = \{t: t <-1.76131\}$

3.The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{25-25}{10/\sqrt{15}}=0$$

4. Since it is observed that $t =0>-1.76131= t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=14$ degrees of freedom, $t=0$ is $p = 0.5,$ and since $p= 0.5>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

5. Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than 25, at the $\alpha = 0.05$ significance level.