Answer:
![p_v = P(z](https://tex.z-dn.net/?f=%20p_v%20%3D%20P%28z%3C-2.30%29%20%3D0.0107)
Now we can decide based on the significance level
. If
we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.
we see that
so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15
Explanation:
For this case they conduct the following system of hypothesis for the ture mean of interest:
Null hypothesis: ![\mu \leq 15](https://tex.z-dn.net/?f=%5Cmu%20%5Cleq%2015)
Alternative hypothesis: ![\mu >15](https://tex.z-dn.net/?f=%5Cmu%20%3E15)
The statistic for this hypothesis is:
![z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B%5Cbar%20X%20-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
And on this case the value is given ![z = -2.30](https://tex.z-dn.net/?f=%20z%20%3D%20-2.30)
For this case in order to take a decision based on the significance level we need to calculate the p value first.
Since we have a lower tailed test the p value would be:
![p_v = P(z](https://tex.z-dn.net/?f=%20p_v%20%3D%20P%28z%3C-2.30%29%20%3D0.0107)
Now we can decide based on the significance level
. If
we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.
we see that
so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15
we see that
so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15