Answer:
Step-by-step explanation:
From the given information:
The null and the alternative hypothesis is:
[tex]H_o: \mu_{antidepressant}= \mu_o[/tex]
[tex]H_1: \mu_{antidepressant}\ne \mu_o[/tex]
The population mean = 25
The sample mean = 23.3
sample size n = 36
standard deviation = 6
From the following conditions given; the condition that is not required is:
[tex]\text{The sample size must be very large}[/tex]
Since the sample size is 30, we do not need the condition and it's enough for the Z test provided that the population is not normally distributed. If the population is distributed normally, then any sample size is valid:
Then, the test statistics can be computed as follows:
[tex]Z = \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}} }[/tex]
[tex]Z = \dfrac{23.3 - 25}{\dfrac{6}{\sqrt{36}} }[/tex]
[tex]Z = \dfrac{23.3 - 25}{\dfrac{6}{6} }[/tex]
[tex]Z = -1.7[/tex]
At the level of significance 0.05
Since the test is two-tailed, the critical value [tex]Z_{\alpha/2} = Z_{0.025} = 1.96[/tex]
Provided that the Z value is not less than the critical value;
Then, we fail to reject the null hypothesis.
We conclude that there is no enough evidence that the drug reduces depression.