Question

If you use a 95% confidence level in a two-tail hypothesis test, what will you decide about your Null Hypothesis if the computed value of the test statistic is Z = 2.57? Why?

Answer 1

Answer:

z_{calc}=2.57[/tex]

Since is a two-sided test the p value would be:  

$p_v =2*P(z>2.57)=0.0101$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from $\mu_o$ at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X$ represent the sample mean

$\sigma$ represent the population standard deviation

$n$ sample size  

$\mu_o$ represent the value that we want to test  

Confidence = 95% or 0.95

$\alpha=1-0.95=0.05$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to an specified value $\mu_o$, the system of hypothesis would be:  

Null hypothesis:$\mu =\mu_o$  

Alternative hypothesis:$\mu \neq \mu_o$  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) and we got a value calculated let's say $z_{calc}=2.57$

P-value  

Since is a two-sided test the p value would be:  

$p_v =2*P(z>2.57)=0.0101$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from $\mu_o$ at 5% of signficance.