Question

For the population, scores on the test are normally distributed with μ = 70 and σ = 15. The sample of n = 25 students had a mean score of M = 75. Are the data sufficient to that the herb has a significant effect on memory? Use a two-tailed test with α = .05. What is the standard error for this study?

Answer 1

Answer:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$      

$p_v =2*P(t_{24}>1.67)=0.108$  

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 FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.    

Explanation:

1) Data given and notation      

$\bar X=75$ represent the sample mean

$\sigma=15$ represent the standard deviation for the population

$n=25$ sample size      

$\mu_o =70$ represent the value that we want to test    

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

t 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 determine if the mean is different from 70, the system of hypothesis would be:      

Null hypothesis:$\mu = 70$      

Alternative hypothesis:$\mu \neq 70$      

We know the population deviation, so for this case 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) the info given like this:      

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

$df=n-1=25-1=24$  

Since is a two tailed test the p value would be:      

$p_v =2*P(t_{24}>1.67)=0.108$  

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 FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.