Question

Using traditional methods, it takes 94 hours to receive a basic flying license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique with 210 students and observed that they had a mean of 95 hours. Assume the standard deviation is known to be 5. A level of significance of 0.05 will be used to determine if the technique performs differently than the traditional method. Is there sufficient evidence to support the claim that the technique performs differently than the traditional method

Answer 1

Answer:

$z=\frac{95-94}{\frac{5}{\sqrt{210}}}=2.898$    

$p_v =2*P(Z>2.898)=0.0038$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ 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 94 at 5% of signficance.  

Step-by-step explanation:

Data given and notation    

$\bar X=95$ represent the sample mean    

$\sigma=5$ represent the population standard deviation    

$n=210$ sample size    

$\mu_o =94$ 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 check if the mean is different from 94, the system of hypothesis are :    

Null hypothesis:$\mu = 94$    

Alternative hypothesis:$\mu \neq 94$    

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) the info given like this:    

$z=\frac{95-94}{\frac{5}{\sqrt{210}}}=2.898$    

P-value    

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

$p_v =2*P(Z>2.898)=0.0038$    

Conclusion    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ 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 94 at 5% of signficance.