Question

Using traditional methods, it takes 10.1 hours to receive a basic driving license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique with 12 students and observed that they had a mean of 10.5 hours with a standard deviation of 1.4. A level of significance of 0.1 will be used to determine if the technique performs differently than the traditional method. Assume the population distribution is approximately normal. State the null and alternative hypotheses.

Answer 1

Answer:

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:  

Null hypothesis:$\mu =10.1$  

Alternative hypothesis:$\mu \neq 10.1$

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$    

$p_v =2*P(t_{(11)}>0.990)=0.343$    

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL  reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=10.5$ represent the sample mean

$s=1.4$ represent the sample standard deviation

$n=12$ sample size  

$\mu_o =10.1$ represent the value that we want to test

$\alpha=0.1$ 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 technique performs differently than the traditional method, the system of hypothesis would be:  

Null hypothesis:$\mu =10.1$  

Alternative hypothesis:$\mu \neq 10.1$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-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:  

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=12-1=11$  

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

$p_v =2*P(t_{(11)}>0.990)=0.343$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL  reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.