Question

Using traditional methods, it takes 10.0 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 10.2 hours with a standard deviation of 1.1. A level of significance of 0.05 will be used to determine if the technique performs differently than the traditional method. Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places.

Answer 1

`df=n-1=12-1=11`

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

`p_v =2*P(t_((11))>0.630)=0.5416`

Explanation:

Data given and notation

`\bar X=10.2` represent the sample mean

`s=1.1` represent the sample standard deviation

`n=12` sample size

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 technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis:
`\mu = 10`

Alternative hypothesis:
`\mu \\eq 10.2`

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=(\bar X-\mu_o)/((s)/(√(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=(10.2-10)/((1.1)/(√(12)))=0.630`

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))>0.630)=0.5416`

Answer 2

The t-statistic is t=0.631.

Explanation:

We have to perform an hypothesis test on the mean.

The claim is that the new method performs different from the actual method.

Then, the null and alternative hypothesis are:

`H_0: \mu=10.0\\\\H_a:\mu\\eq10.0`

The significance level is 0.05.

The sample has a size n=12. The sample mean is 10.2 hours and the sample standard deviation is 1.1.

As the population standard deviation is not known, we will estimate it from the sample standard deviation and use a t-statistic.

The degrees of freedom are:

`df=n-1=12-1=11`

The t-statistic is:

`t=(\bar x-\mu)/(s/√(n))=(10.2-10.0)/(1.1/√(12))=(0.2)/(0.317)= 0.631`

The P-value of this two-tailed test, with t=0.631 and df=11, is

`P-value=2P(t>0.631)=0.541`

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis failed to be rejected.