Question

Using traditional methods, it takes 98 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 70 students and observed that they had a mean of 96 hours. Assume the standard deviation is known to be 6. A level of significance of 0.02 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? What is the conclusion?

Answer 1

`z=(96-98)/((6)/(√(70)))=-2.789`

`p_v =2*P(z<-2.789)=0.0052`

If we compare the p value and the significance level given
`\alpha=0.02` 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 population mean is significant different from 98 at 2% of significance.

Explanation:

Data given

`\bar X= 96` represent the sample mean

`\sigma =6` population standard deviation

n = 70 random sample selected

`\alpha=0.02` represent the significance level

State the null and alternative hypotheses.

The system of hypothesis for this case is given by:

Null hypothesis:
`\mu = 98`

Alternative hypothesis:
`\mu \\eq 98`

The statistic for this case is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

Calculate the statistic

But now the population deviation changes
`\sigma=30`. We can replace in formula (1) the info given like this:

`z=(96-98)/((6)/(√(70)))=-2.789`

P-value

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

`p_v =2*P(z<-2.789)=0.0052`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.02` 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 population mean is significant different from 98 at 2% of significance.