Question

An investigator wishes to determine whether alcohol consumption causes deterioration in the performance of automobile drivers. Before the driving test, subjects drink a glass of orange juice, which, in the case of the treatment group, is laced with two ounces of vodka. Performance is measured by the number of errors made on a driving simulator. A total of 120 volunteer subjects are randomly assigned, in equal numbers, to the two groups. For subjects in the treatment group, the mean number of errors (–X1) equals 26.4, and for subjects in the control group, the mean number of errors (–X2) equals 18.6. The estimated standard error equals 2.4.

1) Use t to test the null hypothesis at the .05 level of significance.

a) what is the null hypothesis?

b) is this a one-tailed or two-tailed test?

c) what are the degrees of freedom?

d) what is the t critical for 0.05 significance?

e) what is the calculated t?

2) Specify the p-value for this test

3) If appropriate, construct a 95 percent confidence interval for the true population mean difference and interpret this interval.

4)If the test result is statistically significant, use Cohen’s d to estimate the effect size, given that the standard deviation, s p , equals 13.15.

5)State how these test results might be reported in the literature, given s 1 5 13.99 and s 2 5 12.15.

Answer 1

Solution: The investigator is interested to know whether alcohol consumption causes deterioration in the performance of automobile drivers. The deterioration in performnace is measured in terms of number of errors while driving.

In statistical theory, a null hypothesis is a general statement that there is no difference between the groups being considered for the study. Hence, in our case the null hypothesis denoted by is

: Alcohol consumtption causes no deterioration in the performance of the automobile drivers.

This hypothesis is tested using a suitable tool and we arrive at a conclusion to either accept it or reject it. If this hypothesis is rejected an alternative statement to the null hypothesis is considered prevalent and it is known as alternative hypothesis.

In our case, the best alternative statement to the null hypothesis is that the consumption of alcohol does reduce the performance of drivers. That is, the alternative hypothesis denoted by is

: Alcohol consumtption causes deterioration in the performance of the automobile drivers.

The performance of the drivers in the study is measured by the average number of errors they commit while driving. Hence the above set of hypotheses can be written equivalently as

: There is significant difference between the averga enumber of errors committed by the drivers who consume alcohol and the drivers who do not consume alcohol.

against

: On an average drivers who consume alcohol committ more errors than the drivers who do not consume alcohol.

To test these hypotheses, the investigator has considred 120 volunteers that are equally divided in to two groups namely treatmentgroup and control group. The volunteers in the treatments group are given a glass of orange juice laced with 2 ounces of vodka while the drinks of control group are not laced with vodka.

It is observed that, the vilunteers in treatment group committed on an average 26.4 errors while the volunteers in control group committed 18.6 errors. The estimated standard error for the difference of these errors is found to be 2.4 .

Let X1 denote the mean number of errors committed by the volunteers in treatment group.

Let X2 denote the mean number of errors committed by the volunteers in control group.

Then, in view of the hypotheses given above,

(a) The null hypothesis can be written as

.

This null hypothesis is then tested agianst the alternative

.

(b) A one-tailed test is a test which corresponds to a critical region either at the left or at the right hand side of the sampling distribution of the test. That is, a one-sided test is usedul when testing for either the hypothesized relationship between treatment and control group is either greater or lesser.

In our case, we are testing the null hypothesis of equality of errors agianst an alterantive of errors in treatment being greater than errors in control which is one-sided. Hence, the test of versus is a one-sided test.

(c) It has been asked to use a two-sample t-test to test versus . Since the volunteers are divided in to nonoverlapping groups of equal size the t-test used here is called independent sample t-test or an unpaired sample t-test. The t-test statistic in this case is given by

.

and distribution with degrees of freedom.

Statistical parameters (for example ) are based on a particular amount of data. The amount of or the number of data items that go in to the estimate of a parameter are called degrees of freedom. In general, the degrees of freedom of an estimate are equal to the number of independent data items that are used to estimate the parameter minus the number of constants estimated in the intermediate steps.

In our case, the desired estimate is , we are using a total of 120 scores corresponding to 120 volunteers to compute this value. However, in order to estimate , we require mean number of errors that these volunteers committ. Hence, the degrees of freedom for here are the number of independene data items used (n=120) minus the constants (mean errors) estimated from these scores in itermediate steps (1).

That is, the degrees of freedom for is .

(d) After computing the value of the it is compared with the tabulated value of t for a specified level of significnace. If , then the null hypothesis is rejected. The tabulated value, corresponding to a degrees of freedom and level of significnace is found using the t-tables as shown in the below picture.

For this problem, the degrees of freedom are 119 and the , hence the .

Here, the and therefore the null hypothesis is rejected and we conclude that "drivers belonging to treatment group committ more errors on an average than drivers in control group" that is, . is accepted.

(e) The calculated value of t is .