Question

In: Statistics and Probability

A researcher randomly assigns five individuals to receive a new experimental procedure, measuring number of decision-making...

A researcher randomly assigns five individuals to receive a new experimental procedure, measuring number of decision-making errors before and after the procedure. Baseline scores were 7, 6, 9, 7, and 6. Scores after the experimental treatment were. 5, 2, 4, 3, and 6. Conduct a two-tailed hypothesis test with a = .05, being sure to specify the null and alternative hypotheses.

Solutions

Expert Solution

Let be the true mean number of decision-making errors before and after the procedure respectively.

The same set of 5 individuals are tested for before and after the procedure errors. Hence this is a paired sample ans we will do the hypotheses test for paired (dependent) sample.

Let be the mean difference in error before and after the procedure.

The hypotheses are

We get the difference d between the before and after procedure as below

Before After d=(before-after)
7 5 2
6 2 4
9 4 5
7 3 4
6 6 0

We calculate the following

n=5 is the sample size

the sample mean is

The sample standard deviation is

We do not know the population standard deviation of mean difference. We estimate it using

the estimated standard error of mean difference is

The sample size is less than 30 and we do not know the population standard deviation. Hence we will use t distribution as the sampling distribution of mean.

The hypothesized value of mean difference is (from the null hypothesis)The test statistics is

This is a 2 tailed test (the alternative hypothesis has "not equal to")

The critical value of t for alpha=0.05 is found for . The degrees of freedom for t is 5-1=4.

Using the t tables, for df=4, and area under the right tail=0.025 (or area under both tails =0.05) we get the right tail critical value as 2.776

That is the critical values are -2.776,+2.776

We will reject the null hypothesis if the test statistics does not lie within the acceptance region -2.776 to +2.776.

Here, the test statistics is 3.354 and it does not lie within the interval -2.776 to +2.776. Hence we reject the null hypothesis.

We conclude that, at 5% level of significance, there is sufficient evidence to support the claim that the mean number of decision-making errors before and after the procedure are different.


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