Several students were tested for reaction times (in thousandths of a second) usi

Question

Several students were tested for reaction times (in thousandths of a second) using their right and left hands. (Each value is the elapsed time between the release of a strip of paper and the instant that it is caught by the subject.) Results from five of the students are included in the graph to the right. Use a 0.02 significance level to test the claim that there is no difference between the reaction times of the right and left hands. Click the icon to view the reaction time data table. What are the hypotheses for this test? Let mu_d be the mean of the differences of the right and left hand reaction times. H_0: mu_d = 0 H_1: mu_d notequalto 0 What is the test statistic? t = (Round to three decimal places as needed.)

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = d = 0, there is no difference between the reaction time.
Alternative hypothesis: 1 - 2 = d 0, there is difference between the reaction time.

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.02. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(21.612/5) + (25.912/5)] = sqrt(3.33 + 9) = 15.08854

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (21.612/5 + 25.912/5)2 / { [ (21.612 / 5)2 / (4) ] + [ (25.912 / 5)2 / (4) ] }
DF = 51830.92 / (2180.82 + 4506.8142) = 152.03/3.757 = 7.75 or 8

t = [ (x1 - x2) - d ] / SE = [ (154.6 - 180.4) - 0 ] / 15.08854 = -1.7099 or -1.710

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than -1.7099; that is, less than -1.7099 or greater than 1.7099.

We use the t Distribution Calculator to find P(t < -1.7099)

The P-Value is 0.127043.
The result is not significant at p < 0.02.

Interpret results. Since the P-value (0.127) is greater than the significance level (0.02), we cannot reject the null hypothesis.

Conclusion. Retain the null hypothesis and the claim is significant that there is no difference between the reaction time.

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