A study was undertaken to compare the respiratory responses of hypnotized and no
ID: 3309604 • Letter: A
Question
A study was undertaken to compare the respiratory responses of hypnotized and nonhypnotized subjects to certain instructions. The 16 male volunteers were allocat- ed at random to an experimental group to be hypnotized or to a control group. Baseline measurements were taken at the start of the experiment. In analyzing the data, the researchers noticed that the baseline breathing patterns of the two groups were different; this was surprising, since all the subjects had been treated the same up to that time. One explanation proposed for this unexpected difference was that the experimental group were more excited in anticipation of the experience of being hypnotized. The accompanying table presents a summary of the baseline measurements of total ventilation (liters of air per
minute per square meter of body area). Parallel dotplots of the data are given in the following graph.36 [Note: For- mula (6.7.1) yields 14 df.]
(a) Use a t test to test the hypothesis of no difference against a nondirectional alternative. Let a = 0.05.
(b) Use a t test to test the hypothesis of no difference against the alternative that the experimental condi- tions produce a larger mean than the control condi- tions. Let a = 0.05.
(c) Which of the two tests, that of part (a) or part (b), is more appropriate? Explain.
EXPERIMENTAL CONTROL 5.32 5.60 5.74 6.06 6.32 6.34 6.79 7.18 4.50 4.78 4.79 4.86 5.41 5.70 6.08 6.21 6.169 0.621 5.291 0.652 7.0 6.5 6.0 E5.5 5.0 4.5 Experimental ControlExplanation / Answer
Solution:-
a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 = 2
Alternative hypothesis: 1 2
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 0.05. 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 = 0.3183
DF = 14
t = [ (x1 - x2) - d ] / SE
t = 2.76
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 14 degrees of freedom is more extreme than -2.76; that is, less than - 2.76 or greater than 2.76.
Thus, the P-value = 0.0154
Interpret results. Since the P-value (0.0154) is less than the significance level (0.05), we cannot accept the null hypothesis.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1< 2
Alternative hypothesis: 1 > 2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.03183
DF = 14
t = [ (x1 - x2) - d ] / SE
t = 2.76
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 population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of 2.76. We use the t Distribution Calculator to find P(t > 2.76).
Therefore, the P-value in this analysis is 0.0077
Interpret results. Since the P-value (0.0077) is less than the significance level (0.05), we have to reject the null hypothesis.
c) The directional test in the part (b) is more appropriate in this case.
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