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Persons having Raynaud\'s syndrome are apt to suffer a sudden impairment of bloo

ID: 3260894 • Letter: P

Question

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm^2/min) was measured. For 50 subjects with the syndrome, the average heat output was 0.91 with a standard deviation of 0.3, and for 50 subjects without the syndrome, the average output was 2.04 with a standard deviation of 0.4. (a) At significance level 0.05, does the data suggest that the difference in average heat output of those with the syndrome and those without is less than -1.0? Be sure to state the relevant hypotheses and find the rejection region. (b) What sample sizes are required if the power of the test is to be 0.9 when actual difference is -1.2 assuming an equal number of subjects with and without the syndrome are to be tested?

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu1-mu2 = -1.0 Vs H1 : mu1-mu2 < -1.0

Assume alpha = level of significance = 0.05

Here sample information is given so we use two sample t-test.

Given that,

Here test statistic follows normal distribution.

We can find two sample t-test in MINITAB.

steps :

Stat --> Basic statistics --> 2-Sample t --> click on summarized data --> Input all the values --> Click on assuming equal variances --> Options --> Confidence level : 95.0 --> Test difference : -1.0 --> Alternative : less than --> ok --> ok


Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 50 0.900 0.300 0.042
2 50 2.040 0.400 0.057


Difference = mu (1) - mu (2)
Estimate for difference: -1.14000
95% upper bound for difference: -1.02258
T-Test of difference = -1 (vs <): T-Value = -1.98 P-Value = 0.025 DF = 98
Both use Pooled StDev = 0.3536

Test statistic = -1.98

P-value = 0.025

P-value < alpha

Reject H0 at 5% level of significance.
Conclusion : There is sufficient evidence to say that the difference in average heat output of those with the syndrome and those without is less than -1.0.

n1 50 X1bar 0.9 s1 0.3 n2 50 X2bar 2.04 s2 0.4 alpha 0.05
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