Question

Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.100.10 significance level to test for a difference between the measurements from the two arms. What can be​ concluded? Right arm 148148 133133 142142 130130 131131 Left arm 182182 164164 176176 149149 134134 In this​ example, mu Subscript dμd is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis​ test? A. Upper H 0H0​: mu Subscript dμdnot equals≠0 Upper H 1H1​: mu Subscript dμdgreater than>0 B. Upper H 0H0​: mu Subscript dμdequals=0 Upper H 1H1​: mu Subscript dμdless than<0 C. Upper H 0H0​: mu Subscript dμdequals=0 Upper H 1H1​: mu Subscript dμdnot equals≠0 D. Upper H 0H0​: mu Subscript dμdnot equals≠0 Upper H 1H1​: mu Subscript dμdequals=0 Identify the test statistic. tequals=nothing ​(Round to two decimal places as​ needed.) Identify the​ P-value. ​P-valueequals=nothing ​(Round to three decimal places as​ needed.) What is the conclusion based on the hypothesis​ test? Since the​ P-value is ▼ greater less than the significance​ level, ▼ reject fail to reject the null hypothesis. There ▼ is not is sufficient evidence to support the claim of a difference in measurements between the two arms.

Answer 1

Null and alternative hypotheses :

Two tailed test.

Test Statistic :

Where, is sample mean of differences.

is sample standard deviation of differences.

n is sample size ( Number of data pairs )

Calculation :

n = 5

So test statistic is ,

Test statistic = t = -4.05   ( rounded to two decimal places )

P-value :

P-value for this two tailed test is ,

P-value = 2*P( t < test statistic ) = 2*P( t < -4.048 )

Using Excel function ,    = T.DIST( t , df , 1 )       , df = n - 1 = 5 - 1 = 4

P( t < -4.048 ) = T.DIST( -4.048 , 4 , 1 ) = 0.00775

So, p-value = 2*0.00775 = 0.016

Conclusion :

Since the​ P-value ( 0.016 ) is less than the significance​ level ( = 0.10 ), so reject null hypothesis.

There is sufficient evidence to support the claim of a difference in measurements between the two arms.