Question

In: Statistics and Probability

Calcium and Blood Pressure Does increasing the amount of calcium in our diet reduce blood pressure?...

Calcium and Blood Pressure Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was strongest for black men. Such observational studies do not establish causation. Researchers therefore designed a randomized comparative experiment. The subjects were 12 healthy black men who volunteered to take part in the experiment. They were randomly assigned to two groups: 10 of the men received a calcium supplement for 12 weeks, while the control group of 11 men received a placebo pill that looked identical. The experiment was double-blind. The response variable is the decrease in systolic (top number) blood pressure for subject after 12 weeks, in millimeters of mercury.

An increase appears as a negative response here are the data:

Group 1 (calcium): 7 -4 18 17 -3 -5 1 10 11 -2

Group 2 (placebo); -1 12 -1 -3 3 -5 5 2 -11 -1 -3

Do the data provide sufficient evidence to conclude that a calcium supplement reduces blood pressure more than a placebo?

Carry out an appropriate test to support your answer.

??:

 ??:

Find the value for Test Statistic. 

Test Static =  

p-value =  

Step 4:   Circle one:    Reject  ?? or Fail to Reject  ??  

Conclusion in context: 

Solutions

Expert Solution

The Null and Alternative Hypotheses

where, is the mean decrease in systolic blood pressure in Group 1 and   is the mean decrease in systolic blood pressure in Group 2

Since the population standard deviation is not known, the t statistic will be used to compare the two groups mean hence the Two-sample t-test is used to compare the means assuming equal variance.

test statistic

The t statistic is obtained using the formula,

From the data values,

Group 1 Group 2
Mean 5 -0.8
Std Dev 8.7433 5.9404

P-value

The P-value for the t statistic is obtained from t distribution table for the degree of freedom = n1+n2-2=18

The corresponding P-value is 0.0499 which is less than 0.05 at the 5% significance level for the one-sided alternative hypothesis. Hence, it can be concluded that the null hypothesis is rejected.


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