Question

In: Statistics and Probability

Salt-free diets are often prescribed to people with high blood pressure. The following data values were...

  1. Salt-free diets are often prescribed to people with high blood pressure. The following data values were obtained from an experiment designed to estimate the reduction in diastolic blood pres- sure as a result of consuming a salt-free diet for 2 weeks. Assume diastolic readings to be normally distributed.

    (a) Calculate a 90% confidence interval for the mean reduction.
    (b) Is there evidence that diastolic blood pressure is reduced by having salt-free diets, use α = 0.05.

Patients A. B. C.    D. E. F. G. H

Before. 93.   106 87 92 102 95 88 110

After. 92. 102 89 92 101 96 88 105

Solutions

Expert Solution

(a) From the Data: = -1 and Sd = 2.3905

= 0.10, the degrees of freedom = n – 1 = 7

The critical value = 1.895

The Confidence interval is given by ME

The Lower Limit = -1 – 1.602 = -2.602

The Upper Limit = -1 + 1.602 = 0.602

The 90% Confidence interval is (-2.602, 0.602)

_______________________________

(b) The Hypothesis:

H0: = 0 : The test preparation has no effect on the scores.

Ha: < 0: The test preparation has an effect on the scores.

This is a left tailed test

The Test Statistic: Since sample size is small, and population std. deviation is unknown, we use the students t test.

The p value: (Left tailed) at t = -1.18, degrees of freedom = n -1 = 8 - 1 = 7 is p value = 0.1383

The Critical values: (Left Tail) for = 0.05, df = 7 is -1.895

The Decision Rule:   If t observed is < -t critical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision:   Since t observed (-1.18) is > t critical (-1.895), We fail to Reject H0.

Also since P value (0.1383) is > (0.05) , We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% significance level to conclude that the diastolic blood pressure is reduced by having salt free diets.

____________________________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.

# Difference Mean (X-Mean)2
1 -1 -1 0
2 -4 -1 9
3 2 -1 9
4 0 -1 1
5 -1 -1 0
6 1 -1 4
7 0 -1 1
8 -5 -1 16
n 8
Sum -8
Mean -1
SS 40
Variance 5.7143
Std Dev 2.3905

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