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stiks02 [169]
3 years ago
14

A recent study examined the association between high blood pressure and increased risk of death from cardiovascular disease. The

re were 2676 men with low blood pressure and 3338 men with high blood pressure. In the low-blood-pressure group, 21 men died from cardiovascular disease; in the high-blood-pressure group, 55 died.
a. Compute the 95% confidence interval for the difference in proportions.

b. Do the study data confirm that death rates are higher among men with high blood pressure? State hypotheses, carry out a significance test, and give your conclusions.
Mathematics
1 answer:
sergey [27]3 years ago
6 0

According to sources, the most probable answer to this query is that thee results seems to support a .05 probability. Meaning, this supports the hypothesis' claim. Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

<h3>What are the mean and the standard error for each sample?</h3>

Considering the data given:

\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285

\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973

<h3>What is the mean and the standard error for the distribution of differences?</h3>

The mean is the subtraction of the means, hence:

\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1

The standard error is the square root of the sum of the variances of each sample, hence:

s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612

<h3>What is the confidence interval?</h3>

It is given by:

\overline{x} \pm zs

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

  • \overline{x} - zs = -1 - 1.96(0.1612) = -1.3
  • \overline{x} + zs = -1 + 1.96(0.1612) = -0.7

More can be learned about the z-distribution at brainly.com/question/25890103

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