Hypothesis testing with significance level

Question

Is the chance of hypertony is the same in normal and overweighted population? How can we decide it using the following evidences with significance level α = 0.01? Out of 4200 normal patients 792, while out of 1000 overweighted ones 249 suffer of hypertony. Next decide, whether the overweight increases the chance of hypertony, significantly.

To be honest, I have no idea with the above problem? I was wondering if someone can help me? Thanks.

Answer 1

Try creating a hypothesis for the difference in proportions: I'll supply a vague outline but you should use correct notation and terms for your hypothesis test.

Null: The chance of getting Hypertony is the same between a normal and overweight population OR $$\text{$\mu $o}=\text{$\mu $n}$$

Alternative: being overweight increases the chance of hypertony OR $$\text{$\mu $o}>\text{$\mu $n}$$

* SIDE NOTE: I should have used "P-hat" and not "M" because we are using proportions and not means/averages

Find the proportion of normal and overweight patients who suffer from hypertony. This is just p = x/n

  x1 = 792
  x2 = 249
  n1 = 4200
  n2 = 1000
  p1 = .18857
  p2 = .249

I really should have rounded p1 to .189, but anyways Can you implement these values in a ti-83 calculator yourself?

Heres how to do it brute force by hand:

$$\text{upper}=.06043 + 2.575 \sqrt{\frac{(1-\text{p1}) \text{p1}}{\text{n1}}+\frac{(1-\text{p2}) \text{p2}}{\text{n2}}}$$

$$\text{lower}=.06043 - 2.575 \sqrt{\frac{(1-\text{p1}) \text{p1}}{\text{n1}}+\frac{(1-\text{p2}) \text{p2}}{\text{n2}}}$$

$$=0.02194,0.09892$$

NOTE this does not supply a p-value but since the whole interval is positive and since we used overweight-normal, then this is evidence that there is a higher chance of having hypertony if overweight.. Only a graphing calculator or other statistical software will supply a P-value

OR use statistical software such as R.

 prop.test(x=c(249,792), n=c(1000,4200), conf.level = .99, alternative = "greater")

2-sample test for equality of proportions with continuity correction

 data:  c(249, 792) out of c(1000, 4200)
 X-squared = 18.045, df = 1, p-value = 1.079e-05
 alternative hypothesis: greater
 99 percent confidence interval:
 0.02503623 1.00000000
 sample estimates:
 prop 1    prop 2 
 0.2490000 0.1885714 

With P-value <.0001 you can safely reject the null hypothesis and conclude that being overweight does increase the chance of hypertony

I hope this helps..