Ayaana Buck

2021-01-30

In a study designed to see whether a controlled diet could retard the process of arteriosclerosis, a total of 846 randomly chosen persons were followed over an eight year period. Half were instructed to eat only certain foods, the other half could eat whatever they wanted. At the end of eight years, sixty-six persons in the diet group were found to have died of either myocardial infarction or cerebral infarction, as compared to ninety-three deaths of a similar nature in the control group. Do the appropriate analysis. Let $\alpha =0.05$

Faiza Fuller

Given:
${x}_{1}=66$
${n}_{1}=\frac{846}{2}=423$
${x}_{2}=93$
${n}_{2}=\frac{846}{2}=423$
$\alpha =0.05$
The sample proportion is the number of successes divided by the sample size:
${\stackrel{^}{p}}_{1}=\frac{{x}_{1}}{{n}_{1}}=\frac{66}{423}\approx 0.1560$
${\stackrel{^}{p}}_{2}=\frac{{x}_{2}}{{n}_{2}}=\frac{93}{423}\approx 0.2199$
${\stackrel{^}{p}}_{p}=\frac{{x}_{1}+{x}_{2}}{{n}_{1}+{n}_{2}}=\frac{66+93}{423+423}\approx 0.1879$
Claim: Diet retards the process
The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.
${H}_{0}:{p}_{1}={p}_{2}$
${H}_{a}:{p}_{1}<{p}_{2}$
The critical values are the values corresponding to a probability of $\alpha =0.05$ in the normal probability table in the appendix:
$z=-1.645$
The rejection region then contains all values below -1.645.
Determine the value of the test statistic:
$z=\frac{{\stackrel{^}{p}}_{1}-{\stackrel{^}{p}}_{2}}{\sqrt{{\stackrel{^}{p}}_{p}\left(1-{\stackrel{^}{p}}_{p}\right)}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}=\frac{0.1560-0.2199}{\sqrt{0.1879\left(1-0.1879\right)}\sqrt{\frac{1}{423}+\frac{1}{423}}}\approx -2.379$
If the value of the test statistic is within the rejection region, then the null hypothesis is rejected:

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