Can anyone please help me
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Answer:
The hypothesis is:
<em>H₀</em>: .
<em>Hₐ</em>: .
Step-by-step explanation:
Let <em>X</em> = number of men who exercise regularly and <em>Y</em> = number of women who exercise regularly.
The information provided is:
Compute the sample proportion of men and women who exercise regularly as follows:
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and .
The random variable <em>Y</em> also follows a Binomial distribution with parameters <em>n</em> = 200 and .
According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
The standard deviation of this sampling distribution of sample proportion is:
So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.
A two proportion <em>z</em>-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.
The hypothesis for this test cab be defined as:
<em>H₀</em>: The proportion of women is same as men who exercise regularly, i.e. .
<em>Hₐ</em>: The proportion of women is more than men who exercise regularly, i.e. .