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lbvjy [14]
3 years ago
10

I need to solve 6 2/3-21/6

Mathematics
1 answer:
lora16 [44]3 years ago
4 0

Step-by-step explanation:

62/3-21

62/18=0.290

290/6

Divide it

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A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w
melamori03 [73]

Answer:

The hypothesis is:

<em>H₀</em>: p_{X}-p_{Y}=0.

<em>Hₐ</em>: p_{X}-p_{Y}.

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:

n_{X}=150\\X=88\\n_{Y}=200\\Y=130

Compute the sample proportion of men and women who exercise regularly as follows:

\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587

\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and \hat p_{X}=0.587.

The random variable <em>Y</em> also follows a Binomial distribution with parameters <em>n</em> = 200 and \hat p_{Y}=0.65.

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:

\hat p=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

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. p_{X}-p_{Y}=0.

<em>Hₐ</em>: The proportion of women is more than men who exercise regularly, i.e. p_{X}-p_{Y}.

6 0
3 years ago
Read 2 more answers
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