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Morgarella [4.7K]
3 years ago
8

According to a​ study, 47 47​% of all males between the ages of 18 and 24 live at home. ​ (unmarried college students living in

a dorm are counted as living at​ home.) suppose that a survey is administered and 99 99 of 208 208 respondents indicated that they live at home.​ (a) use the normal approximation to the binomial to approximate the probability that at least 99 99 respondents live at​ home? (b) do the results from part​ (a) contradict the​ study?
Mathematics
1 answer:
natita [175]3 years ago
6 0
The probability is 0.4325.  This is close to 0.47, so no, the results do not really contradict the study.

To approximate a normal distribution, we use the fact that 

μ = np and

σ² = np(1-p).

This means that σ = √(np(1-p)).

Our formula for a z-score is 

z = (X-μ)/σ; using what we just stated, this makes it

z = (X-np)/(√(np(1-p))

For this problem, n=208, p=0.47, and X=99:

z = (99-(208)(0.47))/√((208)(0.47)(0.53))
z = (99-97.76)/√(51.8128) = 1.24/(√51.8128) = 0.17

Using a z-table (http://www.z-table.com) we see that the area under the curve to the left of, less than, this z-score is 0.5675.  We want the probability of <u>at least</u> 99 people; this is greater than or equal to, not less than.  To find what we need, we subtract from 1:
1 - 0.5675 = 0.4325.
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Answer:

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

\hat p=\frac{18}{100}=0.18 estimated proportion of ounce cups of coffee that were underfilled

p_o=0.1 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion it's higher than 0.1 or 10%:  

Null hypothesis:p\leq 0.1  

Alternative hypothesis:p > 0.1  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a one right tailed test the p value would be:  

p_v =P(Z>2.67)=0.0037  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance thetrue proportion is not significanlty higher than 0.1 or 10% .  

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3 years ago
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