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Brums [2.3K]
3 years ago
8

I need help with my test please someone help me

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
7 0

Answer: whats the question

Step-by-step explanation:

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The data in NutritionStudy include information on nutrition and health habits of a sample of people. One of the variables is Smo
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<em>H₀</em>: <em>p</em> = 0.20.

<em>Hₐ</em>: <em>p</em> ≠ 0.20.

Step-by-step explanation:

The question is:

The data in Nutrition Study include information on nutrition and health habits of a sample of 315 people. One of the variables is Smoke, indicating whether a person smokes or not (yes or no). Use technology to test whether the data provide evidence that the proportion of smokers is different from 20% given that 43 identify themselves as smokers. Clearly state the null and alternative hypotheses

In this case we need to test whether the proportion of smokers is different from 20%.

A one-proportion <em>z</em>-test can be used to determine the conclusion for this test.

The hypothesis defined as:

<em>H₀</em>: The proportion of smokers is 20%, i.e. <em>p</em> = 0.20.

<em>Hₐ</em>: The proportion of smokers is different from 20%, i.e. <em>p</em> ≠ 0.20.

The information provided is:

<em>n</em> = 315

<em>X</em> = number of people who identified themselves as smokers = 43

Compute the sample proportion of smokers as follows:

\hat p=\frac{X}{n}=\frac{43}{315}=0.137

Compute the test statistic as follows:

z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.137-0.20}{\sqrt{\frac{0.20(1-0.20)}{315}}}=-2.80

The test statistic is -2.80.

Compute the <em>p</em>-value as follows:

p-value=2\times P(Z

*Use a <em>z</em>-table.

The <em>p</em>-value is 0.00512.

The <em>p</em>-value is quite small. So, the null hypothesis will be rejected at any significance level.

Thus, it can be concluded that the  proportion of smokers is different from 20%.

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