A common characterization of obese individuals is that their body mass index is at least 30 [BMI = weight/(height)2, where height is in meters and weight is in kilograms]. An article reported that in a sample of female workers, 265 had BMIs of less than 25, 159 had BMIs that were at least 25 but less than 30, and 121 had BMIs exceeding 30. Is there compelling evidence for concluding that more than 20% of the individuals in the sampled population are obese?

According to the definition of an obese person (From 30 to n), the 22.2% of the sampled population are obese. Then, yes more than 20% of the are obese.

MrRoyal MrRoyal · Answer 2 · 2022-04-03T17:46:27+02:00

There is no compelling evidence to conclude that more than 20% of the individuals in the sampled population are obese

How to derive a conclusion?

Start by calculating the sample size (n)

[tex]n = 265 + 159 + 121[/tex]

[tex]n = 545[/tex]

From the complete question, 120 of the sample are obese.

So, we start by calculating the proportion that are obese

[tex]\^p = \frac{120}{545}[/tex]

[tex]\^p = 0.2202[/tex]

The test is a right tailed test.

For 20% of the individuals in the sampled population to be obese, the hypotheses would be:

Null: [tex]H_o : p = 20\%[/tex]

Alternate: [tex]H_a : p \ge 20\%[/tex]

Calculate the test statistic

[tex]z = \frac{\^ p - p }{\sqrt{p(1 - p)/n}}[/tex]

This gives

[tex]z = \frac{0.2202 - 0.20 }{\sqrt{0.20(1 - 0.20)/545}}[/tex]

[tex]z = \frac{0.0202}{0.0171}[/tex]

[tex]z = 1.18[/tex]

The corresponding p-value at significance level of 0.05 is:

[tex]p = 0.1190[/tex]

By comparison:

0.119 is less than 1.18

This means that we fail to reject the null hypothesis

Hence, there is no compelling evidence to conclude that more than 20% of the individuals in the sampled population are obese

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