Math Problem Statement

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, 264 had BMIs of less than 25, 159 had BMIs that were at least 25 but less than 30, and 122 had BMIs exceeding 30. Is there compelling evidence for concluding that more than 20% of the individuals in the sampled population are obese? A button hyperlink to the SALT program that reads: Use SALT. (a) State the appropriate hypotheses with a significance level of 0.05.

H0: p = 0.20 Ha: p > 0.20 H0: p = 0.20 Ha: p ≠ 0.20
H0: p > 0.20 Ha: p = 0.20 H0: p = 0.20 Ha: p < 0.20 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = What can you conclude?

Do not reject the null hypothesis. There is not sufficient evidence that more than 20% of the population of female workers is obese. Reject the null hypothesis. There is sufficient evidence that more than 20% of the population of female workers is obese.
Reject the null hypothesis. There is not sufficient evidence that more than 20% of the population of female workers is obese. Do not reject the null hypothesis. There is sufficient evidence that more than 20% of the population of female workers is obese. (b) Explain in the context of this scenario what constitutes type I error.

A type I error would be declaring that 20% or less of the population of female workers is obese, when in fact more than 20% are actually obese. A type I error would be declaring that 20% or more of the population of female workers is obese, when in fact less than 20% are actually obese.
A type I error would be declaring that less than 20% of the population of female workers is obese, when in fact 20% or more are actually obese. A type I error would be declaring that more than 20% of the population of female workers is obese, when in fact 20% or less are actually obese. Explain in the context of this scenario what constitutes type II error.

A type II error would be declaring that 20% or less of the population of female workers is obese, when in fact more than 20% are actually obese. A type II error would be declaring that 20% or more of the population of female workers is obese, when in fact less than 20% are actually obese.
A type II error would be declaring that less than 20% of the population of female workers is obese, when in fact 20% or more are actually obese. A type II error would be declaring that more than 20% of the population of female workers is obese, when in fact 20% or less are actually obese. (c) What is the probability of not concluding that more than 20% of the population is obese when the actual percentage of obese individuals is 23%? (Round your answer to four decimal places.) just do c

Solution

To solve part (c), we calculate the probability of not rejecting the null hypothesis (Type II error) when the true proportion of obese individuals is 23% (0.23).

Steps:

  1. Determine the null and alternative hypotheses:

    • H0:p=0.20H_0: p = 0.20
    • Ha:p>0.20H_a: p > 0.20 (one-tailed test)

  2. Compute the standard error:

    • Standard error (SE) = p0(1−p0)n\sqrt{\frac{p_0 (1 - p_0)}{n}}
    • p0=0.20p_0 = 0.20, and n=264+159+122=545n = 264 + 159 + 122 = 545
    • SE=0.20(1−0.20)545=0.20(0.80)545=0.0171SE = \sqrt{\frac{0.20(1 - 0.20)}{545}} = \sqrt{\frac{0.20(0.80)}{545}} = 0.0171 (rounded to 4 decimal places)

  3. Find the test statistic for the rejection region:

    • At α=0.05\alpha = 0.05, the critical z-value for a one-tailed test is 1.645.
    • The corresponding sample proportion (p^\hat{p}) for rejecting H0H_0 is: p^=p0+z⋅SE=0.20+1.645⋅0.0171=0.2281\hat{p} = p_0 + z \cdot SE = 0.20 + 1.645 \cdot 0.0171 = 0.2281

  4. Compute the probability of Type II error (β\beta) when p=0.23p = 0.23:

    • Compute the z-score for p^=0.2281\hat{p} = 0.2281 under p=0.23p = 0.23: z=p^−pp(1−p)n=0.2281−0.230.23(1−0.23)545z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.2281 - 0.23}{\sqrt{\frac{0.23(1 - 0.23)}{545}}}
    • SEtrue=0.23(1−0.23)545=0.0182SE_{true} = \sqrt{\frac{0.23(1 - 0.23)}{545}} = 0.0182 (rounded to 4 decimal places)
    • z=0.2281−0.230.0182=−0.10z = \frac{0.2281 - 0.23}{0.0182} = -0.10

  5. Find the cumulative probability corresponding to z=−0.10z = -0.10:

    • From z-tables or a calculator, P(z≤−0.10)=0.4602P(z \leq -0.10) = 0.4602.

Thus, the probability of Type II error (β\beta) is 0.4602.

Final Answer:

The probability of not concluding that more than 20% of the population is obese when the actual percentage is 23% is:

0.4602\boxed{0.4602}

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Type II Error
Standard Error
Proportion Testing

Formulas

Standard Error (SE) = sqrt(p(1-p)/n)
Test Statistic (z) = (p_hat - p_0) / SE
Probability of Type II Error (β) = P(Z <= z)

Theorems

Central Limit Theorem
Properties of the Standard Normal Distribution

Suitable Grade Level

Undergraduate Level (Statistics)

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