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A common characterization of obese individuals is that their body mass index is at least \(30\) (BMI 5 weighty(height)2, where height is in meters and weight is in kilograms). The article 鈥淭he Impact of Obesity on Illness Absence and Productivity in an Industrial Population of Petrochemical Workers鈥 (Annals of Epidemiology, 2008: 8鈥14) reported that in a sample of female workers, \(262\) had BMIs of less than \(25,159\) had BMIs that were at least \(25\) but less than \(30\), and \(120\) had BMIs exceeding \(30\). Is there compelling evidence for concluding that more than \(20\% \) of the individuals in the sampled population are obese? a. State and test appropriate hypotheses with a significance level of \(.05\). b. Explain in the context of this scenario what constitutes type I and II errors. 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 \(25\% \)?

Short Answer

Expert verified

(a) There isn't enough evidence to back up the allegation that more than\(20\% \)of the people in the sample population are fat.

(b) The type I is 鈥淭he sample population is more than\(0.20\)or\(20\% \)are obese, which is incorrect鈥, and the type II is 鈥淭he sample population is more than\(0.20\)or\(20\% \)are not obese, which is incorrect鈥.

(c) The probability of not concluding is \(\beta = 12.10\% \).

Step by step solution

01

Define p-value in hypothesis testing.

The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:p = 0\\{H_a}:p \ne 0\end{array}\)

The formula for the value of the test statistic is given by,\(z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\).

The sample proportion is calculated by dividing the number of successes by the sample size. \(\hat p = \frac{x}{n}\)

02

Test the appropriate hypothesis.

(a)

Let the given be:

\(\begin{aligned}{c}x &= 120\\n &= 262 + 159 + 120\\ &= 541\\\alpha &= 0.05\end{aligned}\)

Claim that the proportion is more than\(0.20\)or\(20\% \).

Sample proportion:

\(\begin{aligned}{c}\hat p &= \frac{x}{n}\\ &= \frac{{120}}{{541}}\\ &\approx 0.2218\end{aligned}\)

The value of the test-statistic:

\(\begin{aligned}{c}z &= \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\\ &= \frac{{0.2218 - 0.2}}{{\sqrt {\frac{{0.2(1 - 0.2)}}{{541}}} }}\\ &\approx 1.27\end{aligned}\)

When the null hypothesis is true, the P-value is the chance of getting the test statistic's value, or a value that is more extreme. Using the normal probability table in the appendix, calculate the P-value.

\(\begin{aligned}{c}P &= P(Z > 1.27)\\ &= 1 - P(Z < 1.27)\\ &= 1 - 0.8980\\ &= 0.1020\end{aligned}\)

Since the P-value is smaller than the significance level\(\alpha \), then reject the null hypothesis:

\(P > 0.05 \Rightarrow {\rm{Fail to reject }}{H_0}\)

There isn't enough evidence to back up the allegation that more than \(20\% \) of the people in the sample population are fat.

03

Describe the type I and type II error.

(b)

Claim that the proportion is more than\(0.20\)or\(20\% \).

The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:p = 0.20\\{H_a}:p > 0.20\end{array}\)

Type I error: When\({H_0}\)is true, reject the null hypothesis\({H_0}\).

Interpretation: The sample population is more than\(0.20\)or\(20\% \)are obese, which is incorrect.

Type Il error: When\({H_0}\)is false, fail to reject the null hypothesis\({H_0}\).

Interpretation: The sample population is more than \(0.20\) or \(20\% \) are not obese, which is incorrect.

04

Determine the probability.

(c)

Let: \({p_A} = 25\% = 0.25\)

Claim that the population mean expense ratio exceeds\(1\% \).

Using the normal probability table in the appendix, find the z-score corresponding to a probability of\(1 - \alpha = 0.95\)(Note: take the complement because the test is right-sided):

\(z = 1.645\)

The population mean (of the hypothesis) is increased by the product of the z-score and the standard deviation to get the sample mean:

\(\begin{aligned}{c}\hat p &= p + z\sqrt {\frac{{p(1 - p)}}{n}} \\ &= 0.20 + 1.645\sqrt {\frac{{0.20(1 - 0.20)}}{{541}}} \\ &\approx 0.2283\end{aligned}\)

The z-value is the sample mean divided by the standard deviation, after subtracting the population mean (alternative mean).

\(\begin{aligned}{c}z &= \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\\ &= \frac{{0.2283 - 0.25}}{{\sqrt {\frac{{0.25(1 - 0.25)}}{{541}}} }}\\ &\approx - 1.17\end{aligned}\)

When the null hypothesis is false, the probability of making a type II error is the probability of not rejecting the null hypothesis. Using the normal probability table in the appendix, calculate the chances of failing to reject the null hypothesis.

\(\begin{aligned}{c}\beta &= P(Z < - 1.17)\\ &= 0.1210\\ &= 12.10\% \end{aligned}\)

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Most popular questions from this chapter

Each of a group of \(20\) intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let \(p\) denote the proportion of all such players who would prefer gut to nylon, and let \(X\) be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most \(50\% \) of all such players prefer gut. We simplify this to \({H_0}:p = .5\), planning to reject \({H_0}\) only if sample evidence strongly favors gut strings.

a. Is a significance level of exactly \(.05\) achievable? If not, what is the largest a smaller than \(.05\) that is achievable?

b. If \(60\% \) of all enthusiasts prefer gut, calculate the probability of a type II error using the significance level from part (a). Repeat if 80% of all enthusiasts prefer gut.

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A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data (from 鈥淓ngineering Properties of Soil,鈥 Soil Science, 1998: 93鈥102).

\(\begin{array}{l}1.10 5.09 0.97 1.59 4.60 0.32 0.55 1.45\\0.14 4.47 1.20 3.50 5.02 4.67 5.22 2.69\\3.98 3.17 3.03 2.21 0.69 4.47 3.31 1.17\\0.76 1.17 1.57 2.62 1.66 2.05\end{array}\)

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b.If a level .01 test is used, what is 尾(30,500)?

c.If a level .01 test is used and it is also required that 尾(30,500) = .05, what sample size n is necessary?

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A normal probability plot shows a reasonably linear pattern.

a. Is there compelling evidence for concluding that the population mean expense ratio exceeds \(1\% \)? Carry out a test of the relevant hypotheses using a significance level of \(.01\).

b. Referring back to (a), describe in context type I and II errors and say which error you might have made in reaching your conclusion. The source from which the data was obtained reported that \(\mu = 1.33\) for the population of all \(762\) such funds. So, did you actually commit an error in reaching your conclusion?

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