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Chapter 7: Problem 29

In a Gallup poll conducted December \(2-4,2000,42 \%\) of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conduct a survey of 150 randomly selected students on your campus and find that 80 of them would prefer a boy. (a) Approximate the probability that, in a random sample of 150 students, at least 80 would prefer a boy, assuming the true percentage is \(42 \%\) (b) Does this result contradict the Gallup poll? Explain.

Short Answer

Expert verified
The probability of at least 80 out of 150 preferring a boy is 0.22%. This significantly contradicts the Gallup poll result of 42%.

Step by step solution

01

Define the Problem

We are given that 42% of survey respondents prefer a boy. We are to approximate the probability that at least 80 out of 150 randomly selected students prefer a boy.
02

Define the Population Proportion and Sample Size

Let the population proportion be denoted as \( p = 0.42 \). The sample size \( n \) is 150, and the sample proportion \( \hat{p} \) is calculated from the sample data: \hat{p} = \frac{80}{150}.
03

Calculate the Sample Proportion

Calculate the sample proportion:\[ \hat{p} = \frac{80}{150} = 0.5333 \]
04

Check Normality Assumption

Check if the sample size is large enough to assume normality using the formulas \( np \geq 10 \) and \( n(1-p) \geq 10 \):\[ np = 150 \cdot 0.42 = 63 \geq 10 \]\[ n(1-p) = 150 \cdot 0.58 = 87 \geq 10 \]Since both conditions are satisfied, we can use the normal approximation.
05

Calculate the Standard Error

The standard error (SE) of the sample proportion is given by:\[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.42 \cdot 0.58}{150}} \approx 0.0398 \]
06

Calculate the Z-Score

We need to find the Z-score for the sample proportion \hat{p} = 0.5333\ using \( p = 0.42 \) and SE calculated:\[ Z = \frac{\hat{p} - p}{SE} = \frac{0.5333 - 0.42}{0.0398} \approx 2.85 \]
07

Find the Probability

Using the Z-score of 2.85, find the probability from the Z-table or using a calculator. The area to the left of Z = 2.85 is approximately 0.9978. Therefore, the probability of at least 80 students preferring a boy is:\[ P(Z > 2.85) = 1 - P(Z < 2.85) = 1 - 0.9978 = 0.0022 \]
08

Result Interpretation

The probability calculated in Step 7 (0.0022) indicates that there is a very low probability (0.22%) that at least 80 out of 150 students would prefer a boy if the true percentage is 42%.
09

Conclusion about the Gallup Poll

Given the low probability (0.22%), it suggests that the survey result (80 out of 150) significantly contradicts the Gallup poll result that 42% would prefer a boy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

population proportion
In probability and statistics, the **population proportion** refers to the percentage of a population that possesses a certain characteristic. For example, if 42% of a population prefers a boy, then the population proportion, denoted by \( p \), is 0.42. When conducting a survey, the population proportion is used to make inferences about the entire population. This helps us understand what portion of the population has a particular attribute, based on data from a specific sample.
Here, the Gallup poll had a population proportion of 42%, which carries over to our exercise.
normal approximation
The **normal approximation** is a technique used in statistics for approximation of a binomial distribution to a normal distribution. This is particularly useful when dealing with large sample sizes, as it allows for simpler calculations.
To use normal approximation, certain conditions must be met. We generally check:
  • \( np \geq 10 \)
  • \( n(1-p) \geq 10 \)

If these conditions are satisfied, the binomial distribution can be approximately represented by a normal distribution, making it easier to compute probabilities.
Z-score
The **Z-score** is a measure of how many standard deviations an element is from the mean. It allows for comparison between different sets of data, standardizing results.
The formula for calculating the Z-score of a sample proportion \( \hat{p} \) when the population proportion \( p \) and standard error \( SE \) are known is given by:
\[ Z = \frac{\hat{p} - p}{SE} \]
The Z-score is then used to find probabilities associated with standard normal distributions. For example, a Z-score of 2.85 means the sample proportion is 2.85 standard deviations above the population proportion.
standard error
The **standard error** (SE) measures the accuracy with which a sample proportion represents the population proportion. It is essentially the standard deviation of the sampling distribution of a statistic.
For a sample proportion, the standard error can be calculated using the formula:
\[ SE = \sqrt{\frac{p(1-p)}{n}} \]
This formula accounts for the variability in the sample proportion around the population proportion. Smaller SE values indicate a more precise estimate of the population proportion from the sample data.

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Most popular questions from this chapter

Find the indicated \(Z\) -score. Be sure to draw a standard normal curve that depicts the solution. Find the \(Z\) -score such that the area under the standard normal curve to the left is 0.2

Find the indicated probability of the standard normal random variable \(Z\). $$P(Z>0.92)$$

Compute \(P(x)\) using the binomial probability formula. Then determine whether the normal distribution can be used as an approximation for the binomial distribution. If so, approximate \(P(x)\) and compare the result to the exact probability. $$n=75, p=0.75, X=60$$

Find the indicated probability of the standard normal random variable \(Z\). $$P(Z<1.93)$$

Find the indicated \(Z\) -score. Be sure to draw a standard normal curve that depicts the solution. Find the \(Z\) -score such that the area under the standard normal curve to the right is 0.25
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