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91Ó°ÊÓ

In a recent year, \(73 \%\) of firstyear college students responding to a national survey identified "being very well-off financially" as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the \(\alpha=0.05\) significance level that the proportion of all first-year students at this university who think being very well-off is important differs from the national value, \(73 \% ?\)

Short Answer

Expert verified
Yes, there is convincing evidence the proportion differs.

Step by step solution

01

Formulate Hypotheses

First, we set up the null and alternative hypotheses. Let \( p \) represent the proportion of first-year students at the university who think being well-off is important. The null hypothesis \( H_0 \) is \( p = 0.73 \), and the alternative hypothesis \( H_a \) is \( p eq 0.73 \).
02

Calculate Sample Proportion

The sample proportion, \( \hat{p} \), is the number of successes (students who think being well-off is important) divided by the total number of students in the sample. So, \( \hat{p} = \frac{132}{200} = 0.66 \).
03

Calculate Standard Error

Find the standard error (SE) for the sample proportion using the formula \( \text{SE} = \sqrt{\frac{p(1-p)}{n}} \), where \( p = 0.73 \) and \( n = 200 \). Plug in the values to get \( \text{SE} = \sqrt{\frac{0.73 \times 0.27}{200}} \approx 0.032 \).
04

Compute Z-score

Use the Z-score formula \( Z = \frac{\hat{p} - p}{\text{SE}} \). Substituting the known values gives \( Z = \frac{0.66 - 0.73}{0.032} \approx -2.19 \).
05

Determine P-value

Look up the P-value for \( Z = -2.19 \) in the standard normal (Z) distribution table or use a calculator. The two-tailed P-value is approximately \( 2 \times 0.0143 = 0.0286 \).
06

Decision Making

Since the P-value of 0.0286 is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis \( H_0 \).
07

Conclusion

There is convincing evidence at the \( \alpha=0.05 \) significance level that the proportion of first-year students at this university who think being very well-off is important differs from 73%.

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Key Concepts

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

Sample Proportion
When conducting a hypothesis test about a population proportion, the sample proportion plays a critical role. It is represented by \( \hat{p} \) and calculated by dividing the number of "successes" in the sample by the total number of observations. In our exercise, we have 132 students out of 200 who view financial success as important. Thus, the sample proportion is \( \hat{p} = \frac{132}{200} = 0.66 \).

The sample proportion serves as our estimate of the true population proportion within the specific sample. It provides a point of comparison against the national value of \( 0.73 \) used in the null hypothesis. This simple yet powerful calculation helps us start our journey in testing the initial assumptions made about the population.
Standard Error
The standard error (SE) is a measure that describes the accuracy of the sample proportion as an estimate of the population proportion. It quantifies how much the sample proportion \( \hat{p} \) is expected to fluctuate from the true population proportion \( p \). In this context, the formula used is \[ \text{SE} = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the hypothesized population proportion (0.73 in this case), and \( n \) is the sample size (200).

Plugging in these values gives us \( \text{SE} = \sqrt{\frac{0.73 \times 0.27}{200}} \approx 0.032 \).

The standard error informs us about the sampling variability and is crucial when interpreting the results of our hypothesis test. It acts as the denominator in the Z-score formula, contributing significantly to the analysis.
Z-score
In hypothesis testing, the Z-score is a statistic that tells us how far away our sample's result, or \( \hat{p} \), is from the population proportion \( p \). This distance is measured in units of the standard error. The Z-score formula is given by: \[ Z = \frac{\hat{p} - p}{\text{SE}} \].

For the current problem, \( Z = \frac{0.66 - 0.73}{0.032} \), which simplifies to approximately \( -2.19 \).

A Z-score helps us understand the likelihood of observing our sample proportion under the null hypothesis. It is a bridge between sample statistics and probability distributions, translating our observations into a language that can be interpreted using statistical tables or software.
P-value
The P-value is an integral part of hypothesis testing, serving as a measure of the evidence against the null hypothesis provided by the sample. It is the probability of observing a test statistic as extreme as the sample result, assuming the null hypothesis is true.

In this instance, with a Z-score of -2.19, we consult a standard normal distribution table or use software to find the P-value. We get approximately \( 2 \times 0.0143 = 0.0286 \) because it's a two-tailed test.

Since this P-value (0.0286) is less than our significance level of 0.05, it indicates strong evidence against the null hypothesis, suggesting that the actual proportion of students valuing being well-off financially at this university indeed differs from the national average of 73%.

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Most popular questions from this chapter

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