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Is college worth it? Part. I. Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, \(48 \%\) said they decided not to go to college because they could not afford school. 40 (a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement. (b) Would you expect a confidence interval for the proportion of American adults who decide not to go to college because they cannot afford it to include 0.5? Explain.

Short Answer

Expert verified
There is insufficient evidence that less than 50% cannot afford college. Likely, a confidence interval includes 0.5.

Step by step solution

01

Define Hypotheses

We begin by defining our null and alternative hypotheses. For part (a), we want to test whether the proportion, \( p \), of Americans who decide not to go to college because they cannot afford it is less than 0.5. \( H_0: p = 0.5 \) (the proportion is 0.5) and \( H_a: p < 0.5 \) (the proportion is less than 0.5).
02

Check Sample Conditions

We need to verify that the sample data meets conditions for the normal approximation to hold. The sample size is 331. Check if \( n\hat{p} \) and \( n(1-\hat{p}) \) are both greater than 10. Here, \( \hat{p} = 0.48 \); thus, \( n\hat{p} = 331 \times 0.48 = 158.88 \) and \( n(1-\hat{p}) = 331 \times 0.52 = 172.12 \), both greater than 10.
03

Calculate Test Statistic

Using the formula for the standard error of the proportion \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( p = 0.5 \), \[ SE = \sqrt{\frac{0.5 \times 0.5}{331}} \approx 0.0275 \]. The test statistic \( z \) is calculated as \( z = \frac{\hat{p} - p}{SE} = \frac{0.48 - 0.5}{0.0275} \approx -0.7273 \).
04

Find P-value

Using the standard normal distribution, find the p-value associated with the calculated z-score. A z-score of approximately -0.7273 corresponds to a p-value of about 0.233. This isn't significant at usual levels (e.g., 0.05).
05

Draw Conclusion for Hypothesis Test

Since the p-value is greater than 0.05, we fail to reject the null hypothesis. There's not enough evidence to support the claim that a minority (less than 50%) of Americans who don't go to college do so because they can't afford it.
06

Confidence Interval Expectation

A confidence interval centered at a sample proportion of 0.48 with standard error calculated should likely not include 0.5 if the interval is narrow at usual confidence levels like 95%. However, based on our test hypothesis, without exact calculations, a wider interval may include 0.5.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate a population parameter based on sample data. It provides a range of plausible values for a parameter such as a mean or proportion. In the context of this exercise, we are interested in the confidence interval for the proportion of Americans who chose not to attend college due to financial constraints. The confidence interval is constructed around the sample proportion, which is 0.48 in this case. This interval accounts for variability in the sample by incorporating the standard error. The interval has a certain level of confidence, often 95%, which indicates that if we were to take many samples, 95% of the calculated intervals would contain the true population proportion. To calculate the confidence interval, use the formula: \[ \hat{p} \pm z^* \cdot SE \] where \( \hat{p} \) is the sample proportion (0.48), \( z^* \) is the critical value from the standard normal distribution for the desired confidence level, and \( SE \) is the standard error of the sample proportion. Understanding how the confidence interval relates to our hypothesis test helps infer whether certain values, like 0.5, might reasonably represent the true proportion of Americans in this context.
Sample Proportion
The sample proportion, often denoted as \( \hat{p} \), represents the fraction of individuals in a sample with a particular trait. In this problem, it refers to the proportion of sampled American adults who did not attend college due to unaffordability. Here, our sample proportion is 0.48, indicating that 48% of the surveyed participants cited cost as their reason for not attending college. The sample proportion serves as an estimate for the true population proportion and plays a crucial role in both hypothesis testing and constructing confidence intervals. A higher or lower sample proportion would directly affect subsequent calculations and interpretations. It's important to understand that the sample proportion reflects only the data observed in the particular study and might vary if a different sample were taken. Calculating its standard error helps gauge how much the sample proportion might differ from the actual population proportion, thus facilitating our hypothesis testing analysis.
P-value
The p-value is a measure used in hypothesis testing to determine the strength of the evidence against the null hypothesis. It represents the probability of observing a test statistic at least as extreme as the one obtained, assuming that the null hypothesis is true. In this exercise, the null hypothesis claims that 50% of Americans who forgo college do so chiefly because it is unaffordable. Our calculated p-value of approximately 0.233 indicates how likely it is to obtain a sample proportion of 0.48 or less purely by chance. A p-value less than 0.05 typically suggests significant evidence against the null hypothesis, prompting rejection. However, in our case, a p-value of 0.233 means we lack sufficient evidence to reject the hypothesis that the actual proportion isn't less than 50%. Therefore, the data does not strongly support the claim that fewer than half of the people avoid college due to cost.

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Most popular questions from this chapter

Young Americans, Part. II. About \(25 \%\) of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random samples of size 12 is right skewed. (b) In order for the the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump to be approximately normal, we need random samples where the sample size is at least 40 . (c) A random sample of 50 young Americans where \(20 \%\) have delayed starting a family due to the continued economic slump would be considered unusual. (d) A random sample of 150 young Americans where \(20 \%\) have delayed starting a family due to the continued economic shmp would be considered unusual. (e) Tripling the sample size will reduce the standard error of the sample proportion by one-third.

Offshore drilling, Part II. Results of a poll evaluating support for drilling for oil and natural gas off the coast of California were introduced in Exercise \(3.29 .\) \begin{tabular}{lcc} & \multicolumn{2}{c} { College Grad } \\ \cline { 2 - 3 } & Yes & No \\ \hline Support & 154 & 132 \\ Oppose & 180 & 126 \\ Do not know & 104 & 131 \\ \hline Total & 438 & 389 \end{tabular} (a) What percent of college graduates and what percent of the non-college graduates in this sample support drilling for oil and natural gas off the Coast of California? (b) Conduet a hypothesis test to determine if the data provide strong evidence that the proportion of college graduates who support off-shore drilling in California is different than that of noncollege graduates.

Sleep deprived transportation workers. The National Sleep Foundation conducted a survey on the sleep habits of randomly sampled transportation workers and a control sample of non-transportation workers. The results of the survey are shown below. \begin{tabular}{lccccc} & \multicolumn{4}{c} { Transportation Professionals } \\ \cline { 3 - 6 } & & & Truck & Train & Bux/1axi/Limo \\ & Control & Pilots & Drivers & Operators & Drivers \\ \hline Less than 6 hours of sleep & 35 & 19 & 35 & 29 & 21 \\ 6 to 8 hours of sleep & 193 & 132 & 117 & 119 & 131 \\ More than 8 hours & 64 & 51 & 51 & 32 & 58 \\ \hline Total & 292 & 202 & 203 & 180 & 210 \end{tabular} Conduct a hypothesis test to evaluate if these data provide evidence of a difference between the proportions of truck drivers and non-transportation workers (the control group) who get less than 6 hours of sleep per day, i.e. are considered sleep deprived.

Shipping holiday gifts. A December 2010 survey asked 500 randomly sampled Los Angeles residents which shipping carrier they prefer to use for shipping holiday gifts. The table below shows the distribution of responses by age group as well as the expected counts for each cell (shown in parentheses). Method \begin{tabular}{l|cc|cc|cc|c} & \multicolumn{6}{|c|} { Age } & \\ \cline { 2 - 5 } & \multicolumn{2}{|c|} {\(18-34\)} & \multicolumn{2}{|c|} {\(35-54\)} & \multicolumn{2}{|c|} {\(55+\)} & Total \\ \hline USPS & 72 & \((\mathrm{~B} 1)\) & 97 & (102) & 76 & (62) & 245 \\ UPS & 52 & (83) & 76 & \((6 \mathrm{~b})\) & 34 & (41) & 162 \\ FedEx & 31 & (21) & 24 & (27) & 9 & (16) & 64 \\ Something else & 7 & (5) & 6 & (7) & 3 & (4) & 16 \\ Not sure & 3 & (5) & 6 & (5) & 4 & (3) & 13 \\ \hline Total & \multicolumn{2}{|c|} {165} & \multicolumn{2}{|c|} {209} & \multicolumn{2}{|c|} {126} & 500 \end{tabular} (a) State the null and alternative hypotheses for testing for independence of age and preferred shipping method for holiday gifts among Los Angeles residents. (b) Are the conditions for inference using a chi-square test satisfied?

Study abroad. A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey between April 25 and April 30, 2007 shows that \(55 \%\) of high school seniors are fairly certain that they will participate in a study abroad program in college. (a) Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning. (b) Let's suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a \(90 \%\) confidence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context. (c) What does "90\% confidence" mean? (d) Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college?

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