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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

Vegetarian college students. Suppose that \(8 \%\) of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since \(n \geq 30\). (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where \(12 \%\) are vegetarians would be considered unusual. (d) A random sample of 250 college students where \(12 \%\) are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250 .

Prenatal vitamins and Autism. Researchers studying the link between prenatal vitamin use and autism surveyed the mothers of a random sample of children aged \(24-60\) months with autism and conducted another separate random sample for children with typical development. The table below shows the number of mothers in each group who did and did not use prenatal vitamins during the three months before pregnancy (periconceptional period). 40 \begin{tabular}{llccc} & \multicolumn{4}{c} { Autism } \\ \cline { 3 - 4 } & & Autism Typical development & Total \\ \cline { 2 - 5 } Periconceptional & No vitamin & 111 & 70 & 181 \\ prenatal vitamin & Vitamin & 143 & 159 & 302 \\ \cline { 2 - 5 } & Total & 254 & 229 & 483 \end{tabular} (a) State appropriate hypotheses to test for independence of use of prenatal vitamins during the three months before pregnancy and autism. (b) Complete the hypothesis test and state an appropriate conclusion. (Reminder: verify any necessary conditions for the test.) (c) A New York Times article reporting on this study was titled "Prenatal Vitamins May Ward Off Autism", Do you find the title of this article to be appropriate? Explain your answer. Additionally, propose an alternative title.

True or false, Part. II. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(X^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

3.29 Offshore drilling, Part 1. A 2010 survey asked 827 randomly sampled registered voters in California "Do you support? Or do you oppose? Drilling for oil and natural gas off the Coast of California? Or do you not know enough to say?" Below is the distribution of responses, separated based on whether or not the respondent graduated from college. (a) What percent of college graduates and what percent of the non-college graduates in this sample do not know enough to have an opinion on drilling for oil and natural gas off the Coast of California? \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} the (b) Conduct a hypothesis test to determine if data provide strong evidence that the proportion of college graduates who do not have an opinion on this issue is different than that of non-college graduates.

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