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According to an article in Forbes magazine of April 3, 2014 , \(57 \%\) of students said that they did not attend the college of their first choice due to financial concerns (www.forbes.com). In a recent poll of 1600 students, 864 said that they did not attend the college of their first choice due to financial concerns. Using a \(1 \%\) significance level. perform a test of hypothesis to determine whether the current percentage of students who did not attend the college of their first choice due to financial concerns is lower than \(57 \%\). Use both the \(p\) -value and the critical-value approaches.

Short Answer

Expert verified
Given the information, the calculated Z score and corresponding P-value (use statistical software or online calculator) can tell us whether to reject or not reject the null hypothesis. If P <= 0.01 or Z <= - Critical Value, the conclusion will be to reject the null hypothesis, otherwise, we do not reject null hypothesis. There is insufficient data here to compute the final result. With the right values, following these steps will enable the testing of the hypothesis.

Step by step solution

01

Set the Null and Alternate Hypotheses

The null hypothesis \(H_0\) is usually a statement of no change or no effect. The alternate hypothesis \(H_a\) is what we'd conclude if the null hypothesis doesn't hold. In this case: \n- Null hypothesis \(H_0: P = 0.57\) (The percentage of students who did not attend the college of their first choice due to financial concerns is 57%). \n- Alternate hypothesis \(H_a: P < 0.57\) (The percentage of students who did not attend the college of their first choice due to financial concerns is less than 57%).
02

Calculate the Test Statistic

The test statistic is calculated using the null hypothesis. We are trying to test if the proportion from the sample (864 out of 1600) is significantly less than 0.57. We are using sample proportion (\(p\)), standard deviation (\(\sigma\)), proportion from the null hypothesis (\(P\)) and the sample size (n).\n The formula is: \( Z = (p - P) / \sqrt{P(1 - P)/n} \)\n Here p = 864/1600 = 0.54\n The calculated Z value gives us the number of standard deviations our actual calculated value (proportion from our random sample of students), 0.54, is away from the hypothesized population proportion (in the null hypothesis), 0.57.
03

Find the P-value and Critical Value

The P-value is the smallest level at which you would still reject the null hypothesis. A p-value less than the significance level of 1% means we reject the null hypothesis. The critical value is a cut-off at which we reject the Null hypothesis. These are values that are being compared to the test statistic to make a decision about the hypotheses.
04

Draw Conclusion

If the calculated P-value is less than our significance level of 1%, then we can reject the null hypothesis or if our calculated Z test statistic is less than the critical value then also we can reject the null hypothesis. We conclude that there is enough evidence supporting the alternative hypothesis, which is the percentage of students who did not attend the college of their first choice due to financial concerns is indeed less than 57%.

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Key Concepts

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

Significance Level
In hypothesis testing, the significance level is a critical concept used to decide if the results of an experiment are statistically significant. When conducting a hypothesis test, you need to decide how confident you want to be in your results. This confidence level is inversely related to the significance level. The significance level, often represented by α (alpha), dictates the probability of rejecting a true null hypothesis. Commonly used significance levels are 1%, 5%, and 10%. In this exercise, we use a 1% significance level.
This implies a 1% risk of rejecting the null hypothesis when it is actually true. Choosing an appropriate significance level is crucial because it affects the likelihood of making a Type I error, which occurs when you wrongly reject the null hypothesis.
Null and Alternate Hypotheses
The null hypothesis (\(H_0\)) and the alternate hypothesis (\(H_a\)) are foundational concepts in hypothesis testing. They provide the framework for statistical testing. The null hypothesis is usually a statement of no effect or no change, whereas the alternative hypothesis is what you might believe to be true or hope to prove.
For example, consider the exercise at hand:
  • Null Hypothesis \((H_0)\): The percentage of students who didn't attend their first-choice college due to financial concerns is 57% \((P = 0.57)\).
  • Alternate Hypothesis \((H_a)\): The percentage of these students is less than 57% \((P < 0.57)\).
The aim of the hypothesis test is to determine if there is enough evidence in the sample data to conclude that the alternate hypothesis is true.
P-value and Critical Value
The P-value and critical value are essential tools in hypothesis testing. They help decide whether to reject the null hypothesis.
The P-value measures the strength of the evidence against the null hypothesis. A small P-value (e.g., less than the significance level of 1%) suggests that the null hypothesis is unlikely, leading to its rejection. On the other hand, the critical value acts as a threshold. It is specific to the selected significance level and is used to compare against the test statistic.
To decide whether to reject the null hypothesis:
  • Compare the P-value to the significance level. If the P-value is smaller, reject the null hypothesis.
  • If the test statistic exceeds the critical value in the expected direction (left or right based on the hypothesis), reject the null hypothesis.
Both approaches ultimately help verify if the alternative hypothesis presents a more credible explanation given the data.
Z Test Statistic
The Z test statistic is used to determine how far the sample statistic is from the null value in terms of standard deviations. It is particularly useful when dealing with large sample sizes and proposes if an effect is statistically significant.
In the exercise, to calculate the Z statistic, you use the formula: \[ Z = \frac{(p - P)}{\sqrt{\frac{P(1 - P)}{n}}} \] where:
  • \(p\) is the sample proportion (e.g., 864 out of 1600 students, so 0.54),
  • \(P\) is the proportion assumed by the null hypothesis (0.57),
  • \(n\) is the sample size (1600).
This formula conveys how many standard deviations the observed proportion is from the expected proportion under the null hypothesis. It provides a means to contextualize the significance of the observed difference, further informing the decision to reject or fail to reject the null hypothesis. A Z value that is far from 0 in the relevant direction lends support to the alternative hypothesis.

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Most popular questions from this chapter

In Las Vegas, Nevada, and Atlantic City, New Jersey, tests are performed often on the various gaming devices used in casinos. For example, dice are often tested to determine if they are balanced. Suppose you are assigned the task of testing a die, using a two-tailed test to make sure that the probability of a 2 -spot is \(1 / 6\). Using a \(5 \%\) significance level, determine how many 2 -spots you would have to obtain to reject the null hypothesis when your sample size is a. 120 b. 1200 c. 12,000 Calculate the value of \(\hat{p}\) for each of these three cases. What can you say about the relationship between (1) the difference between \(\hat{p}\) and \(1 / 6\) that is necessary to reject the null hypothesis and (2) the sample size as it gets larger?

For each of the following examples of tests of hypothesis about the population proportion, show the rejection and nonrejection regions on the graph of the sampling distribution of the sample proportion. a. A two-tailed test with \(\alpha=.10\) b. A left-tailed test with \(\alpha=.01\) c. A right-tailed test with \(\alpha=.05\)

According to Moebs Services Inc., the average cost of an individual checking account to major U.S. banks was \(\$ 380\) in 2013 (www. moebs.com). A bank consultant wants to determine whether the current mean cost of such checking accounts at major U.S. banks is more than \(\$ 380\) a year, A recent random sample of 150 such checking accounts taken from major U.S. banks produced a mean annual cost to them of \(\$ 390\). Assume that the standard deviation of annual costs to major banks of all such checking accounts is \(\$ 60 .\) a. Find the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, would you reject the null hypothesis if the maximum probability of Type I error is to be \(05 ?\) What if the maximum probability of Type I error is to be \(.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05\). Would you reject the null hypothesis? What if \(\alpha=.01 ?\) What if \(a=0\) ?

Consider \(H_{0}: \mu=72\) versus \(H_{1}: \mu>72 . \Lambda\) random sample of 16 observations taken from this population produced a sample mean of \(75.2\). The population is normally distributed with \(\sigma=6\). a. Calculate the \(p\) -value. b. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at a significance level of \(.01\) ? c. Considering the \(p\) -value of part a, would you reject the null hypothesis if the test were made at a significance level of \(.025 ?\)

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean number of hours spent working per week by college students who hold jobs is different from 20 hours b. To test whether or not a bank's ATM is out of service for an average of more than 10 hours per month c. To test if the mean length of experience of airport security guards is different from 3 years d. To test if the mean credit card debt of college seniors is less than \(\$ 1000\) e. To test if the mean time a customer has to wait on the phone to speak to a representative of a mail-order company about unsatisfactory service is more than 12 minutes

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