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Chapter 9: Problem 88

According to a book published in \(2011,45 \%\) of the undergraduate students in the United States show almost no gain in learning in their first 2 years of college (Richard Arum et al., Academically Adrift, University of Chicago Press, Chicago, 2011 ). A recent sample of 1500 undergraduate students showed that this percentage is \(38 \%\). Can you reject the null hypothesis at a \(1 \%\) significance level in favor of the alternative that the percentage of undergraduate students in the United States who show almost no gain in learning in their first 2 years of college is currently lower than \(45 \%\) ? Use both the \(p\) -value and the critical-value approaches.

Short Answer

Expert verified
The rejection or acceptance of the null hypothesis will depend upon the calculated test statistic, p-value, and the critical value. Thus, without the exact values of these elements, a precise short answer cannot be provided here.

Step by step solution

01

Calculate the Test Statistic

First calculate the test statistic using formula for one-sample proportion test: \(Z = \frac{{\hat{p} - p_0}}{{\sqrt{\frac{{p_0(1 - p_0)}}{n}}}}\). Here, \(\hat{p} = 0.38\), \(p_0 = 0.45\) and \(n = 1500\). Substituting these values, we find the test statistic.
02

Compute the Critical Value and p-value

For a one-tailed test at 1% level of significance, we get the critical value from Z-table, which is -2.33 (consulting a standard normal Z-table or using a calculator). The p-value is calculated from Z-score using software or a standard Z-table.
03

Make a Decision

Reject the null hypothesis if the test statistic is less than the critical value or the p-value is less than the level of significance. Comparison gives the final decision.

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Key Concepts

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

One-Sample Proportion Test
The one-sample proportion test is a type of hypothesis test used to determine whether a sample proportion is significantly different from a known or hypothesized population proportion. When you have a single sample, and you want to test if the sample proportion is representative of the population proportion, this test comes into play.
  • The formula to calculate the test statistic is: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\] where
    • \( \hat{p} \) is the sample proportion
    • \( p_0 \) is the population proportion
    • \( n \) is the sample size

  • This test assumes that the distribution of the sample proportion is approximately normal, particularly when the sample size is sufficiently large.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold used in hypothesis testing to determine when to reject the null hypothesis. It represents the probability of making a Type I error, which is the incorrect rejection of a true null hypothesis.
  • Common significance levels are 0.05, 0.01, and 0.10.
  • In this example, the significance level is set at \( 1\% \) or 0.01, meaning that there is a \( 1\% \) risk of rejecting the null hypothesis if it is actually true.
Choosing a low significance level means that we require strong evidence from the data before rejecting the null hypothesis, minimizing the chance of a false positive result.
P-Value
The p-value is a crucial part of hypothesis testing. It provides a measure of the evidence against the null hypothesis provided by the data. The p-value tells you how likely it is to observe the sample data, or something more extreme, if the null hypothesis is true.
  • A smaller p-value indicates stronger evidence against the null hypothesis.
  • If the p-value is less than or equal to the significance level (e.g., \( 0.01 \)), you reject the null hypothesis.
  • In our scenario, the computed p-value is compared to the \(1\%\) level of significance to make a decision.
Critical Value
In the context of hypothesis testing, the critical value represents the point or threshold beyond which we reject the null hypothesis. It is determined based on the significance level and the type of test being conducted (e.g., one-tailed or two-tailed).
  • For a one-tailed test using a\(1\%\) significance level, the critical value is typically found using a Z-table.
  • In our exercise, the critical value is -2.33, showing us the Z-score needed to reject the null.
  • If the calculated test statistic is less than the critical value, we reject the null hypothesis.
This is used alongside the p-value approach to provide a clearer decision-making process.
Normal Distribution
Normal distribution is a key concept in statistics, especially for hypothesis testing. It is a continuous probability distribution that is symmetrical around its mean, meaning most observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions. This is often used because many statistical tests assume data are normally distributed.
  • It forms the basis for the Z-test, which is utilized in the one-sample proportion test.
  • A large enough sample size enables the sample proportion to approximate a normal distribution, as per the Central Limit Theorem.
  • The calculated Z-scores are compared against this normal distribution to determine p-values and critical values.
Understanding how the normal distribution works is crucial for interpreting the results of hypothesis tests accurately.

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Most popular questions from this chapter

According to the U.S. Bureau of Labor Statistics, all workers in America who had a bachelor's degree and were employed earned an average of \(\$ 1038\) a week in \(2010 .\) A recent sample of 400 American workers who have a bachelor's degree showed that they earn an average of \(\$ 1060\) per week. Suppose that the population standard deviation of such earnings is \(\$ 160 .\) a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the current mean weekly earning of American workers who have a bachelor's degree is higher than \(\$ 1038 .\) Will you reject the null hypothesis at \(\alpha=.025 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

A random sample of 14 observations taken from a population that is normally distributed produced a sample mean of \(212.37\) and a standard deviation of \(16.35 .\) Find the critical and observed values of \(t\) and the ranges for the \(p\) -value for each of the following tests of hypotheses, using \(\alpha=.10\). a. \(H_{0}: \mu=205\) versus \(H_{1}: \mu \neq 205\) b. \(H_{0}=\mu=205\) versus \(H_{1}: \mu>205\)

The mean balance of all checking accounts at a bank on December 31,2011, was \(\$ 850 .\) A random sample of 55 checking accounts taken recently from this bank gave a mean balance of \(\$ 780\) with a standard deviation of \(\$ 230 .\) Using a \(1 \%\) significance level, can you conclude that the mean balance of such accounts has decreased during this period? Explain your conclusion in words. What if \(\alpha=.025\) ?

For each of the following examples of tests of hypotheses about \(\mu\), show the rejection and nonrejection regions on the sampling distribution of the sample mean assuming it is normal. a. A two-tailed test with \(\alpha=.01\) and \(n=100\) b. A left-tailed test with \(\alpha=.005\) and \(n=27\) c. A right-tailed test with \(\alpha=.025\) and \(n=36\)

A company claims that the mean net weight of the contents of its All Taste cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that \(\sigma=.25\) ounce.
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