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Chapter 9: Problem 77

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

Short Answer

Expert verified
Yes, the data supports the claim that the mean time is longer than 4.5 years at the 1% significance level.

Step by step solution

01

Understand the Hypotheses

To test the claim, we set up the null hypothesis as \(H_0: \mu = 4.5\), which states that the mean time to finish the degree is 4.5 years. The alternative hypothesis, which we want to test, is \(H_a: \mu > 4.5\), indicating that the mean time is longer than 4.5 years.
02

Determine the Test Statistic

Use the following formula for the test statistic for a sample mean: \[ z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \] where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the given values: \(\bar{x} = 5.1\), \(\mu_0 = 4.5\), \(s = 1.2\), \(n = 49\), we calculate: \[ z = \frac{5.1 - 4.5}{\frac{1.2}{\sqrt{49}}} = \frac{0.6}{0.1714} \approx 3.50 \].
03

Find the Critical Value

With a 1% significance level for a one-tailed test looking for a mean greater than 4.5 years, the critical value of \(z\) is approximately 2.33 (from z-tables). If the calculated \(z\) is greater than 2.33, we reject the null hypothesis.
04

Compare and Draw Conclusion

The calculated \(z\)-score of 3.50 exceeds the critical value of 2.33. Therefore, the data provides enough evidence at the 1% level to reject the null hypothesis in favor of the alternative hypothesis, supporting the claim that students take, on average, longer than 4.5 years to graduate.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that there is no effect or difference in the population, and it serves as the starting point for statistical testing. It is usually denoted by \( H_0 \). In our example, the null hypothesis states that the average time students take to complete their degrees is 4.5 years: \( H_0: \mu = 4.5 \). This serves as the default assumption that there is no deviation from established facts.
To evaluate whether this assumed value holds true, we collect data via a survey or experiment. The goal is often to see if there is substantial evidence to reject the null hypothesis in favor of the alternative hypothesis.
  • The null hypothesis is often a statement of "no effect" or "no difference."
  • It is tested against an alternative hypothesis that specifies an effect or difference.
  • Rejection of the null hypothesis suggests that the evidence is strong enough to support the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis presents the idea that there is a new effect or deviation from the norm. In our scenario, it suggests that the mean time to finish undergraduate degrees is greater than 4.5 years: \( H_a: \mu > 4.5 \). This is the hypothesis we aim to support through statistical testing.
The alternative hypothesis contrasts with the null hypothesis and is what you hope to support.
  • It asserts that there is a statistically significant effect or difference.
  • In our example, it claims that students, on average, take longer than 4.5 years to graduate.
  • The outcome of testing may lead to rejecting the null hypothesis, thereby supporting the alternative hypothesis.
Z-Score
The z-score is a statistical measure that indicates how many standard deviations an element is from the mean of its data set. In hypothesis testing, it is used to compare the observed sample mean to the expected population mean under the null hypothesis.
In our exercise, we calculated a z-score using the formula:\[ z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \]Substituting the sample mean \(\bar{x} = 5.1\), hypothesized population mean \(\mu_0 = 4.5\), standard deviation \(s = 1.2\), and sample size \(n = 49\), the z-score was found to be approximately 3.50.
  • A z-score allows you to determine how extreme a sample result is compared to what was expected under the null hypothesis.
  • A higher absolute z-score indicates a greater deviation from the null hypothesis.
  • The calculated z-score is compared against the critical value to draw conclusions about the hypotheses.
Critical Value
In hypothesis testing, the critical value defines the threshold at which you decide whether to reject the null hypothesis. It is derived from the significance level of the test, which in this case is 1% for a one-tailed test.
Using a z-table, we found that the critical value for our specific test is approximately 2.33. This value represents the point beyond which we would consider the sample result too extreme if the null hypothesis were true.
  • The critical value helps in making a decision regarding the hypotheses.
  • If the z-score is beyond the critical value, it suggests the data provide enough evidence to reject the null hypothesis.
  • In our example, the calculated z-score of 3.50 exceeds the critical value of 2.33, leading us to reject \( H_0 \).

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Most popular questions from this chapter

The mean age of graduate students at a University is at most 31 y ears with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1\(\%\) level? The p-value is \(0.0264 .\) State the null and alternative hypotheses and interpret the \(p\) -value.

Draw the general graph of a left-tailed test.

The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why?

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis. State the null hypothesis, Ho, and the hypothesis. \(H_{a},\) in terms of the appropriate parameter \((\mu \text { or } p)\) a. The mean number of years Americans work before retiring is \(34 .\) b. At most 60\(\%\) of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least \(\$ 100,000\) per year. d. Twenty-nine percent of high school seniors get drunk each month. e. Fewer than 5 \(\%\) of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in her lifetime is not more than ten. g. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11\(\%\) for women. j. Private universities' mean tuition cost is more than \(\$ 20,000\) per year.

Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal. What symbol represents the random variable for this test?
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