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Chapter 10: Problem 41

In recent years the interest rate on home mortgages has declined to less than 6.0 percent. However, according to a study by the Federal Reserve Board the rate charged on credit card debit is more than 14 percent. Listed below is the interest rate charged on a sample of 10 credit cards. $$\begin{array}{|llllllllll|}\hline 14.6 & 16.7 & 17.4 & 17.0 & 17.8 & 15.4 & 13.1 & 15.8 & 14.3 & 14.5 \\\\\hline\end{array}$$ Is it reasonable to conclude the mean rate charged is greater than 14 percent? Use the .01 significance level.

Short Answer

Expert verified
Yes, it is reasonable to conclude that the mean rate is greater than 14 percent.

Step by step solution

01

Formulate the Hypotheses

First, we set up the null and alternative hypotheses. The null hypothesis \(H_0\) is that the mean interest rate \(\mu\) is 14 percent: \(H_0: \mu = 14\). The alternative hypothesis \(H_a\) is that the mean interest rate is greater than 14 percent: \(H_a: \mu > 14\).
02

Collect and Summarize the Data

Next, calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\) of the given interest rates. The sample rates are \([14.6, 16.7, 17.4, 17.0, 17.8, 15.4, 13.1, 15.8, 14.3, 14.5]\).
03

Calculate the Sample Mean

To find the sample mean \(\bar{x}\): \(\bar{x} = \frac{14.6 + 16.7 + 17.4 + 17.0 + 17.8 + 15.4 + 13.1 + 15.8 + 14.3 + 14.5}{10} = 15.66\).
04

Calculate the Sample Standard Deviation

Calculate the sample standard deviation \(s\) using the formula: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\). This gives \(s \approx 1.55\).
05

Perform the Hypothesis Test

Use a one-sample t-test to determine if the mean rate is greater than 14 percent. The test statistic is calculated as \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\), where \(\mu = 14\), \(\bar{x} = 15.66\), and \(n = 10\). Calculate \(t \approx 3.39\).
06

Determine the Critical Value and Decision-Making

Using a t-distribution table at a significance level of \(\alpha = 0.01\) with \(n-1 = 9\) degrees of freedom, find the critical value (\(t_{\alpha}\)) for a one-tailed test, which is approximately 2.821. Since our test statistic \(t \approx 3.39\) is greater than 2.821, we reject the null hypothesis.
07

Conclusion

Since the test statistic is greater than the critical value, it is reasonable to conclude that the mean interest rate charged is greater than 14 percent at the 0.01 significance level.

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Key Concepts

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

One-sample t-test
A one-sample t-test is a statistical test used to determine whether the mean of a single sample differs significantly from a known or hypothesized population mean. This test is particularly useful when the sample size is small and the population standard deviation is unknown. In the case of our exercise, we're trying to assess if the mean interest rate of the credit card sample is greater than 14 percent. The process involves:
  • Setting up a null hypothesis (\(H_0\): that the mean rate is 14%).
  • Forming an alternative hypothesis (\(H_a\): that the mean rate is greater than 14%).
  • Calculating the test statistic which, in this scenario, involves using the formula \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
A crucial part of this test, which differentiates it from a Z-test, is the usage of a t-distribution, which accounts for increased variability in smaller samples. This distribution changes based on sample size, which is adjusted for by degrees of freedom.
Significance Level
The significance level, often denoted by \(\alpha\), indicates the probability of rejecting the null hypothesis when it is true. This is known as the Type I error rate. Common values for \(\alpha\) include 0.05, 0.01, and 0.10. An \(\alpha\) of 0.01 was used in our exercise, suggesting rigorous criteria for determining statistical significance. At a 0.01 significance level:
  • You have a 1% risk of concluding that a difference exists when there is no actual difference.
  • This stringent level reduces the chance of making a Type I error, making your test conclusions more robust.
  • This was appropriate for our analysis given we're testing whether the credit card interest rates are significantly greater than 14%.
When performing hypothesis tests, choosing a significance level is vital and often balances the need for accuracy with practical considerations. In our exercise, this rigorous significance level gives confidence in our decision to reject the null hypothesis.
Critical Value
In hypothesis testing, the critical value is the threshold that the test statistic must exceed to reject the null hypothesis. For a one-sample t-test, the critical value is determined based on the significance level and degrees of freedom of the data鈥攚hich are computed as \(n - 1\), where \(n\) is the sample size.In this exercise:
  • The sample size was 10, giving us 9 degrees of freedom.
  • Using our significance level (\(\alpha = 0.01\)), we looked up the critical t-value in the t-distribution table.
  • The critical value here was approximately 2.821 for a one-tailed test.
  • If the calculated test statistic (3.39) exceeds this critical value, we reject the null hypothesis.
This understanding of critical values is essential because it determines the boundary for decision-making in the context of hypothesis testing. Surpassing the critical value thus supports the evidence of the alternative hypothesis.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of values. It's particularly important in hypothesis testing because it helps determine the variability within a sample you're studying. In the context of our exercise:
  • The sample standard deviation $s$ was calculated as 1.55 for the credit card interest rates.
  • It is used in the formula for the test statistic $t$ since the population standard deviation is unknown.
  • This calculation involves finding the average of the squared differences from the sample mean and then taking the square root, which reflects how spread out the credit card rates are from the sample mean.
Understanding sample standard deviation is crucial since it directly affects the test statistic value. Variability impacts how confident we are about our sample representing the population鈥攈ence a pivotal component for accurate hypothesis testing.

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Most popular questions from this chapter

Tina Dennis is the comptroller for Meek Industries. She believes that the current cash-flow problem at Meek is due to the slow collection of accounts receivable. She believes that more than 60 percent of the accounts are in arrears more than three months. A random sample of 200 accounts showed that 140 were more than three months old. At the .01 significance level, can she conclude that more than 60 percent of the accounts are in arrears for more than three months?

The postanesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently enlarged. The hope was that with the enlargement the mean number of patients per day would be more than \(25 .\) A random sample of 15 days revealed the following numbers of patients. $$\begin{array}{lllllllllllllll}25 & 27 & 25 & 26 & 25 & 28 & 28 & 27 & 24 & 26 & 25 & 29 & 25 & 27 & 24 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude that the mean number of patients per day is more than 25 ? Estimate the \(p\) -value and interpret it.

Research at the University of Toledo indicates that 50 percent of the students change their major area of study after their first year in a program. A random sample of 100 students in the College of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program? Test at the .05 level of significance.

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, "You can average more than \(\$ 80\) a day in tips." Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was \(\$ 84.85,\) with a standard deviation of \(\$ 11.38\). At the .01 significance level, can Ms. Brigden conclude that she is earning an average of more than \(\$ 80\) in tips?

The Rocky Mountain district sales manager of Rath Publishing, Inc., a college textbook publishing company, claims that the sales representatives make an average of 40 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 28 sales representatives reveals that the mean number of calls made last week was \(42 .\) The standard deviation of the sample is 2.1 calls. Using the .05 significance level, can we conclude that the mean number of calls per salesperson per week is more than \(40 ?\)
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