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Chapter 10: Problem 8

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?

Short Answer

Expert verified
Beth is earning more than \( \$80 \) on average in tips.

Step by step solution

01

Identify Hypotheses

We start by establishing the null and alternative hypotheses: - Null Hypothesis (\( H_0 \)): \( \mu = 80 \).- Alternative Hypothesis (\( H_a \)): \( \mu > 80 \). Here, we're testing whether Beth is earning on average more than \( \$80 \).
02

Determine the Significance Level

The significance level given in the problem is \( \alpha = 0.01 \). This will help us determine the critical value against which we will compare the test statistic.
03

Calculate the Test Statistic

We use the formula for the test statistic for a single sample mean: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where \( \bar{x} = 84.85 \), \( \mu = 80 \), \( \sigma = 11.38 \), and \( n = 35 \). Substitute the values:\[ z = \frac{84.85 - 80}{\frac{11.38}{\sqrt{35}}} \approx \frac{4.85}{1.92} \approx 2.53 \]
04

Find Critical Value and Decision Rule

For a significance level of \( \alpha = 0.01 \) in a one-tailed test, the critical value of \( z \) is approximately 2.33. If the calculated \( z \) statistic is greater than 2.33, we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.
05

Make a Decision

The calculated test statistic is \( z = 2.53 \), which is greater than the critical value of 2.33. Thus, we reject the null hypothesis.
06

Conclusion

Since we rejected the null hypothesis, we conclude that at the 0.01 significance level, there is enough evidence to support the claim that Beth is earning more than \( \$80 \) on average in tips.

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Key Concepts

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

Significance Level
When performing a hypothesis test, the significance level is a key concept that helps us determine how strong our evidence needs to be to support our hypothesis. The significance level is typically denoted by \( \alpha \) and reflects the probability of rejecting the null hypothesis when it is actually true, which is known as a Type I error.
This means you are setting a threshold for how much risk you're willing to take in making an incorrect decision.
  • A significance level of \( \alpha = 0.05 \) means you're willing to accept a 5% risk of concluding that a difference exists when there is none.
  • In Beth's example, we use \( \alpha = 0.01 \), indicating she is only willing to take a 1% risk of incorrectly stating she earns more than \( \$80\) on average.
The choice of \( \alpha \) depends on how confident you want to be in your conclusions. A smaller \( \alpha \) leads to more stringent criteria for rejecting the null hypothesis.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. The purpose of the test statistic is to help you decide whether to reject the null hypothesis. This value is determined by comparing observed data with data expected under the null hypothesis.
In Beth's case, the test statistic is calculated using the formula for a single sample mean:\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Here:
  • \( \bar{x} = 84.85 \) is Beth's mean tips per day,
  • \( \mu = 80 \) is the hypothesized mean,
  • \( \sigma = 11.38 \) is the standard deviation,
  • \( n = 35 \) is the number of days.
Plugging these values into the formula gives a test statistic \( z \approx 2.53 \). This calculated statistic will then be compared to the critical value to make a decision.
Critical Value
The critical value is a point on the test statistic scale beyond which we reject the null hypothesis. It represents the cutoff of our acceptance region for the null hypothesis and is determined by the significance level.
  • For a test with \( \alpha = 0.01 \) and a one-tailed test, the critical value for \( z \) is approximately 2.33.
  • The choice of critical value depends on the significance level and whether it is a one-tailed or two-tailed test.
In this scenario, the calculated test statistic (\( z = 2.53 \)) is compared to the critical value of 2.33.
If the test statistic exceeds the critical value, the null hypothesis is rejected, suggesting there is enough evidence to support the alternative hypothesis that Beth earns more than \( \$80 \) in tips on average.
Null Hypothesis
The null hypothesis is a fundamental part of statistical hypothesis testing. It is a statement that there is no effect or no difference, and it serves as the default assumption that you seek to test.
For instance, in Beth's example:
  • The null hypothesis \( (H_0) \) states that the average amount of tips is \( \\(80 \) (\( \mu = 80 \)).
  • The purpose of hypothesis testing is to determine if there is enough statistical evidence to reject \( H_0 \) in favor of the alternative hypothesis \( (H_a) \), which claims the average amount of tips is greater than \( \\)80 \).
In many cases, the null hypothesis is what is to be tested, and the aim is to see if there is enough empirical evidence to support rejecting it.
The outcome in Beth's case shows rejecting the null hypothesis, because her average tips were significantly greater than the threshold after considering the test statistic and the critical value.

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Most popular questions from this chapter

From past experience a television manufacturer found that 10 percent or less of its sets needed any type of repair in the first two years of operation. In a sample of 50 sets manufactured two years ago, 9 needed repair. At the .05 significance level, has the percent of sets needing repair increased? Determine the \(p\) -value.

Given the following hypothesis: $$H_{0}: \mu \geq 20$$$H_{1}: \mu<20\( A random sample of five resulted in the following values: \)18,15,12,19,\( and \)21 .\( Using the .01 significance level, can we conclude the population mean is less than \)20 ?\( a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the \)p$ -value.

A spark plug manufacturer claimed that its plugs have a mean life in excess of 22,100 miles. Assume the life of the spark plugs follows the normal distribution. A fleet owner purchased a large number of sets. A sample of 18 sets revealed that the mean life was 23,400 miles and the standard deviation was 1,500 miles. Is there enough evidence to substantiate the manufacturer's claim at the .05 significance level?

A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds with a standard deviation of 2.8 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the \(p\) -value.

Most air travelers now use e-tickets. Electronic ticketing allows passengers to not worry about a paper ticket, and it costs the airline companies less to handle than a paper ticketing. However, in recent times the airlines have received complaints from passengers regarding their e-tickets, particularly when connecting flights and a change of airlines were involved. To investigate the problem an independent watchdog agency contacted a random sample of 20 airports and collected information on the number of complaints the airport had with e-tickets for the month of March. The information is reported below. $$\begin{array}{|rrrrrrrrrr}\hline 14 & 14 & 16 & 12 & 12 & 14 & 13 & 16 & 15 & 14 \\\12 & 15 & 15 & 14 & 13 & 13 & 12 & 13 & 10 & 13 \\\\\hline\end{array}$$ At the . 05 significance level can the watchdog agency conclude the mean number of complaints per airport is less than 15 per month? a. What assumption is necessary before conducting a test of hypothesis? b. Plot the number of complaints per airport in a frequency distribution or a dot plot. Is it reasonable to conclude that the population follows a normal distribution? c. Conduct a test of hypothesis and interpret the results.
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