/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A grocery store chain has as one... [FREE SOLUTION] | 91Ó°ÊÓ

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A grocery store chain has as one standard of service that the mean time customers wait in line to begin checking out not exceed 2 minutes. To verify the performance of a store the company measures the waiting time in 30 instances, obtaining mean time 2.17 minutes with standard deviation 0.46 minute. Use these data to test the null hypothesis that the mean waiting time is 2 minutes versus the alternative that it exceeds 2 minutes, at the \(10 \%\) level of significance.

Short Answer

Expert verified
The mean waiting time exceeds 2 minutes at the 10% significance level.

Step by step solution

01

State the Hypotheses

The null hypothesis, denoted as \( H_0 \), states that the mean waiting time \( \mu \) is equal to 2 minutes: \( H_0: \mu = 2 \). The alternative hypothesis, denoted as \( H_a \), states that the mean waiting time \( \mu \) is greater than 2 minutes: \( H_a: \mu > 2 \).
02

Determine the Significance Level and Test Type

The significance level is given as \( \alpha = 0.10 \). We are conducting a one-tailed test since the alternative hypothesis suggests a 'greater than' condition.
03

Calculate the Test Statistic

Use the formula for the test statistic in a one-sample t-test: \[ t = \frac{\bar{x} - \mu_0}{\left( \frac{s}{\sqrt{n}} \right)} \]where \( \bar{x} = 2.17 \) is the sample mean, \( \mu_0 = 2 \) is the population mean under the null hypothesis, \( s = 0.46 \) is the sample standard deviation, and \( n = 30 \). Substituting these values in:\[ t = \frac{2.17 - 2}{\left( \frac{0.46}{\sqrt{30}} \right)} \approx 2.086 \]
04

Determine the Critical Value

Using a t-distribution table, we look up the critical t-value for \( df = n-1 = 29 \) at \( \alpha = 0.10 \) in a one-tailed test. This value is approximately \( t_{critical} = 1.311 \).
05

Compare the Test Statistic to the Critical Value

The calculated test statistic \( t = 2.086 \) is greater than the critical value \( t_{critical} = 1.311 \). This indicates that the test statistic falls into the rejection region for the null hypothesis.
06

Conclusion

Since the test statistic exceeds the critical value, we reject the null hypothesis \( H_0 \). There is sufficient evidence at the 10% level of significance to support the claim that the mean waiting time exceeds 2 minutes.

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Key Concepts

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

One-Sample T-Test
The one-sample t-test is a statistical method used to determine whether the mean of a single sample differs significantly from a known or hypothesized population mean. In this scenario, we are examining whether the mean waiting time at a grocery store exceeds the standard of 2 minutes. This test is perfect when sample sizes are small (typically less than 30) and when the population standard deviation is unknown.
In the exercise, we gathered a sample of 30 instances where customers waited in line. We calculated the mean waiting time for these instances as 2.17 minutes, with a standard deviation of 0.46 minutes. The purpose of the one-sample t-test here is to test whether this sample mean is significantly greater than the 2-minute expectation set by the grocery store chain.
  • This method is useful when the sample size is limited and the population's standard deviation is unknown, using only the data available from the sample.
  • It helps determine if any differences between the sample mean and the population mean are due to random chance or a more significant reason.
In such tests, the result tells us if what we observe in the sample also holds true for the population from which the sample was taken.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by the researcher which defines the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it measures the risk you are willing to accept for incorrectly rejecting a true null hypothesis, known as a Type I error.
In the provided exercise, the significance level is set at 10% or 0.10. This means there is a 10% risk of concluding that the mean waiting time exceeds 2 minutes when it might not. The significance level directly impacts the critical value we refer to when deciding whether to reject the null hypothesis. A higher \( \alpha \) means more flexibility to declare a significant result, while a lower \( \alpha \) demands more stringent evidence.
  • It helps you control how conservative you want your test to be – typically, common levels are 0.05 (5%) and 0.01 (1%).
  • In business or real-life applications, the significance level might be chosen based on the severity of consequences from wrong decisions, like in medical tests or financial forecasts.
Setting the significance level is a important step as it influences the power and precision of your test results.
Null and Alternative Hypotheses
Whenever performing hypothesis testing, it's important to establish the null and alternative hypotheses. These hypotheses set the stage for statistical testing as we attempt to draw conclusions from our data.
The null hypothesis, represented as \( H_0 \), suggests that there is no effect or no difference, and any observed effect is due to sampling or experimental error. In the problem at hand, the null hypothesis asserts that the mean waiting time is equal to 2 minutes (\( H_0: \mu = 2 \)).
The alternative hypothesis, denoted as \( H_a \), proposes that there is indeed an effect or a difference. For the given exercise, it states the belief that the mean waiting time is greater than 2 minutes (\( H_a: \mu > 2 \)). It is this hypothesis we aim to test and substantiate.
  • The decision to reject or fail to reject the null hypothesis depends on the calculated test statistic and the critical value derived from the significance level.
  • Formulating clear hypotheses helps in conducting a focused analysis and avoiding ambiguity in interpreting results.
These hypotheses guide the analysis and interpretation by clearly establishing what we are trying to confirm or refute through our data.

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Most popular questions from this chapter

Compute the value of the test statistic for the indicated test, based on the information given. a. Testing \(H_{0}: \mu=342\) vs. \(H a: \mu<342, \sigma=11.2, n=40, x-=339, s=10.3\) b. Testing \(H_{0}: \mu=105\) vs. Ha: \(\mu>105, \sigma=5.3, n=80, x-=107, s=5.1\) c. Testing \(H_{0}: \mu=-13.5\) vs. Ha: \(\mu \neq-13.5, \sigma\) unknown, \(n=32, x-=-13.8, s=1.5\) d. Testing \(H_{0}: \mu=28\) vs. \(H a: \mu \neq 28, \sigma\) unknown, \(n=68, x-27.8, s=1.3\)

An automobile manufacturer recommends oil change intervals of 3,000 miles. To compare actual intervals to the recommendation, the company randomly samples records of 50 oil changes at service facilities and obtains sample mean 3,752 miles with sample standard deviation 638 miles. Determine whether the data provide sufficient evidence, at the \(5 \%\) level of significance, that the population mean interval between oil changes exceeds 3,000 miles.

The average amount of time that visitors spent looking at a retail company's old home page on the world wide web was 23.6 seconds. The company commissions a new home page. On its first day in place the mean time spent at the new page by 7,628 visitors was 23.5 seconds with standard deviation 5.1 seconds. a. Test at the \(5 \%\) level of significance whether the mean visit time for the new page is less than the former mean of 23.6 seconds, using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(5 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

A report five years ago stated that \(35.5 \%\) of all state-owned bridges in a particular state were "deficient." An advocacy group took a random sample of 100 state-owned bridges in the state and found 33 to be currently rated as being "deficient." Test whether the current proportion of bridges in such condition is \(35.5 \%\) versus the alternative that it is different from \(35.5 \%,\) at the \(10 \%\) level of significance.

In a recent year the fuel economy of all passenger vehicles was \(19.8 \mathrm{mpg}\). A trade organization sampled 50 passenger vehicles for fuel economy and obtained a sample mean of 20.1 mpg with standard deviation \(2.45 \mathrm{mpg}\). The sample mean 20.1 exceeds \(19.8,\) but perhaps the increase is only a result of sampling error. a. Perform the relevant test of hypotheses at the \(20 \%\) level of significance using the critical value approach. b. Compute the observed significance of the test. c. Perform the test at the \(20 \%\) level of significance using the \(p\) -value approach. You need not repeat the first three steps, already done in part (a).

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