/*! 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 9 In the past the average length o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the past the average length of an outgoing telephone call from a business office has been 143 seconds. A manager wishes to check whether that average has decreased after the introduction of policy changes. A sample of 100 telephone calls produced a mean of 133 seconds, with a standard deviation of 35 seconds. Perform the relevant test at the \(1 \%\) level of significance.

Short Answer

Expert verified
The average call length has decreased.

Step by step solution

01

Hypotheses Formulation

To determine if the average length of calls has decreased, we set up the null hypothesis (H0) as the mean call length being equal to 143 seconds. The alternative hypothesis (H1) is that the mean call length is less than 143 seconds: - H0: \( \mu = 143 \)- H1: \( \mu < 143 \)
02

Significance Level

The problem specifies a \(1\%\) level of significance, which means \( \alpha = 0.01 \). This is the probability of rejecting the null hypothesis when it is true.
03

Test Statistic Calculation

We need to calculate the test statistic using the sample mean, population mean, sample standard deviation, and the number of calls in the sample. The formula for the test statistic \( t \) is:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Where:- \( \bar{x} = 133 \) seconds is the sample mean,- \( \mu = 143 \) seconds is the hypothesized population mean,- \( s = 35 \) seconds is the sample standard deviation, and- \( n = 100 \) is the sample size.
04

Calculation of Test Statistic Value

Calculate the test statistic value by substituting the values into the formula:\[ t = \frac{133 - 143}{\frac{35}{\sqrt{100}}} = \frac{-10}{3.5} = -2.857 \]
05

Critical Value and Decision Rule

For a \(1\%\) level of significance, we check the critical value for a one-tailed \( t \)-test with \( n-1 = 99 \) degrees of freedom. The critical value \( t_{critical} \approx -2.33 \) (from t-distribution table). If the test statistic \( t \) is less than \( t_{critical} \), we reject the null hypothesis.
06

Conclusion

Since the calculated test statistic \( t = -2.857 \) is less than the critical value \( t_{critical} = -2.33 \), we reject the null hypothesis. There is enough evidence to conclude that the average length of outgoing telephone calls has decreased.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Significance Level
The significance level is a crucial component in hypothesis testing. It represents the probability we are willing to accept for making a Type I error, which is rejecting the null hypothesis when it is actually true. In our problem, we are using a significance level of 1%, denoted as \( \alpha = 0.01 \). This means we are allowing for only a 1% chance of incorrectly concluding that the average length of phone calls has decreased. This low significance level reflects the manager's desire for high confidence in the decision to implement the new policy based on the data at hand.
Test Statistic Calculation
The test statistic allows us to make statistical comparisons. It is a standardized value that helps determine the difference between the sample statistic and the population parameter if the null hypothesis is true.We calculate the test statistic using the following formula:\[t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]where:
  • \( \bar{x} \) is the sample mean (133 seconds)
  • \( \mu \) is the population mean (143 seconds)
  • \( s \) is the sample standard deviation (35 seconds)
  • \( n \) is the sample size (100 calls)
The test statistic reflects how far the sample mean deviates from the population mean when expressed in terms of the standard error. In our example, the test statistic value calculated is -2.857.
Critical Value
The critical value determines the threshold our test statistic must exceed to reject the null hypothesis. It is found using statistical tables, given the desired significance level and the degrees of freedom in our sample.In this case, with a 1% significance level and 99 degrees of freedom (\( n-1 \)), the critical value is approximately \( t_{critical} = -2.33 \) for a one-tailed test. The critical value acts as a cutoff point: if the test statistic is less than the critical value, we have enough statistical evidence to reject the null hypothesis. Here, because our test statistic of -2.857 is less than -2.33, this supports the manager's claim that the call duration has decreased.
Null and Alternative Hypotheses
Formulating the null and alternative hypotheses is the first step of hypothesis testing. These hypotheses set the basis for what we are testing.- The **null hypothesis** (\( H_0 \)): asserts that there is no change or effect. In our scenario, it claims that the average call duration remains 143 seconds, i.e., \( \mu = 143 \).- The **alternative hypothesis** (\( H_1 \)): suggests a change or different effect. Here, it states that the average duration has decreased, i.e., \( \mu < 143 \).Understanding these hypotheses guides us through setting up and interpreting the statistical test, as they frame our investigation into whether policy changes have indeed shortened call durations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average time workers spent commuting to work in Verona five years ago was 38.2 minutes. The Verona Chamber of Commerce asserts that the average is less now. b. The mean salary for all men in a certain profession is \(\$ 58,291\). A special interest group thinks that the mean salary for women in the same profession is different. c. The accepted figure for the caffeine content of an 8 -ounce cup of coffee is \(133 \mathrm{mg}\). A dietitian believes that the average for coffee served in a local restaurants is higher. d. The average yield per acre for all types of corn in a recent year was 161.9 bushels. An economist believes that the average yield per acre is different this year. e. An industry association asserts that the average age of all self-described fly fishermen is 42.8 years. A sociologist suspects that it is higher.

The mean household income in a region served by a chain of clothing stores is $$\$ 48,750 .$$ In a sample of 40 customers taken at various stores the mean income of the customers was $$\$ 51,505$$ with standard deviation $$\$ 6,852 .$$ a. Test at the \(10 \%\) level of significance the null hypothesis that the mean household income of customers of the chain is $$\$ 48,750$$ against that alternative that it is different from $$\$ 48,750.$$ b. The sample mean is greater than $$\$ 48,750,$$ suggesting that the actual mean of people who patronize this store is greater than $$\$ 48,750 .$$ Perform this test, also at the \(10 \%\) level of significance. (The computation of the test statistic done in part (a) still applies here.)

Large Data Set 1 lists the GPAs of 1,000 students. a. Regard the data as arising from a census of all freshman at a small college at the end of their first academic year of college study, in which the GPA of every such person was measured. Compute the population mean \(\mu\). b. Regard the first 50 students in the data set as a random sample drawn from the population of part (a) and use it to test the hypothesis that the population mean is less than \(2.50,\) at the \(10 \%\) level of significance. (The null hypothesis is that \(\mu=2.50 .)\) c. Is your conclusion in part \((b)\) in agreement with the true state of nature (which by part (a) you know), or is your decision in error? If your decision is in error, is it a Type I error or a Type II error?

A magazine publisher tells potential advertisers that the mean household income of its regular readership is $$\$ 61,500.$$ An advertising agency wishes to test this claim against the alternative that the mean is smaller. A sample of 40 randomly selected regular readers yields mean income $$\$ 59,800$$ with standard deviation $$\$ 5,850.$$ Perform the relevant test at the \(1 \%\) level of significance.

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \cdot \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average July temperature in a region historically has been \(74.5^{\circ} \mathrm{F}\). Perhaps it is higher now. b. The average weight of a female airline passenger with luggage was 145 pounds ten years ago. The FAA believes it to be higher now. c. The average stipend for doctoral students in a particular discipline at a state university is \(\$ 14,756\). The department chairman believes that the national average is higher. d. The average room rate in hotels in a certain region is \(\$ 82.53 .\) A travel agent believes that the average in a particular resort area is different. e. The average farm size in a predominately rural state was 69.4 acres. The secretary of agriculture of that state asserts that it is less today.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.