/*! 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 37 Many grocery stores and large re... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 37

Many grocery stores and large retailers such as Kroger and Walmart have installed self-checkout systems so shoppers can scan their own items and cash out themselves. How do customers like this service and how often do they use it? Listed below is the number of customers using the service for a sample of 15 days at a Walmart location. \(\begin{array}{rrrrrrrrr}120 & 108 & 120 & 114 & 118 & 91 & 118 & 92 & 104 & 104 \\ 112 & 97 & 118 & 108 & 117 & & & & & \end{array}\) Is it reasonable to conclude that the mean number of customers using the self- checkout system is more than 100 per day? Use the .05 significance level.

Short Answer

Expert verified
The mean number of customers using self-checkout is not significantly more than 100 per day.

Step by step solution

01

State the Hypotheses

Begin by defining the null and alternative hypotheses. - Null Hypothesis \( (H_0) \): The mean number of customers using self-checkout per day is 100, i.e., \( \mu = 100 \).- Alternative Hypothesis \( (H_a) \): The mean number of customers using self-checkout per day is more than 100, i.e., \( \mu > 100 \).
02

Collect Sample Data

List the sample data from the 15 days: \( 120, 108, 120, 114, 118, 91, 118, 92, 104, 104, 112, 97, 118, 108, 117 \).
03

Calculate the Sample Mean

Calculate the sample mean using the formula:\[\bar{x} = \frac{\sum x_i}{n}\]Substitute the sum of the data and \( n = 15 \):\[ \bar{x} = \frac{1413}{15} = 94.2 \]
04

Calculate the Sample Standard Deviation

Calculate the sample standard deviation using:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]First, find the squared deviations and then their sum: \( \sum (x_i - \bar{x})^2 = 4970.8 \).Then, calculate \[ s = \sqrt{\frac{4970.8}{14}} \, \approx 17.82 \].
05

Perform the t-test

Use the t-test for testing the mean:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]Substitute the known values:\[ t = \frac{94.2 - 100}{17.82/\sqrt{15}} \approx -1.44 \].
06

Determine the Critical t-value

For a one-tailed test at \( \alpha = 0.05 \), and degrees of freedom \( df = n - 1 = 14 \), find the critical t-value from the t-distribution table. The critical t-value is approximately 1.761.
07

Compare the t-value and Critical Value

Compare the calculated t-value with the critical t-value.\(-1.44 < 1.761\)Since the calculated t-value is not greater than the critical t-value, do not reject the null hypothesis.
08

Conclusion

State the conclusion based on the comparison. There is not enough evidence to conclude that the mean number of customers using the self-checkout system is more than 100 per day at the \( \alpha = 0.05 \) significance level.

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 91影视!

Key Concepts

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

t-test
The t-test is a statistical test used to compare the sample mean to a known value (often the population mean).
The purpose of performing a t-test in this context is to determine if the observed mean number of customers using the self-checkout system per day significantly differs from the expected value (100, in this exercise).
  • This test is particularly useful when the sample size is small, typically less than 30.
  • It helps in assessing whether any observed difference is just due to random sampling variability.
In this Walmart scenario, the t-test helped determine if the lower than expected mean usage of self-checkout systems was statistically significant or not.
The result of the t-test was a calculated t-value of approximately -1.44, which was compared against a critical value from the t-distribution table for the specific significance level and degrees of freedom.
null hypothesis
The null hypothesis is a baseline statement that there is no effect or difference under investigation.
In hypothesis testing, the null hypothesis ( H_0) serves as the starting assumption. Here:
  • The null hypothesis was that the mean number of customers using the self-checkout system per day equals 100, 渭 = 100.
  • It posits that any observed variance is due to natural sampling variation without implying a meaningful deviation from the expected average number of users.
Setting a null hypothesis is essential for conducting a t-test, as it presents a scenario to test against.
The objective is to determine if there is enough evidence to reject this null hypothesis.
In this case, the t-test led to the conclusion that there wasn鈥檛 sufficient evidence to reject the null hypothesis that the mean is 100 customers per day.
sample mean
The sample mean is a key statistical measure representing the central tendency of the data.
It is calculated by summing all data points and dividing by the number of observations:\[\bar{x} = \frac{\sum x_i}{n}\]
  • In this exercise, the sample mean of customer usage per day was determined from 15 days' worth of data.
  • The computed sample mean was 94.2, providing an estimate of the average number of customers utilizing the self-checkout systems per day during the sampled period.
The sample mean is crucial because it serves as a point of comparison against the hypothesized population mean (in this case, 100 customers).
By calculating the sample mean, retailers and analysts can make informed discussions about trending customer behavior and usage patterns.
significance level
The significance level, denoted by\(\alpha\), is a threshold used to decide whether to reject the null hypothesis.
It defines the probability of committing a Type I error 鈥� that is, erroneously rejecting a true null hypothesis.
  • A typical significance level used in hypothesis testing is 0.05, signifying a 5% risk of concluding a difference exists when in fact, it does not.
  • In this exercise, a 0.05 significance level means there's a 5% chance the test incorrectly suggests the mean number of customers is greater than 100 when it isn鈥檛 truly.
Selecting an appropriate significance level is vital as it influences the test's sensitivity and the rigor of the conclusions drawn.
In the case examined, because the computed t-value did not exceed the critical value at this alpha level, the null hypothesis was not rejected.

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

(a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The waiting time for customers at MacBurger Restaurants follows a normal distribution with a population standard deviation of 1 minute. At the Warren Road MacBurger, the quality assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the . 05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

For a recent year, the mean fare to fly from Charlotte, North Carolina, to Chicago, Illinois, on a discount ticket was \(\$ 267 .\) A random sample of 13 round-trip discount fares on this route last month shows: \(\begin{array}{llllllllll}\$ 321 & \$ 286 & \$ 290 & \$ 330 & \$ 310 & \$ 250 & \$ 270 & \$ 280 & \$ 299 & \$ 265 & \$ 291 & \$ 275 & \$ 281\end{array}\) At the .01 significance level, can we conclude that the mean fare has increased? What is the \(p\) -value?

A recent survey by nerdwallet.com indicated Americans paid a mean of \(\$ 6,658\) interest on credit card debt in 2017 . A sample of 12 households with children revealed the following amounts. At the .05 significance level, is it reasonable to conclude that these households paid more interest? \(\begin{array}{lllllllll}7,077 & 5,744 & 6,753 & 7,381 & 7,625 & 6,636 & 7,164 & 7,348 & 8,060 & 5,848 & 9,275 & 7,052\end{array}\)

The American Water Works Association reports that the per capita water use in a single-family home is 69 gallons per day. Legacy Ranch is a relatively new housing development. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Thirty-six owners responded, and the sample mean water use per day was 64 gallons with a standard deviation of 8.8 gallons per day. At the .10 level of significance, is that enough evidence to conclude that residents of Leqacy Ranch use less water on average?

According to the local union president, the mean gross income of plumbers in the Salt Lake City area follows the normal probability distribution with a mean of \(\$ 45,000\) and a standard deviation of \(\$ 3,000 .\) A recent investigative reporter for KYAK TV found, for a sample of 120 plumbers, the mean gross income was \(\$ 45,500\). At the .10 significance level, is it reasonable to conclude that the mean income is not equal to \(\$ 45,000 ?\) Determine the \(p\) -value.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation
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.