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Chapter 10: Problem 41

Watch Corporation of Switzerland claims that its watches on average will neither gain nor lose time during a week. A sample of 18 watches provided the following gains (+)or losses(-)in seconds per week. $$ \begin{array}{|rrrrrrrrr|}\hline-0.38 & -0.20 & -0.38 & -0.32 & +0.32 & -0.23 & +0.30 & +0.25 & -0.10 \\\\-0.37 & -0.61 & -0.48 & -0.47 & -0.64 & -0.04 & -0.20 & -0.68 & +0.05 \\\\\hline\end{array}$$ Is it reasonable to conclude that the mean gain or loss in time for the watches is \(0 ?\) Use the .05 significance level. Estimate the \(p\) -value.

Short Answer

Expert verified
Perform a t-test; if the p-value < 0.05, reject the claim that the mean gain/loss is 0.

Step by step solution

01

Formulate Hypotheses

The null hypothesis \(H_0\) is that the mean gain or loss in time (\(\mu\)) is 0, i.e., \(H_0: \mu = 0\). The alternative hypothesis \(H_a\) is that the mean is not 0, i.e., \(H_a: \mu eq 0\).
02

Determine the Sample Statistics

First, we need to calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\). The sample mean is calculated by summing all the gains and losses and dividing by the number of watches (18). The sample standard deviation is calculated by taking the square root of the variance.
03

Calculate the Test Statistic

Use the t-test for the sample mean. The test statistic \( t \) is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean (0 in this case), \(s\) is the sample standard deviation, and \(n\) is the sample size (18).
04

Find the Critical Value and p-Value

With 17 degrees of freedom (n-1), we find the critical value for a 0.05 significance level using a t-distribution table. Compute the p-value based on the t-statistic calculated in Step 3. If \( |t| \) is greater than the critical value, the p-value will confirm or refute \(H_0\).
05

Make a Decision

Compare the calculated p-value with the significance level (0.05). If the p-value is less than 0.05, reject \(H_0\). Otherwise, do not reject \(H_0\). This will indicate if it is reasonable to conclude that the mean gain or loss is zero.

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Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In the context of the Watch Corporation exercise, we use a one-sample t-test.
This is because we are comparing the sample mean of watch gains/losses to a known value, which is the hypothesized population mean, zero in this case.

The goal is to see if the observed sample mean significantly deviates from this hypothesized mean. To perform a t-test:
  • We calculate the sample mean, which is the average of the observed data.
  • Next, compute the sample standard deviation, which measures the variation or dispersion of sample values.
  • Finally, use these values to calculate the t-statistic using the formula \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
By analyzing the t-statistic, you can determine whether the sample mean significantly differs from the hypothesized mean.
null hypothesis
In hypothesis testing, the null hypothesis, often symbolized as \(H_0\), is a starting assumption.
It proposes that there is no effect or no difference, which means any observed effect is due to sampling or experimental error.

For the Watch Corporation's claim, the null hypothesis is that the mean gain or loss in watch time is zero, represented as \(H_0: \mu = 0\).
The purpose of setting the null hypothesis is to have a baseline to test against, with any deviation suggesting a need to consider the alternative hypothesis.
  • If statistical tests indicate sufficient evidence against this assumption, we may reject the null hypothesis.
  • Otherwise, we maintain that there's no statistically significant deviation from what was hypothesized.
Thus, the null hypothesis serves as a critical component as it anchors the analysis and helps determine the outcome of the hypothesis test.
significance level
The significance level, denoted by \(\alpha\), is a threshold used to determine when to reject the null hypothesis.
Choosing a significance level is crucial because it represents the risk of falsely rejecting the null hypothesis when it is true (Type I error).

In the watch example, the significance level is set at 0.05. This means that:
  • There is a 5% risk of concluding that the mean watch gain or loss is different from zero when it is, in fact, zero.
  • The decision rule is straightforward: if the p-value obtained from the test is less than 0.05, we reject \(H_0\).
  • Thus, it acts as a benchmark to guide whether observed data is statistically significant against the null hypothesis.
By setting a significance level, researchers can control the probability of making incorrect conclusions and ensure the reliability of their results.
p-value
The p-value is a crucial concept in hypothesis testing, indicating the probability of obtaining results at least as extreme as those observed, under the assumption that the null hypothesis is true.
In simpler terms, it helps us measure the strength of the evidence against the null hypothesis.

For the Watch Corporation case, once we compute the t-statistic, we look up or calculate the corresponding p-value:
  • If the p-value is smaller than the chosen significance level (0.05), it suggests that the observed sample mean is unlikely to occur if the null hypothesis were true, leading to its rejection.
  • Conversely, a larger p-value indicates that the data is compatible with the null hypothesis.
  • Thus, the p-value provides a means to quantitatively assess the evidence and make informed decisions about the hypotheses.
Interpreting p-values allows us to understand the result's reliability and the statistical significance of our test outcomes.

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Most popular questions from this chapter

An urban planner claims that, nationally, 20 percent of all families renting condominiums move during a given year. A random sample of 200 families renting condominiums in the Dallas Metroplex revealed that 56 had moved during the past year. At the .01 significance level, does this evidence suggest that a larger proportion of condominium owners moved in the Dallas area? Determine the \(p\) -value.

A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Its records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, the agency believes that the mean selling time is now greater than 90 days. A statewide survey of 100 farms sold recently revealed that the mean selling time was 94 days, with a standard deviation of 22 days. At the . 10 significance level, has there been an increase in selling time?

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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 $$\begin{array}{|lllllllllllllll|}\hline 25 & 27 & 25 & 26 & 25 & 28 & 28 & 27 & 24 & 26 & 25 & 29 & 25 & 27 & 24 \\\\\hline\end{array}$$ 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.

A United Nations report shows the mean family income for Mexican migrants to the United States is 27,000$ per year. A FLOC (Farm Labor Organizing Committee) evaluation of 25 Mexican family units reveals a mean to be 30,000 with a sample standard deviation of 10,000 . Does this information disagree with the United Nations report? Apply the 0.01 significance level.
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