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A company claims that its battery lasts no less than 42.5 hours in continuous use in a specified toy. A simple random sample of batteries yields a sample mean life of 41.89 hours with a standard deviation of 4.75 hours. A computer calculates a test statistic of \(t=-1.09\) and a \(p\) -value of \(0.139 .\) If the test uses df \(=71,\) what is the best estimate of the sample size?

Short Answer

Expert verified
The best estimate of the sample size is \(n = 72\).

Step by step solution

01

Understanding the given values

The given values in the problem are: test statistic \(t = -1.09\), \(p\)-value \(= 0.139\), degrees of freedom (df) \(= 71\). The problem states the test uses this degrees of freedom.
02

Formulate the relationship between sample size and degree of freedom

The formula to calculate degrees of freedom in a t-test is \(df = n - 1\), where \(n\) is the sample size.
03

Calculate the sample size

Rearrange the formula to solve for \(n\), the sample size. This results in \(n = df + 1\). Substitute the given df value into the formula to find the sample size: \(n = 71 + 1.\)

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Key Concepts

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

Sample Size
When conducting a t-test, one of the key figures is the *sample size*, which refers to the number of observations or data points collected in a study. In the context of hypothesis testing, the sample size is crucial because:
  • It influences the degrees of freedom.
  • It affects the precision and reliability of the test results.
  • Larger sample sizes usually lead to more accurate estimations of the population parameters.
For example, in the provided exercise, the sample size can be calculated using the degrees of freedom. Given that the degrees of freedom ( (df = 71), we determine the sample size ( ) by adding 1: = 71 + 1 = 72. Hence, in this case, the sample includes 72 batteries. Understanding how to determine the sample size is essential as it provides the foundation for calculating other statistical figures.
Degrees of Freedom
In a t-test, the term "degrees of freedom" refers to the number of independent values or observations that are free to vary when computing a statistic. It is an indicator of the sample size used in the statistical analysis.
Degrees of freedom can be calculated as:
  • Degrees of freedom (df) = - 1
  • For one-sample t-tests, the formula is particularly straightforward.
Given the example, where (df given is 71, we can trace back to find the actual sample size of 72 by applying the formula: = df + 1. This concept helps to determine how much information operates freely within the data set and is a critical part of hypothesis testing and ensuring the validity of a statistical test.
Test Statistic
The test statistic in a t-test is a calculated value that allows us to test a hypothesis about the population means. It helps determine whether to reject the null hypothesis. This value is computed by comparing the sample mean to the hypothesized population mean, scaled by the standard deviation and square root of the sample size. In the example problem, the test statistic is given as (t = -1.09). This negative value suggests that the sample mean is slightly less than the hypothesized mean of 42.5 hours. Despite being lower, whether this difference is significant enough to reject the null hypothesis depends on the subsequent analysis involving the p-value. Understanding the test statistic's role helps quantify the difference between the expected and observed data.
P-value
The p-value is a critical concept in statistics, often used to determine the significance of the test results. It measures the probability that the observed data would occur if the null hypothesis were true.
  • A low p-value (typically < 0.05) suggests we reject the null hypothesis, indicating a statistically significant difference.
  • A high p-value implies insufficient evidence to reject the null hypothesis.
In the problem posed, the p-value is 0.139, which is higher than the usual significance level of 0.05. This result suggests that the observed difference between the sample mean and the claimed mean isn't statistically significant.
Thus, the null hypothesis, which states that the battery life lasts at least 42.5 hours, cannot be rejected based on the sample evidence. The p-value provides a clear numerical interpretation of our test results, assisting in making data-driven decisions with greater confidence.

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Most popular questions from this chapter

Lung cancer is the leading cause of cancer deaths in both women and men in the United States. According to Centers for Disease Control and Prevention 2005 statistics, lung cancer accounts for more deaths than breast cancer, prostate cancer, and colon cancer combined. Overall, only about \(16 \%\) of all people who develop lung cancer survive 5 years. Suppose you want to see if this survival rate is still true. How large a sample would you need to take to estimate the true proportion surviving 5 years after diagnosis to within \(1 \%\) with \(95 \%\) confidence? (Use the \(16 \%\) as the initial value of \(p .\) )

A commercial farmer harvests his entire field of a vegetable crop at one time. Therefore, he would like to plant a variety of green beans that mature all at one time (small standard deviation between maturity times of individual plants). A seed company has developed a new hybrid strain of green beans that it believes to be better for the commercial farmer. The maturity time of the standard variety has an average of 50 days and a standard deviation of 2.1 days. A random sample of 30 plants of the new hybrid showed a standard deviation of 1.65 days. Does this sample show a significant lowering of the standard deviation at the 0.05 level of significance? Assume that maturity time is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Getting a college education today is almost as important as breathing and it's expensive! It is not just the tuition, room, and board; textbooks are expensive too. It is very important for students, and their parents, to have an accurate estimate of total textbook costs. The total cost of required textbooks for nine freshman- or sophomore-level classes at 10 randomly selected New York public colleges was collected: $$\begin{array}{lllll}582.19 & 806.40 & 913.44 & 915.75 & 932.35 \\\957.45 & 960.92 & 996.24 & 1070.44 & 1223.44\end{array}$$ a. Construct a histogram and find the mean and standard deviation. b. Demonstrate how this set of data satisfies the assumptions for inference. c. Find the \(95 \%\) confidence interval for \(\mu,\) the mean total cost of required textbooks. d. Interpret the meaning of the confidence interval.

Determine the \(p\) -value for each of the following hypothesis-testing situations. a. \(H_{o}: p=0.5, H_{a}: p \neq 0.5, z \star=1.48\) b. \(H_{o}: p=0.7, H_{a}: p \neq 0.7, z \star=-2.26\) c. \(H_{o}: p=0.4, H_{a}: p>0.4, z \star=0.98\) d. \(H_{o}: p=0.2, H_{a}: p<0.2, z \star=-1.59\)

One of the objectives of a large medical study was to estimate the mean physician fee for cataract removal. For 25 randomly selected cases, the mean fee was found to be \(3550\)dollar with a standard deviation of \(275\)dollar. Set a \(99 \%\) confidence interval on \(\mu,\) the mean fee for all physicians. Assume fees are normally distributed.

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