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A special cable has a breaking strength of 800 pounds. The standard deviation of the population is 12 pounds. A researcher selects a random sample of 20 cables and finds that the average breaking strength is 793 pounds. Can he reject the claim that the breaking strength is 800 pounds? Find the \(P\) -value. Should the null hypothesis be rejected at \(\alpha=0.01 ?\) Assume that the variable is normally distributed.

Short Answer

Expert verified
Reject the null hypothesis; the \( P \)-value is 0.009, which is less than \( \alpha = 0.01 \).

Step by step solution

01

Identify the null and alternative hypotheses

The null hypothesis, denoted as \(H_0\), is that the population mean breaking strength of the cables is 800 pounds: \(H_0: \mu = 800\). The alternative hypothesis, denoted as \(H_a\), is that the population mean breaking strength is different from 800 pounds: \(H_a: \mu eq 800\). This will be a two-tailed test.
02

Calculate the standard error of the mean

The standard error of the mean (SEM) is calculated using the formula \( SEM = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 12 \) pounds is the standard deviation of the population, and \( n = 20 \) is the sample size. Thus, \( SEM = \frac{12}{\sqrt{20}} \approx 2.68 \).
03

Calculate the test statistic

The test statistic \( z \) for a sample mean is calculated using the formula \( z = \frac{\bar{x} - \mu}{SEM} \), where \( \bar{x} = 793 \) pounds is the sample mean. Substitute the values to get \( z = \frac{793 - 800}{2.68} \approx -2.61 \).
04

Determine the critical value and make a decision

To determine the \( P \)-value for a two-tailed test, we find the probability that \( z \) is less than \(-2.61\) or greater than \(2.61\). Consulting standard \( z \)-tables or using a calculator, the \( P \)-value is approximately 0.009. Since this \( P \)-value is less than \( \alpha = 0.01 \), we reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a foundational concept. It's a statement that we assume to be true unless we have strong evidence against it. In our exercise, the null hypothesis, denoted as \(H_0\), claims that the average breaking strength of the cables is 800 pounds. This is expressed as \(H_0: \mu = 800\). It serves as the benchmark we test against.

The null hypothesis assumes no effect or no difference, hence the term "null." Our task in testing is to see if the sample data introduce significant doubt to this assumption. If the evidence is strong enough (i.e., the data show a substantial deviation from the null hypothesis claim), we may decide to reject the null hypothesis. Otherwise, we do not reject it, essentially maintaining the assumption of no difference or no effect.
Standard Error
Standard error is a crucial measure in statistics. It represents the extent to which sample means would vary if you took multiple samples from the same population. In simpler terms, it’s an indicator of how accurately your sample mean represents the population mean.

For the calculation, we use the formula:
  • \(SEM = \frac{\sigma}{\sqrt{n}}\)
where \(\sigma = 12\) pounds is the population’s standard deviation and \(n = 20\) is the sample size. Here, the standard error works out to be approximately 2.68 pounds.

A smaller standard error suggests that the sample mean is a more accurate representation of the true population mean. This measure helps us understand the reliability of the sample mean and plays a vital role in subsequent computations, like the test statistic.
Test Statistic
The test statistic is a value that we calculate from our sample data to test against the null hypothesis. In our scenario, we use the \(z\)-test, a type of test used when dealing with sample means. Our formula for the test statistic \(z\) is:
  • \(z = \frac{\bar{x} - \mu}{SEM}\)
where \(\bar{x} = 793\) pounds is the sample mean, and \(\mu = 800\) pounds is the population mean according to the null hypothesis. Using these values, we find \(z \approx -2.61\).

The test statistic tells us how far our sample mean is from the null hypothesis mean in terms of standard errors. A higher absolute value of \(z\) indicates a greater deviation, increasing the likelihood of rejecting the null hypothesis, especially if \(z\) falls into the critical region determined by the significance level.
P-value
The \(P\)-value is a statistical measure that helps us determine the significance of our test results. It represents the probability of observing test results as extreme as the ones recorded, under the assumption that the null hypothesis is true.

For our exercise, a \(P\)-value of approximately 0.009 means there's a 0.9% probability of seeing the sample data, or something more extreme, if the breaking strength truly were 800 pounds.

In hypothesis testing, we often compare the \(P\)-value to a pre-defined significance level \(\alpha\), which is the threshold for deciding whether to reject the null hypothesis. Here, our significance level is \(\alpha = 0.01\). Since the \(P\)-value is less than \(\alpha\), we can reject the null hypothesis, suggesting that the true average breaking strength is likely not 800 pounds.

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Most popular questions from this chapter

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. \(\begin{array}{llll}12.03 & 12.10 & 12.02 & 11.98 \\ 12.00 & 12.05 & 11.97 & 11.99\end{array}\)

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Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. The manager of a large company claims that the standard deviation of the time (in minutes) that it takes a telephone call to be transferred to the correct office in her company is 1.2 minutes or less. A random sample of 15 calls is selected, and the calls are timed. The standard deviation of the sample is 1.8 minutes. At \(\alpha=0.01\), test the claim that the standard deviation is less than or equal to 1.2 minutes. Use the \(P\) -value method.

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