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Consider the following null and alternative hypotheses: $$ H_{0}: \mu=60 \text { versus } H_{1}: \mu>60 $$ Suppose you perform this test at \(\alpha=.01\) and fail to reject the null hypothesis. Would you state that the difference between the hypothesized value of the population mean and the observed value of the sample mean is "statistically significant" or would you state that this difference is "statistically not significant"? Explain.

Short Answer

Expert verified
The difference between the hypothesized value of the population mean and the observed value of the sample mean is 'statistically not significant' at the 0.01 level. This is because we failed to reject the null hypothesis, indicating there isn't enough evidence to prove a significant difference.

Step by step solution

01

Understanding the null and alternative hypotheses

The null hypothesis \(H_0\) states that the population mean \(\mu\) is 60. On the other hand, the alternative hypothesis \(H_1\) suggests that the population mean \(\mu\) is greater than 60. We perform a test to see if we have enough evidence to reject \(H_0\) and accept \(H_1\)
02

Understand the meaning of the significance level

The significance level, also denoted as \(\alpha\) is the probability of rejecting the null hypothesis, given that it is true. Here, \(\alpha = 0.01\) implies that there's a 1% risk of concluding that a difference exists when there is indeed no actual difference. It's a measure of the strength of the evidence that must be present in our sample data before we reject the null hypothesis.
03

Interpret the decision not to reject the null hypothesis

Failing to reject the null hypothesis does not prove that the null hypothesis is true, but merely that there is not enough evidence against \(H_{0}\) and in favor of \(H_{1}\) given the data and the chosen level of significance. So, in this context, the difference between the hypothesized population mean and the observed value of the sample mean is not 'statistically significant' at the 0.01 level, suggesting that there is not a significant difference between the two.

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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 starting point.It reflects a status quo or a baseline scenario that there is no effect or difference.In the context of a population mean (\(\mu\)), the null hypothesis (\(H_0\)) might state that the population mean is a specific value.For instance, it could claim that \(\mu = 60\).

The purpose of formulating the null hypothesis is to test if there is enough evidence in your data to disprove it.Think of it as a claim you are skeptical of, and you are looking for data to challenge it.Continuing our exercise, if after calculating you find no sufficient evidence against \(H_0\), you fail to reject the null hypothesis.This doesn't prove \(H_0\) to be true, but rather indicates that your data does not provide strong enough evidence against it.

In practice:
  • The null hypothesis is often assumed true until proven otherwise.
  • The test involves data from a sample to make conclusions about the entire population.
  • A 'fail to reject' decision indicates data is insufficient to declare a significant effect or difference.
Understanding this concept is vital as every statistical test begins with a null hypothesis.
Alternative Hypothesis
The alternative hypothesis (\(H_1\)) is what we consider if we have enough evidence to reject the null hypothesis.It represents a new theory or a change from the status quo suggested by \(H_0\).In our exercise example, the alternative hypothesis states that the population mean \(\mu\) is greater than 60.

Selecting \(H_1\) involves deciding the scenario you plan to support with your research or test.This decision will drive how you perform your hypothesis test.The goal is to obtain enough statistical evidence to accept \(H_1\) and reject \(H_0\).

A few pointers:
  • \(H_1\) is formulated to directly challenge \(H_0\).
  • When evidence strongly supports \(H_1\), \(H_0\) can be rejected.
  • Choosing the right \(H_1\) is crucial for meaningful analysis and valid conclusions.
Overall, the alternative hypothesis helps in validating or discovering new patterns.
Significance Level
The significance level, symbolized by \(\alpha\), is a threshold for determining how much evidence you need to reject the null hypothesis.It quantifies the risk you're willing to take of saying there's a difference when none exists.In the exercise, \(\alpha = 0.01\) signifies a 1% risk.

Significance level selection is crucial as it affects the outcome of your hypothesis tests.A smaller \(\alpha\) means stricter criteria for rejecting \(H_0\).Here, with \(\alpha = 0.01\), you require robust evidence to consider the observed difference significant.

Why it matters:
  • Helps define the boundary of statistical test's confidence.
  • Aids in controlling Type I errors – rejecting a true \(H_0\) mistakenly.
  • Commonly set at 0.05, but can be adjusted based on context.
With the right significance level, results ensure high validity and lesser chances of false positives.
Statistically Significant
The term 'statistically significant' refers to the likelihood that a result or relationship is caused by something other than chance.It's evaluated based on the significance level \(\alpha\).In the exercise, failing to reject the null hypothesis means the difference isn't statistically significant.

For results to be considered statistically significant, they must meet or surpass the evidence threshold set by \(\alpha\).If a test result surpasses this level, it suggests that there is enough evidence to conclude a substantial effect or difference.In other words, the data "speaks loudly enough."

Key points:
  • Significance involves both the statistical test and chosen \(\alpha\).
  • Doesn't imply a large effect, just that it's unlikely due to chance.
  • Helps determine whether further investigation or action is warranted.
When interpreting 'statistically significant,' it's critical to remember that it does not equate to practical significance.It merely indicates a statistical likelihood of validity.

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Most popular questions from this chapter

Mong Corporation makes auto batteries. The company claims that \(80 \%\) of its LL.70 batteries are good for 70 months or longer. A consumer agency wanted to check if this claim is true. The agency took a random sample of 40 such batteries and found that \(75 \%\) of them were good for 70 months or longer. a. Using the \(1 \%\) significance level, can you conclude that the company's claim is false? b. What will your decision be in part a if the probability of making a Type I error is zero? Explain.

Shulman Steel Corporation makes bearings that are supplied to other companies. One of the machines makes bearings that are supposed to have a diamcter of 4 inches. The bearings that have a diameter of either more or less than 4 inches are considered defective and are discarded. When working properly, the machine does not produce more than \(7 \%\) of bearings that are defective. The quality control inspector selects a sample of 200 bearings each week and inspects them for the size of their diameters. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p \geq\) 07, where \(p\) is the proportion of bearings that are defective. He always uses a \(2 \%\) significance level. If the null hypothesis is rejected, the machine is stopped to make any necessary adjustments. One sample of 200 bearings taken recently contained 22 defective bearings. a. Using the \(2 \%\) significance level, will you conclude that the machine should be stopped to make necessary adjustments? b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in part a?

The past records of a supermarket show that its customers spend an average of \(\$ 95\) per visit at this store. Recently the management of the store initiated a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whether this is true, the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. \(\begin{array}{rrrrrrr}109.15 & 136.01 & 107.02 & 116.15 & 101.53 & 109.29 & 110.79 \\ 94.83 & 100.91 & 97.94 & 104.30 & 83.54 & 67.59 & 120.44\end{array}\) Assume that the money spent by all customers at this supermarket has a normal distribution. Using the \(5 \%\) significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than \(\$ 95 ?\) (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections \(3.1 .1\) and \(3.2 .2\) of Chapter \(3 .\) Then make the test of hypothesis about \(\mu .\) )

Make the following tests of hypotheses. a. \(H_{0}: \mu=25, \quad H_{1}: \mu \neq 25, \quad n=81, \quad \bar{x}=28.5, \quad \sigma=3, \quad \alpha=.01\) b. \(H_{0}: \mu=12, \quad H_{1}: \mu<12, \quad n=45, \quad \bar{x}=11.25, \quad \sigma=4.5, \quad \alpha=.05\) c. \(H_{0}: \mu=40, \quad H_{1}: \mu>40, \quad n=100, \quad \bar{x}=47, \quad \sigma=7, \quad \alpha=.10\)

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all long- distance calls made by residential customers is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does your conclusion change if \(\alpha=.05 ?\)

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