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Consider \(H_{0}: \mu=40\) versus \(H_{1}: \mu>40\) a. A random sample of 64 observations taken from this population produced a sample mean of 43 and a standard deviation of \(5 .\) Using \(\alpha=.025\), would you reject the null hypothesis? b, Another random sample of 64 observations taken from the same population produced a sample mean of 41 and a standard deviation of 7 . Using \(\alpha=.025\), would you reject the null hypothesis?

Short Answer

Expert verified
For both the samples, calculate the test statistic and compare it with the critical value. Depending on whether the test statistic is greater than the critical value or not, decide whether to reject or fail to reject the null hypothesis.

Step by step solution

01

Computation of Test Statistic for First Sample

The test statistic is computed as \[Z = \frac{\bar{x}- \mu_{0}}{\frac{\sigma}{\sqrt{n}}}\]. For this sample, the mean, \(\bar{x}\) is 43, the hypothesised population mean, \(\mu_{0}\), is 40, the standard deviation, \(\sigma\), is 5, and the sample size, \(n\), is 64. Compute the value of the test statistic \(Z\).
02

Determination of Critical Value for First Sample

Next is to determine the critical value at \(\alpha = .025\) for a one-tailed test. Standard Z-tables or statistical software can be used for this purpose.
03

Comparison and Conclusion for First Sample

Compare the absolute value of test statistic with the critical value. If the test statistic is greater than the critical value, reject the null hypothesis.
04

Computation of Test Statistic for Second Sample

Repeat the same process for the second sample. Here the mean, \(\bar{x}\) is 41, the hypothesised population mean, \(\mu_{0}\), is 40, the standard deviation, \(\sigma\), is 7, and the sample size, \(n\), is 64. Compute the value of the test statistic \(Z\).
05

Determination of Critical Value for Second Sample

Next is to determine the critical value at \(\alpha = .025\) for a one-tailed test. Standard Z-tables or statistical software can be used for this purpose.
06

Comparison and Conclusion for Second Sample

Compare the absolute value of test statistic with the critical value. If the test statistic is greater than the critical value, reject the null hypothesis.

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Key Concepts

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

Test Statistic
The test statistic is an essential part of hypothesis testing. It measures how far your sample statistic is from the null hypothesis value, in units of standard error. The formula for the test statistic in a Z-test is:
  • \( Z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}} \)
where:
  • \( \bar{x} \) is the sample mean
  • \( \mu_{0} \) is the population mean according to the null hypothesis
  • \( \sigma \) is the standard deviation of the population
  • \( n \) is the sample size
To compute the test statistic, you simply plug in these values. The analysis involves comparing the test statistic to a critical value to make a decision about the null hypothesis. If the test statistic falls into a certain range (determined by the critical value), we can reject the null hypothesis.
Critical Value
The critical value is a crucial threshold in hypothesis testing that determines the boundary for rejecting the null hypothesis. It is based on the significance level \( \alpha \), which represents the probability of rejecting the null hypothesis when it is actually true.

In a Z-test, this critical value can be found using a standard Z-table or statistical software. For a one-tailed test with \( \alpha = 0.025 \), you would look up the corresponding critical value in these resources. The critical value divides the distribution of the test statistic into two regions:
  • The region where you fail to reject the null hypothesis
  • The region where you reject the null hypothesis
If your calculated test statistic exceeds the critical value in a positive one-tailed test, you reject the null hypothesis.
One-tailed Test
A one-tailed test in hypothesis testing is used when you want to determine if there is a significant effect in only one direction. In this context, you're testing whether the sample mean is significantly greater than the population mean stated in the null hypothesis. This gives more power to detect an effect in that specified direction.

For example, if your null hypothesis is \( H_{0}: \mu = 40 \), and your alternative hypothesis is \( H_{1}: \mu > 40 \), you are conducting a one-tailed test. You are interested only in the possibility of the mean being greater than 40, not less.

By focusing on one tail, you can use the entire \( \alpha \) level (e.g., 0.025) to look for significance in one direction, making it more likely to detect a meaningful increase.
Sample Mean
The sample mean (denoted as \( \bar{x} \)) is the average value of the observations in your sample. It is a point estimate of the population mean, \( \mu \), which you are trying to make inferences about. In hypothesis testing, the sample mean is compared to the hypothesized population mean to see if there is enough evidence to reject the null hypothesis.
  • For instance, if you have a sample mean of 43, you might test this against a hypothesized population mean of 40.
  • The difference between the sample mean and the hypothesized mean is then analyzed using the test statistic.
This sample mean is fundamental in hypothesis testing since it supplies the primary data to draw conclusions under uncertainty. The larger the sample size, the more reliable the sample mean becomes as an estimate of the population mean.
Null Hypothesis
The null hypothesis is a statement in hypothesis testing that there is no effect or no difference, and it provides a baseline for comparison with the sample data. It is usually denoted as \( H_{0} \). In the context of the exercise, the null hypothesis is that the population mean \( \mu = 40 \).

When performing hypothesis testing, this null hypothesis is tested against an alternative hypothesis. The alternative hypothesis denotes that there is an effect or a difference, in this case, \( H_{1}: \mu > 40 \). If the evidence from the sample data is strong enough, we can reject the null hypothesis in favor of the alternative hypothesis.
  • Rejection of \( H_{0} \) suggests that the true population mean is greater than 40.
  • Failing to reject \( H_{0} \) indicates insufficient evidence to support that claim.
This null hypothesis acts as the default assumption that you either reject or fail to reject based on your data's strength.

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Most popular questions from this chapter

Acme Bicycle Company makes derailleurs for mountain bikes. Usually no more than \(4 \%\) of these parts are defective, but occasionally the machines that make them get out of adjustment and the rate of defectives exceeds \(4 \%\). To guard against this, the chief quality control inspector takes a random sample of 130 derailleurs each week and checks each one for defects. If too many of these parts are defective, the machines are shut down and adjusted. To decide how many parts must be defective to shut down the machines, the company's statistician has set up the hypothesis test $$ H_{0}: p \leq .04 \text { versus } H_{1}: p>.04 $$ where \(p\) is the proportion of defectives among all derailleurs being made currently. Rejection of \(H_{0}\) would call for shutting down the machines. For the inspector's convenience, the statistician would like the rejection region to have the form, "Reject \(H_{0}\) if the number of defective parts is \(C\) or more." Find the value of \(C\) that will make the significance level (approximately) \(.05\).

A tool manufacturing company claims that its top-of-the-line machine that is used to manufacture bolts produces an average of 88 or more bolts per hour. A company that is interested in buying this machine wants to check this claim. Suppose you are asked to conduct this test. Briefly explain how you would do so when \(\sigma\) is not known.

The standard therapy used to treat a disorder cures \(60 \%\) of all patients in an average of 140 visits. A health care provider considers supporting a new therapy regime for the disorder if it is effective in reducing the number of visits while retaining the cure rate of the standard therapy. A study of 200 patients with the disorder who were treated by the new therapy regime reveals that 108 of them were cured in an average of 132 visits with a standard deviation of 38 visits. What decision should be made using a \(.01\) level of significance?

As reported on carefair.com (November 15,2006\(), 40 \%\) of women aged 30 years and older would rather get Botox injections than spend a week in Paris. In a recent survey of 400 women aged 65 years and older, 108 women would rather get Botox injections than spend a week in Paris. Using a \(10 \%\) significance level, perform a test of hypothesis to determine whether the current percentage of women aged 65 years or older who would rather get Botox injections than spend a week in Paris is less than \(40 \%\). Use both the \(p\) -value and the critical-value approaches.

A highway construction zone has a posted speed limit of 40 miles per hour. Workers working at the site claim that the mean speed of vehicles passing through this construction zone is at least 50 miles per hour. A random sample of 36 vehicles passing through this zone produced a mean speed of 48 miles per hour. The population standard deviation is known to be 4 miles per hour. a. Do you think the sample information is consistent with the workers' claim? Use \(\alpha=.025\). b. What is the Type I error in this case? Explain. What is the probability of making this error? c. Will your conclusion of part a change if the probability of making a Type I error is zero? d. Find the \(p\) -value for the test of part a. What is your decision if \(\alpha=.025\) ?

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