/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Consider the following hypothesi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 50 \\ H_{\mathrm{a}}: \mu>50 \end{array} \\] A sample of 60 is used and the population standard deviation is \(8 .\) Use the critical value approach to state your conclusion for each of the following sample results, Use \\[\alpha=.05\\] a. \(\quad \bar{x}=52.5\) b. \(\bar{x}=51\) c. \(\quad \bar{x}=51.8\)

Short Answer

Expert verified
Reject \( H_0 \) for \( \bar{x} = 52.5 \) and \( \bar{x} = 51.8 \); do not reject for \( \bar{x} = 51 \).

Step by step solution

01

Identify the Test Type and Conditions

We have a one-tailed hypothesis test for the population mean with the hypotheses: \( H_{0}: \mu \leq 50 \) and \( H_{a}: \mu > 50 \). The significance level is \( \alpha = 0.05 \), and the standard deviation \( \sigma = 8 \) is known. The sample size is 60, which is sufficiently large for a z-test.
02

Calculate the Critical Value for the Test

Since this is a right-tailed test (\( H_a: \mu > 50 \)), we need to find the critical z-value. For \( \alpha = 0.05 \) in a standard normal distribution, the critical z-value is approximately 1.645.
03

Compute the Standard Error

The standard error (SE) is calculated using the formula: \( \text{SE} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 8 \) and \( n = 60 \). Therefore, \( \text{SE} = \frac{8}{\sqrt{60}} \approx 1.033 \).
04

Calculate the Test Statistic for Each Sample Mean

The test statistic is calculated using the formula: \( z = \frac{\bar{x} - \mu_0}{\text{SE}} \). For each sample mean received:a. For \( \bar{x} = 52.5 \):\[ z = \frac{52.5 - 50}{1.033} \approx 2.42 \]b. For \( \bar{x} = 51 \):\[ z = \frac{51 - 50}{1.033} \approx 0.97 \]c. For \( \bar{x} = 51.8 \):\[ z = \frac{51.8 - 50}{1.033} \approx 1.74 \]
05

Compare Each Test Statistic to the Critical Value

Compare the calculated z-values to the critical value of 1.645 to determine if the null hypothesis \( H_0 \) should be rejected.- For \( \bar{x} = 52.5 \), \( z = 2.42 \), which is greater than 1.645, therefore reject \( H_0 \).- For \( \bar{x} = 51 \), \( z = 0.97 \), which is less than 1.645, therefore do not reject \( H_0 \).- For \( \bar{x} = 51.8 \), \( z = 1.74 \), which is greater than 1.645, therefore reject \( H_0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-Test
The z-test is a statistical method used to determine if there is a significant difference between the sample mean and the population mean. This test is applicable when the sample size is large enough, typically 30 or more, and the population standard deviation is known. In hypothesis testing, the z-test helps us decide if we should reject the null hypothesis, which in our scenario states that the population mean is less than or equal to 50.
A one-tailed z-test is employed when the alternative hypothesis is directional, like when we're checking if the mean is greater than a given value, as in this example. The z-test involves calculating a test statistic based on the sample data and comparing it to a critical value.
  • This test statistic reflects how many standard deviations the sample mean is from the population mean under the null hypothesis.
  • It's calculated using the formula: \( z = \frac{\bar{x} - \mu_0}{\text{SE}} \).
Once calculated, the z-value can be compared to a critical value to draw a conclusion about the hypothesis.
Critical Value Approach
The critical value approach in hypothesis testing is used to decide whether to accept or reject the null hypothesis. It's a threshold we compare our test statistic to, helping us determine significance.
In the context of a z-test, the critical value is a point on the standard normal distribution that depends on the chosen significance level \( \alpha \).
The right-tailed test, as used here, means we look towards the upper tail of the distribution since it involves checking if the mean is significantly greater than a certain value.
  • For a 5% significance level (\( \alpha = 0.05 \)), the critical z-value is around 1.645.
  • If the calculated z-value is greater than 1.645, the null hypothesis is rejected.
  • This signifies that there is enough statistical evidence to support the alternative hypothesis.
The critical value essentially acts as the boundary. If crossed, it suggests that the sample provides enough evidence against the null hypothesis.
Standard Error
Standard error (SE) measures the variability or dispersion of a sampling distribution. It's different from standard deviation, which measures variability within a single sample or population.
In hypothesis testing, SE serves as a baseline for evaluating how much the sample mean \( \bar{x} \) deviates from the population mean \( \mu \). Smaller SE values indicate less variability in the estimate of the population mean.
The formula for standard error is:\[\text{SE} = \frac{\sigma}{\sqrt{n}} \]where \( n \) is the sample size and \( \sigma \) is the population standard deviation.
In our exercise:
  • Since \( \sigma = 8 \) and \( n = 60 \), we calculate \( \text{SE} \approx 1.033 \).
  • This value indicates how much difference you might expect by chance between the sample mean and the true population mean.
The standard error is a pivotal part of calculating the z-test statistic, which is then used to make informed decisions regarding our hypothesis.
Significance Level
The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. Also known as the level of risk you're willing to take for making a Type I error.
In hypothesis testing, the significance level helps determine the critical value and region for making decisions. Common significance levels include 0.05, 0.01, and 0.10, with 0.05 being the most universally used. Here, \( \alpha = 0.05 \).
The choice of significance level affects:
  • The critical value, which is used as a benchmark for test statistics.
  • The width of the confidence interval around the sample mean.
For our context with \( \alpha = 0.05 \):
  • We have a 5% chance of wrongly rejecting the null hypothesis.
  • The critical z-value is set to 1.645, which corresponds to the point where there's only a 5% probability of observing a test statistic this extreme if the null hypothesis were true.
In summary, setting the significance level is a crucial step in hypothesis testing that influences the reliability and interpretation of results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A radio station in Myrtle Beach announced that at least \(90 \%\) of the hotels and motels would be full for the Memorial Day weekend. The station advised listeners to make reservations in advance if they planned to be in the resort over the weekend. On Saturday night a sample of 58 hotels and motels showed 49 with a no-vacancy sign and 9 with vacancies. What is your reaction to the radio station's claim after secing the sample evidence? Use \(a=.05\) in making the statistical test. What is the \(p\) -value?

According to the federal government, \(24 \%\) of workers covered by their company's health care plan were not required to contribute to the premium (Statistical Abstract of the United States: 2006 . A recent study found that 81 out of 400 workers sampled were not required to contribute to their company's health care plan. a. Develop hypotheses that can be used to test whether the percent of workers not required to contribute to their company's health care plan has declined. b. What is a point estimate of the proportion receiving free company-sponsored health care insurance? c. Has a statistically significant decline occurred in the proportion of workers receiving free company-sponsored health care insurance? Use \(\alpha=.05\).

The label on a 3 -quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

A production line operation is designed to fill cartons with laundry detergent to a mean weight of 32 ounces. A sample of cartons is periodically selected and weighed to determine whether underfilling or overfilling is occurring. If the sample data lead to a conclusion of underfilling or overfilling, the production line will be shut down and adjusted to obtain proper filling. a. Formulate the null and alternative hypotheses that will help in deciding whether to shut down and adjust the production line. b. Comment on the conclusion and the decision when \(H_{0}\) cannot be rejected. c. Comment on the conclusion and the decision when \(H_{0}\) can be rejected.

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.