/*! 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 9 You are given the following hypo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are given the following hypotheses: $$\begin{array}{l} H_{0}: \mu=60 \\ H_{A}: \mu \neq 60 \end{array}$$ We know that the sample standard deviation is 8 and the sample size is \(20 .\) For what sample mean would the p-value be equal to \(0.05 ?\) Assume that all conditions necessary for inference are satisfied.

Short Answer

Expert verified
The sample means are 63.743 or 56.257 for a p-value of 0.05.

Step by step solution

01

Identify the test statistic

We will be using the t-test statistic for this hypothesis test because the population standard deviation is unknown, and the sample size is small. The test statistic is given by the formula: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.
02

Find the critical t-values

Since we are looking for a sample mean corresponding to a p-value of 0.05 in a two-tailed test, we need the critical t-values at \(\alpha/2 = 0.025\) in each tail. With \(n - 1 = 19\) degrees of freedom, we can use a t-table or calculator to find these values. For \(df = 19\), and \(\alpha/2 = 0.025\), the critical t-values are approximately \(\pm 2.093\).
03

Solve for the sample mean

We will set up the equation for the test statistic, substituting the critical t-value and solving for the sample mean \(\bar{x}\):\[\pm 2.093 = \frac{\bar{x} - 60}{8/\sqrt{20}}\]Solving for \(\bar{x}\):\[\bar{x} - 60 = \pm 2.093 \times \frac{8}{\sqrt{20}}\]\[\bar{x} - 60 = \pm 3.743\]\[\bar{x} = 60 \pm 3.743\]Thus, \(\bar{x} = 63.743\) or \(\bar{x} = 56.257\).
04

Verify using the p-value

For each calculated sample mean (63.743 and 56.257), we should verify that the p-values are indeed 0.05. Using these sample means in the test statistic formula:For \(\bar{x} = 63.743\), \[ t = \frac{63.743 - 60}{8/\sqrt{20}} \approx 2.093 \]For \(\bar{x} = 56.257\), \[ t = \frac{56.257 - 60}{8/\sqrt{20}} \approx -2.093 \]Both correspond to \(t\) values where the p-value is 0.05 in a two-tailed test.

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.

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It starts with forming two hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_A \)). In our example, \( H_0: \mu = 60 \) assumes no effect or difference, while \( H_A: \mu eq 60 \) suggests there is an effect or difference.

These hypotheses help differentiate between what you expect based on standard assumptions (\( H_0 \)) and what you may observe if your alternative is true (\( H_A \)). There are typically two types of errors in hypothesis testing: Type I error (rejecting \( H_0 \) when it's true) and Type II error (failing to reject \( H_0 \) when \( H_A \) is true). Understanding these errors is crucial to evaluating the reliability of the test results.

Hypothesis testing typically involves:
  • Setting up a significance level (\( \alpha \)), which is the probability of making a Type I error.
  • Computing a test statistic to compare against critical values or to derive a p-value.
  • Making a decision to reject or not reject the null hypothesis based on this comparison.
P-value
The p-value is a crucial component in hypothesis testing, as it quantifies the evidence against the null hypothesis. It represents the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true.

In simpler terms, the p-value tells us how likely it is that we could have observed our data (or more extreme) if the null hypothesis were true. A low p-value indicates strong evidence against \( H_0 \), suggesting you should consider rejecting it. Conversely, a high p-value suggests the data is consistent with the null hypothesis.

Key points about p-values:
  • A p-value less than the significance level (\( \alpha \)—commonly 0.05) leads us to reject \( H_0 \).
  • P-values do not measure the probability that \( H_0 \) is true.
  • They provide a measure to compare with \( \alpha \) to make a decision in hypothesis testing.
Critical T-values
Critical t-values play a vital role in determining whether to reject the null hypothesis in a t-test. These values act as thresholds on the t-distribution, delineating the rejection region for \( H_0 \). To find critical t-values, you consider the desired significance level (\( \alpha \)) and degrees of freedom (\( df = n - 1 \) for a sample of size \( n \)).

In our example, with \( n = 20 \), the degrees of freedom are \( df = 19 \). For a two-tailed test with \( \alpha = 0.05 \), we divide the significance level between both tails (\( \alpha/2 = 0.025 \)). Using a t-table or calculator, the critical t-values for \( df = 19 \) and \( \alpha/2 = 0.025 \) are approximately \( \pm 2.093 \).

These critical values help to:
  • Set the thresholds where we would reject the null hypothesis.
  • Determine the range within which we would fail to reject \( H_0 \).
Sample Mean
The sample mean (\( \bar{x} \)) is an important statistic used to estimate the population mean (\( \mu \)) based on a sample. It acts as a central measure of the data collected and helps in identifying whether there is a significant difference from the hypothesized population mean under \( H_0 \).

In our hypothesis testing example, we were tasked to find the sample mean that corresponds to a p-value of 0.05. By utilizing the t-test formula, we determined that the sample means \( \bar{x} = 63.743 \) and \( \bar{x} = 56.257 \) both lead to p-values of 0.05, indicating that these sample means are at the threshold of differing significantly from \( \mu_0 = 60 \).

The procedure to solve for the sample mean involves:
  • Setting up the equation from the test statistic formula and inserting critical t-values.
  • Solving for \( \bar{x} \) to find the precise values on either side of \( \mu_0 \)
  • Verifying that these means correspond to the specified p-values in the context of a t-distribution.

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

Air quality measurements were collected in a random sample of 25 country capitals in 2013, and then again in the same cities in 2014. We would like to use these data to compare average air quality between the two years. Should we use a paired or non-paired test? Explain your reasoning.

An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical \(t\) -value \(\left(\mathrm{t}^{\star}\right)\) for the given sample size and confidence level. (a) \(n=6, \mathrm{CL}=90 \%\) (b) \(n=21, \mathrm{CL}=98 \%\) (c) \(n=29, \mathrm{CL}=95 \%\) (d) \(n=12, \mathrm{CL}=99 \%\)

An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at \(\alpha=0.01\). (a) \(n=26, T=2.485\) (b) \(n=18, T=0.5\)

Georgianna claims that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons. We have a random sample of 30 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. (a) Use a hypothesis test to determine if there is sufficient evidence against Georgianna's claim. (b) Construct a \(95 \%\) confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. (c) Do your results from the hypothesis test and the confidence interval agree? Explain your reasoning.

For a given confidence level, \(t_{d f}^{\star}\) is larger than \(z^{\star}\). Explain how \(t_{d f}^{*}\) being slightly larger than \(z^{*}\) affects the width of the confidence interval.

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.