/*! 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 117 Test \(H_{0}: \mu=10\) vs \(H_{a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 117

Test \(H_{0}: \mu=10\) vs \(H_{a}: \mu>10\) using the sample results \(\bar{x}=13.2, s=8.7,\) with \(n=12\)

Short Answer

Expert verified
The hypothesis is tested by computing the t-score and comparing it with the critical value. The null hypothesis \(H_{0}: \mu=10\) is then either rejected or not rejected based on this comparison.

Step by step solution

01

State the null and alternative hypothesis

The null hypothesis \(H_{0}:\) states that the population mean \(\mu = 10\), and the alternative hypothesis \(H_{a}:\)states that the population mean \(\mu > 10\).
02

Compute test statistic

The test statistic for this problem follows a t-distribution, which can be calculated using the formula: \(t = \frac{\(\bar{x}\) - \(\mu_{0}\)}{s / \sqrt{n}}\). Plug the given values into the formula: \(t = \frac{13.2 - 10}{8.7 / \sqrt{12}} \)
03

Determine the critical value for t

The critical value of t is found in the t-table. Because this is a one-tailed test and the significance level (伪) is not given, let's assume 伪 is 0.05. For 11 degrees of freedom (n-1), the critical value for t is approximately 1.796.
04

Decision - Reject or Fail to reject \(H_{0}\)

If the calculated t-score is greater than the critical value, then reject the null hypothesis in favor of the alternative. If it's less, then fail to reject the null hypothesis.

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 91影视!

Key Concepts

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

Understanding the t-distribution
The t-distribution is invaluable for hypothesis testing, especially when the sample size is small and the population standard deviation is unknown.
It's similar to the normal distribution but has heavier tails, which makes it suitable for small sample sizes as it accounts for the added variability.
The shape of the t-distribution depends on the degrees of freedom, typically calculated as the sample size minus one (i.e., \( n-1 \)).
When we increase the sample size, the t-distribution looks more like a normal distribution, becoming almost identical when the sample size is larger than 30.

Key points about the t-distribution in hypothesis testing:
  • Used when the population standard deviation \( \sigma \) is unknown.
  • Relies on the sample standard deviation \( s \) for calculations.
  • Appropriate for small sample sizes (usually less than 30).
  • Degrees of freedom are a crucial factor, affecting the shape of the distribution.
Exploring the null hypothesis
The null hypothesis \( H_0 \) is a statement that indicates there is no effect or no difference in the context of the problem we're analyzing.
In hypothesis testing, it's the default position that we test against.
The burden of proof lies in providing evidence against the null hypothesis through statistical data analysis.

Key characteristics of a null hypothesis in hypothesis testing:
  • Usually stated as an equality \( \mu = 10 \) in this case, implying no change or no difference.
  • The alternative hypothesis, represented as \( H_a \), opposes \( H_0 \), suggesting some effect or difference, here as \( \mu > 10 \).
  • If we find strong enough evidence from the data (e.g., a test statistic), we might reject \( H_0 \).
  • Failing to reject \( H_0 \) does not prove it true but indicates not enough evidence to support the alternative.
Identifying the critical value
The critical value is pivotal in hypothesis testing, acting as a threshold against which we compare our test statistic.
It's determined by the significance level \( \alpha \), which represents the probability of rejecting the null hypothesis when it's actually true.
The critical value splits the distribution into regions where we decide whether or not to reject \( H_0 \).

Steps to determine the critical value for a t-test:
  • First, decide the significance level \( \alpha \); a common choice is 0.05.
  • Check the degrees of freedom, which is \( n-1 \) for a sample size \( n \).
  • Use a t-distribution table to find the critical t-value for your specific \( \alpha \) and degrees of freedom.
  • A one-tailed test considers one side of the distribution, as in \( H_a: \mu > 10 \).

The critical value tells us where the boundary of rejection is.
If the calculated test statistic exceeds this value, it leads us to reject the null hypothesis in favor of the alternative.

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

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. Exercises 6.166 to 6.169 ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Raisins after 15 Days After 15 days, 320 of the 500 fruit flies eating organic raisins are still alive, while 300 of the 500 eating conventional raisins are still alive.

In Exercise 6.93 on page \(430,\) we see that the average number of close confidants in a random sample of 2006 US adults is 2.2 with a standard deviation of \(1.4 .\) If we want to estimate the number of close confidants with a margin of error within 卤0.05 and with \(99 \%\) confidence, how large a sample is needed?

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=556.9, s_{d}=\) \(143.6, n_{d}=100\)

Metal Tags on Penguins and Survival Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. One variable examined is the survival rate 10 years after tagging. The scientists observed that 10 of the 50 metal tagged penguins survived, compared to 18 of the 50 electronic tagged penguins. Construct a \(90 \%\) confidence interval for the difference in proportion surviving between the metal and electronic tagged penguins \(\left(p_{M}-p_{E}\right)\). Interpret the result.

\(\mathbf{6 . 2 2 0}\) Diet Cola and Calcium Exercise B.3 on page 349 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in \(\mathrm{mg}\) ) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8 .\) Figure 6.20 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise \(\mathrm{B} .3\), we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation
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.