/*! 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 43 Give as much information as you ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(\mathrm{df}=9, t=0.73\) b. Upper-tailed test, \(\mathrm{df}=10, t=-0.5\) c. Lower-tailed test, \(n=20, t=-2.1\) d. Lower-tailed test, \(n=20, t=-5.1\) e. Two-tailed test, \(n=40, t=1.7\)

Short Answer

Expert verified
The P-values for the given situations are approximately: a. 1, b. 0.69, c. 0.025, d. 0.001, e. 0.096

Step by step solution

01

Determine P-value for Two-tailed test, df=9, t=0.73

To find the p-value, we must look at a t-distribution table for the given degrees of freedom (df = 9). In this case, we find that t=0.73 corresponds to a probability in the t-distribution table of about 0.77, which is the one-tailed p-value. Because the test is two-tailed, we must double this to get the two-tailed P-value: 0.77*2 = 1.54. Since a probability cannot exceed 1, we interpret this as a P-value of 1, indicating weak evidence against the null hypothesis.
02

Determine P-value for Upper-tailed test, df=10, t=-0.5

Looking at the t-distribution table for df = 10, we find that t = -0.5 corresponds to a probability in the table of about 0.69, which is the two-tailed p-value. Because the test is upper-tailed (or one-tailed), the P-value is equal to this value: P = 0.69, suggesting weak evidence against the null hypothesis.
03

Determine P-value for Lower-tailed test, n=20, t=-2.1

When n=20, this implies df=20-1=19. Given df = 19 and t = -2.1, we have a one-tailed p-value of around 0.025 according to the t-distribution table. Since this is a lower-tailed (or one-tailed) test, the P-value is the same as the one-tailed p-value: P = 0.025, suggesting moderate evidence against the null hypothesis.
04

Determine P-value for Lower-tailed test, n=20, t=-5.1

Given df = 19 (from n=20) and t = -5.1, we find a one-tailed p-value close to 0.001 (or possibly even less) in the t-distribution table. Again, because it's a lower-tailed test, the P-value equals the one-tailed p-value: P ≈ 0.001, suggesting strong evidence against the null hypothesis.
05

Determine P-value for Two-tailed test, n=40, t=1.7

Given df = 40-1=39 and t = 1.7, we find a one-tailed p-value of about 0.048 in the t-distribution table. Because this is a two-tailed test, we must double this value to get the P-value: 0.048*2 = 0.096, indicating weak to moderate evidence against 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 91Ó°ÊÓ!

Key Concepts

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

t-distribution
The t-distribution is a key statistical distribution used in hypothesis testing, especially when dealing with small sample sizes. It's a kind of bell-shaped curve, much like the normal distribution, but it differs in that it has heavier tails.
This means the t-distribution is more forgiving to outliers in data, making it more appropriate for smaller datasets where sample sizes are limited. As the sample size increases, the t-distribution becomes more like the normal distribution.
  • Origin of the t-distribution: It was introduced by William Sealy Gosset for small sample datasets, who published under the pseudonym "Student," hence the name Student's t-distribution.
  • Application: Commonly used for estimating the mean in smaller sample sizes or when the standard deviation is unknown.
Whenever conducting a t-test, you will use the t-distribution tables to find probabilities associated with various t-scores and degrees of freedom.
degrees of freedom
Degrees of freedom (df) are a crucial consideration in statistical analyses, especially when using the t-distribution. In the context of the t-test, degrees of freedom determine the exact shape of the t-distribution.

To calculate the degrees of freedom for a t-test, use the formula: \[ df = n - 1 \] where "n" is the sample size.
  • Importance: Degrees of freedom impact the reliability of statistical tests and influence the decision threshold for statistical significance.
  • Effect: More degrees of freedom mean the t-distribution will have thinner tails and be closer to a normal distribution.
The degrees of freedom adjust as sample sizes increase, reflecting higher data reliability. When aligning sample statistics with population parameters, df plays an essential role in accurate assessments.
two-tailed test
A two-tailed test is used in hypothesis testing when we are interested in checking for effects in both directions, either an increase or a decrease.

Here's how it works:
  • Symmetric interest: If the null hypothesis states that a parameter has a specific value, a two-tailed test checks for any deviation, whether higher or lower.
  • Example: If we are testing if a mean is `10`, the two-tailed test will check if it's significantly different from `10`, whether it is larger or smaller.
  • P-value calculation: The p-value in a two-tailed test is calculated by doubling the one-tailed p-value because it considers the probability of deviation in both directions.
Hence, it is vital when both outcomes matter in statistical conclusions, ensuring that extremes on either side of the distribution contribute to rejecting the null hypothesis.
lower-tailed test
A lower-tailed test, also known as a left-tailed test, investigates if a sample statistic is significantly less than a stated value in the null hypothesis. This is useful when the primary concern is a reduction or minimization effect.

Key aspects of a lower-tailed test include:
  • Direction: It focuses on statistically significant lower values. We are testing if the observed statistic falls significantly below the hypothesized parameter.
  • P-value: The p-value is calculated as the area under the curve to the left of the calculated test statistic in a t-distribution.
  • Application: Useful for testing hypotheses where the interest is whether the effect might be less rather than greater than expected, such as testing if a new drug decreases blood pressure.
This type of test is one-sided, meaning it only values deviations in one specific direction, which can increase the test's power in detecting decreases.
upper-tailed test
An upper-tailed test is designed to determine if a sample statistic is significantly higher than a value specified in the null hypothesis. It is particularly useful in cases where a positive increase is the focus.

Here are some key features:
  • Focus: The test aims at identifying cases where the observed statistic is likely higher than the hypothesized statistic.
  • P-value calculation: P-values are obtained by looking at the right-hand tail of the distribution, indicating higher values.
  • Common Usage: If you are testing whether a new training method increases employees' productivity over a standard method, the interest is in effects on the upper end of the distribution.
Like a lower-tailed test, this one-sided test increases sensitivity when checking for potential increases, aiding in pinpoint precision for hypothesis validation.

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 television manufacturer claims that (at least) \(90 \%\) of its TV sets will need no service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample of \(n=100\) purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let \(\hat{p}\) be the sample proportion of responses indicating no repair (so that no repair is identified with a success). Let \(p\) denote the actual proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that \(p<.9 .\) The appropriate hypotheses are then \(H_{0}: p=.9\) versus \(H_{a}: p<.9\). a. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=.10\) or one that uses \(\alpha=.01 ?\) Explain.

Many consumers pay careful attention to stated nutritional contents on packaged foods when making purchases. It is therefore important that the information on packages be accurate. A random sample of \(n=12\) frozen dinners of a certain type was selected from production during a particular period, and the calorie content of each one was determined. (This determination entails destroying the product, so a census would certainly not be desirable!) Here are the resulting observations, along with a boxplot and normal probability plot: \(\begin{array}{llllllll}255 & 244 & 239 & 242 & 265 & 245 & 259 & 248\end{array}\) \(\begin{array}{llll}225 & 226 & 251 & 233\end{array}\) a. Is it reasonable to test hypotheses about mean calorie content \(\mu\) by using a \(t\) test? Explain why or why not. b. The stated calorie content is \(240 .\) Does the boxplot suggest that true average content differs from the stated value? Explain your reasoning. c. Carry out a formal test of the hypotheses suggested in Part (b).

Pairs of \(P\) -values and significance levels, \(\alpha,\) are given. For each pair, state whether the observed \(P\) -value leads to rejection of \(H_{0}\) at the given significance level. a. \(\quad P\) -value \(=.084, \alpha=.05\) b. \(\quad P\) -value \(=.003, \alpha=.001\) c. \(P\) -value \(=.498, \alpha=.05\) d. \(\quad P\) -value \(=.084, \alpha=.10\) e. \(\quad P\) -value \(=.039, \alpha=.01\) f. \(P\) -value \(=.218, \alpha=.10\)

The article "Theaters Losing Out to Living Rooms" (San Luis Obispo Tribune, June 17,2005\()\) states that movie attendance declined in \(2005 .\) The Associated Press found that 730 of 1000 randomly selected adult Americans preferred to watch movies at home rather than at a movie theater. Is there convincing evidence that the majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a .05 significance level.

A certain television station has been providing live coverage of a particularly sensational criminal trial. The station's program director wishes to know whether more than half the potential viewers prefer a return to regular daytime programming. A survey of randomly selected viewers is conducted. Let \(p\) represent the proportion of all viewers who prefer regular daytime programming. What hypotheses should the program director test to answer the question of interest?

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.