A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements arecentimeters on a pain scale before and after hypnosis. Assume that the paired sample data are simple random samplesand that the differences have a distribution that is approximately normal. Construct a 95% confidence interval for themean of the "before-after" differences. Does hypnotism appear to be effective in reducing pain?BeforeAfter8.85.76.19.510.69.44.31.06.62.57.48.38.76.93.42.5Construct a 95% confidence interval for the mean of the "before-after" differences.]

A. A confidence interval is a range of values that estimates the true population parameter at a…

Answered: A study was conducted to measure the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Scores on an IQ test are normally distributed. A sample of 18 IQ scores had standard deviation s = 8. Construct a 95% confidence interval for the population variance σ².
Suppose that we are interested in comparing the academic success of college students who belong to fraternal organizations and the academic success of those who do not belong to fraternal organizations. Random samples of size 40 are taken from each population where the variances and standard deviations are known. Sample Size Mean GPA Variance (2) 2.03 2.21 Fraternity member (f): 40 Nonmembers (n): 40 .46 .35 Construct a 96% confidence interval estimate for the difference between the mean GPAS (uf- un). Interpret this interval estimate. What can you conclude about the mean GPAs of the two groups?
Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 77 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.7 kilograms with a sample standard deviation of 7.16 kilograms. A second sample of 79 subjects went on a low-fat diet. Their sample mean weight loss was 2.6 kilograms with a standard deviation of 5.90 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let 4, denote the mean weight lost on the low-carb diet and u, denote the mean weight lost on the low-fat diet. Use the a = 0.01 level and the TI-84 Plus calculator. Compute the -value. Round the answer to four decimal places. Determine whether to reject H0 State a conclusion
Researchers were interested in the impact of texting on student learning. A group of 99 college students received texts from the researcher during a prerecorded psychology lecture. At the end of the 20-minute lecture, students answered a 17-question quiz about the material that had just been presented; scores were compared to a population mean of 15 on the test, though the population standard deviation was unknown. Which statistical test should the researchers use to analyze their data? a. dependent-samples t test b. one-sample t test c. independent-samples t test d. z test
Fear of heights - A psychologist has developed a new treatment for acrophobia (the extreme fear of heights) and wants to compare the results of this new treatment to the results of the standard treatment. A study is conducted where a random sample of 16 patients undergo the new treatment and their improvement scores on a diagnostic test are compared to a control sample of 13 patients who received the standard treatment. Summary scores for both groups are shown in the table below. Treatment Sample Mean Sample Standard Deviation New 50.9 12 Standard 47.2 11.4 The degrees of freedom for this problem is d f 26.287709. 1. Select the hypotheses that should be used to assess if the new treatment results in higher average improvement scores than the standard treatment. А. Но : И1 = µ2 vs. Ha : H1 # µ2 OB. Ho : µ1 = µ2 vS. Ha : H1 H2 2. Calculate the test statistic. ? V = 3. Calculate the p-value. p-value =
Do cars really get better mileage per gallon on the highway? The table shows results from a study of the MPG (miles per gallon) of cars both in the city and on the highway. Assume that the two samples are randomly selected, independent, the population standard deviations are not know and not considered equal. At the 0.1 significance level, test the claim that the mpg on the highway is better than in the city. MPG on the Highway 35.6 34.3 32.2 33.9 31.1 27.1 33.3 33.4 29.3 33.5 31.4 33.2 33.5 30.8 33.8 MPG in the City 26.4 25.3 18.6 25.6 24.7 24.6 25.1 22.4 29.3 23.7 23.4 22 24 23.6 25.5 What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer p x̄₁ p₁ σ₁² μ₁ μ₂ μ μ(Highway) x̄₂ p̂₁ s₁² p₂ ? ≤ ≠ < ≥ = > Select an answer p₁ p₂ p̂₁ μ(City) μ μ₁ μ₂ x̄₁ x̄₂ s₁² σ₁² p H1: Select an answer p₂ μ(Highway) p̂₂ σ₂² x̄₁ x̄₂ s₂² μ₁ μ₂ μ p₁ p ? < ≠ = ≥ ≤ > Select an answer p₂ p₁ μ₁ σ₁²…
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
Two stores sell watermelons. At the first store the melons weigh an average of 22 pounds, with a standard deviation of 2.5 pounds. At the second store the melons are smaller. with a mean of 18 pounds and a standard deviation of 2 pounds. You take and SRS of 40 melons from the first store and SRS of 34 from the second store Find the mean of each sampling distribution
Researchers examined the bill color (in hue degree) of male and female zebra finches. Here are the summary statistics from the study. Sex N Mean Standard deviation Males 59 2.91 1.46 Females 60 7.42 2.48 How different are male and female zebra finches in bill color? The margin of error for a 95% confidence interval for the difference in mean bill color μF – μM (in hue degree) is
A Two-Sample Hypothesis TestThe mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? (Assume a 5% level of significance.)Useful tools:Normal Distribution Calculatort-Distribution Calculator 1. Which of the following null and alternative hypotheses match this scenario? A---H0:Δμ=0H0:Δμ=0Ha:Δμ≠0Ha:Δμ≠0 B---H0:Δμ≠0H0:Δμ≠0Ha:Δμ=0Ha:Δμ=0 C---H0:Δμ=0H0:Δμ=0Ha:Δμ>0 2. Which type of test should be applied? A---The alternative hypothesis indicates a right-tailed test. B---The alternative hypothesis indicates a left-tailed test. C---The alternative hypothesis indicates a two-tailed test. 3. Which type of distribution should be…
Match to the appropriate answer. You will not use all options and you may not use each option more than once. Dummy variable Coefficient (beta) Correlation coefficient Heteroscedasticity p-value R² + Residual A. Used with categorical variable B. Lower value means statistical significance at a higher confidence level. C. When the distribution of residuals does change as the independent variable changes. D. Higher value means statistical significance at a higher confidence level. E. Difference between the predicted and observed values. F. When the distribution of residuals does not change as the independent variable changes. G. Min=-1, Max=1 H. Min-0, Max=1 1. Used to determine how much dependent variables changes if independent variable changed.
True or False A 95% confidence level in a mean number of minutes that it takes for an amublance to arrive after a 911 call is (4.2, 7.4). Then there is 95% chance that a randomly selected emergency will have an ambulance response time between 4.2 and 7.4 minutes.

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