Let µ denote some population mean of interest and suppose a 95% confidence interval for u is calculated to be (0.65, 0.71) and a 99% confidence interval for u is calculated to be (0.63,0.73). Also, suppose that we want to test H0: µ = 0.74 vs. Ha: µ z 0.74 What can you say about the corresponding p-value? Choose from following letters. a. The corresponding p-value will be smaller than 0.05 but larger than 0.01. b. The corresponding p-value will be larger than 0.05. c. I can't say anything about the corresponding p-value until I run the test d. The correspondong p value will be less than 0.01

Let µ denote some population mean of interest and suppose a 95% confidence interval for u is calculated to be (0.65, 0.71) and a 99% confidence interval for u is calculated to be (0.63,0.73). Also, suppose that we want to test H0: µ = 0.74 vs. Ha: µ z 0.74 What can you say about the corresponding p-value? Choose from following letters. a. The corresponding p-value will be smaller than 0.05 but larger than 0.01. b. The corresponding p-value will be larger than 0.05. c. I can't say anything about the corresponding p-value until I run the test d. The correspondong p value will be less than 0.01

Name: Hypothesis testingLet u denote some population mean ofinterest and su
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Hypothesis testingLet u denote some population mean ofinterest and suppose a 95% confidenceinterval for u is calculated to be (0.65, 0.71)and a 99% confidence interval for u iscalculated to be (0.63,0.73). Also, supposethat we want to testH0: µ = 0.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A sample of 33 lights in Austin gives an average of 46 cars/min while the city SD=3.4 A sample of 34 Dallas lights give a mean of 44 cars/mim with the city SD=1.4 At the 91% level, do the towns differ? 1. Is the test (A-Left Tailed, B-Right Tailed, C-2-Tailed) (separate with comma if there are 2) 2. Critical Value 3. Test statistic 4. 3-decimal p-Value
Suppose you had obtained a test statistic of 2.45 (this is incorrect). What would the P-value be? Draw a conclusion based on a = 0.01. Select one: O a. 0.01 < P-value < 0.02, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O b. 0.005 < P-value < 0.01, we fail to reject Ho and do not have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O c. 0.01 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet for 6 weeks is not the same for men and women. O d. 0.005 < P-value < 0.02, we reject Ho and have sufficient evidence to conclude that the mean difference in LDL from switching from an omnivorous to a plant-based diet…
Consider the scenario ai from the previous question. Suppose you take an SRS of 50 avocados and find that the mean is 1.1 inches, which yields a p-value of 0.0542. What conclusion would you draw at the =0.05 level? At the =0.10level?
The next six questions refer to the following scenario: A random sample of 25 students were asked to self-identify as "statistics-lovers" or "statistics-adorers." 16 students self-identified as "lovers" and 9 students as "adorers." Below table summarizes the mean of a statistics test and standard deviation that was given after the self-identification. Test the following hypothesis at alpha=5% (0.05). HO: Mean Statistics Lovers = Mean Statistics Adorers HA: Mean Statistics Lovers Mean Statistics Adorers You can assume equal variance.
Use z scores to compare the given values.Based on sample data , newborn males have weights with a mean of 3242.5 g and a standard deviation of 687.2 g . Newborn females have weights with a mean of 3085.4 and a standard deviation of 829.1 g . Who has the weight that is more extreme relative to the group from which they came : a male who weighs 1500 g or a female who weighs 1500 ?
If you chose the α level of .01, then which of the following p value will allow you to conclude that your result is statistically significant? a. 0.005 b. 0.05 c. 0.10 d. 0.03
A sample of n=4 scores has a mean of M=75. Find the z-score for this sample: a. If it was obtained from a population with u= 80 and o=10. b. If it was obtained from a population with u=80 and o=20.
Claim: The mean systolic blood pressure of all healthy adults is less than 125 mm Hg. Sample data: For 261 healthy adults, the mean systolic blood pressure level is 121.18 mm Hg and the standard deviation is 15.97 mm Hg. a. Express the original claim in symbolic form. Let the parameter represent a value with respect to systolic blood pressure of a healthy adult. B. Identify the null and alternative hypotheses.
please make your answers clear
Use the following information for questions 3 and 4. A study of class lengths at two schools was done using independent samples. 31 classes were chosen from each. The first school had a mean length of 50.1 minutes and a standard deviation of 1.6 minutes. The second school had a mean length of 48.9 and a standard deviation of 2.2 minutes. We want to determine at α = 0.01 if the means are different. Assume all test conditions are satisfied. What are the null and alternate hypotheses (write in equation/inequality form)?
These are the numbers I am working with: Sample mean= 78 Population mean= 75 Standard deviation:=10 Significance level: .05 Critical value: Z score > 1.96 Z score= 1.34 How do I figure the confidence limits for the 95% confidence interval. Z-score= 1.34 P-value= .1802
What percentage of scores should fall within ±1 SD for TADHD? Choices: 1. A higher percentage of scores than fell within ±1 SD for ADHD 2. A lower percentage of scores than fell within ±1 SD for ADHD 3. The same percentage of scores that fell within ±1 SD for ADHD What is true of a score of 65.55 for TADHD? Choices: 1. It falls within –1.5 SD of the mean and is thus unlikely to be due to chance 2. It falls within 1.5 SD of the mean and is thus unlikely to be due to chance 3. It falls outside –1.5 SD of the mean but is likely to be due to chance 4. It falls outside 1.5 SD of the mean but is likely to be due to chance