Question 8MO22. A hypothesis will be used to test that a population mean equals 10 against the alternative that the population mean is lessthan 10 with known variance a. What is the critical value for the test statistic Z, for the significance level of 0.050? Round youranswer to two decimal places i.e., 0.00. Provide all the necessary solutions.Zo = Blank 1Blank 1Addyour answer

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Answered: MO22. A hypothesis will be used to test…

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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You wish to test the following claim (Ha) at a significance level of a=0.02. H0: u=83.9 Ha:u<83.9 You believe the population is normally distributed but you do not know the standard deviation. You obtain a sample of size n=119 with the meanM=79.4 and a standard deviation of SD=12.9. What is the test statistic foe this sample? What is the P value for this sample? The p value is less than or greater than a? This test statistic leads to a decision to: reject the null, accept the null, or fail to reject the null?
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Currently patrons at the library speak at an average of 69 decibels. Will this average decline after the installation of a new computer plug in station? After the plug in station was built, the librarian randomly recorded 41 people speaking at the library. Their average decibel level was 67 and their standard deviation was 5. What can be concluded at the the a = 0.01 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ? v| Select an answer v H1: ? v Select an answer v c. The test statistic ? v = (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ? v a f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is that ... O The data suggest the populaton mean has significantly decreased from 69 at a = 0.01, so there is statistically significant evidence to conclude that the population mean decibel…
Aspirin is packed in 120 tab bottles with weight normally distributed with mean 600g and standard deviation 6g. Rejected bottles are those whose weight differs from the mean weight by 15g or more. That is, rejected bottles have weight either too small or too large. We want to find out how many bottles will be rejected from a daily production of 5000 bottles. 1,Do we have to calculate one z-score or two z-scores? 2. The negative z-score is 3. The positive z-score is 4.How many bottles, of the 5000, do we anticipate to have to reject?
A new material will replace the old material only if there is significant evidence that the mean strength of the new material is more than 2 units higher than the mean strength of the old material. Sample from new material: mean is 98.064 , SD is 5.6 and sample size is 13 Sample from old material: mean is 95.282 , SD is 5.8 and sample size is 78 We do not know population standard deviations. Consider the test ?0:????−????=2H0:μnew−μold=2 vs. ?0:????−????>2H0:μnew−μold>2 What is the p-value? (Hint: best to use software, or else the Desmos Calculator for the 2 sample T and Z test) Give as a decimal between 0 and 1 accurate to at least 4 decimal places.
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…
You are conducting a study to see if the accuracy rate for fingerprint identification is significantly different from 0.51. Thus you are performing a two-tailed test. Your sample data produce the test statistic - 2.153. Find the p-value accurate to Z = - 4 decimal places. p-value
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 14 samples is 4.9 ppm with a variance of 1.2 Does the data support the claim at the 0.01 level? Assume the population distribution is approximately normal. Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.
"A fitness coach woud like to determine whether her aerobic fitness intervention leads to a significant increase or decrease in the one rep max (i.e., heaviest possible weight that can be lifted once) for deadlifts in her group of athletes (n = 13). She finds that the mean difference = 3 lbs, with a sum of squared deviation (SS) = 84. Calculate the t-statistic the fitness coach would observe given this data. (Be sure to round your answer to the second decimal place)."
Solve it on paper! Please use t tests on this question. Do not use z tests!
Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level). One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p Between-group (treatment) 2 469.1827 2677.864 2.30 0.104 Within-group (error) 118 203.6918 11745.05 Total 120

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