8- A research wishes to see if the mean of days that a basic,low-price,small automobile sits on adealers lot is 29.A sample of 30 automobile dealers has a mean of 30.1 days for basic,low-price,small automobiles. At Alpha 0.05, test the claim that the mean time is greater than 29 days. Thestandard deviation of the population is 3.8 days. Traditional Method?(a)Identify the null and alternative hypotheses for this test.(a) What is the C.V number?(b) Identify the test statistic for this hypothesis test(c) Decision. Accept or reject. Use bell curve

Answered: 8- A research wishes to see if the mean…

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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The mean age when smokers first start is 13 years old with a population standard deviation of 1.9 years. A researcher thinks that smoking age has significantly changed since the invention of ENDS-electronic nicotine delivery systems. A survey of smokers of this generation was done to see if the mean age has changed. The sample of 31 smokers found that their mean starting age was 12 years old. Do the data support the claim at the 1% significance level? What are the correct hypotheses? Ho: LU Test Statistic z = #v Critical Values =+ O OF 13 H₁: u Based on the hypotheses, find the following: 13 o years (Round to 4 decimal places.) (Round to 2 decimal places.) Based on the above we choose to Reject the null hypothesis years The correct summary would be: There is enough evidence to support the claim smokers first start is different than 13. that the mean age
Does the average Presbyterian donate less than the average Catholic in church on Sundays? The 58 randomly observed members of the Presbyterian church donated an average of $24 with a standard deviation of $15. The 42 randomly observed members of the Catholic church donated an average of $31 with a standard deviation of $7. What can be concluded at the αα = 0.10 level of significance? For this study, we should use Select an answer t-test for the difference between two dependent population means z-test for a population proportion t-test for a population mean t-test for the difference between two independent population means z-test for the difference between two population proportions The null and alternative hypotheses would be: H0:H0: Select an answer p1 μ1 Select an answer ≠ > = < Select an answer p2 μ2 (please enter a decimal) H1:H1: Select an answer p1 μ1 Select an answer < = ≠ > Select an answer p2 μ2 (Please enter a decimal) The test statistic ?…
A particular brand of cereal advertises it packages weigh 510 g.You wish to determine if on the average the weight is less than 510g. A simple randomsample of 81 packages from a large lot are weighed. The average weight is found to be509.5 g with a standard deviation of 0.4 g. 1. Perform a test at a level of significance 0.05. 2. What is the power of the test if you wish to detect an average departure of 0.2 gunderweight (can be stated as weighing 509.8g). 3. What sample size is needed to have a power of 90%? Please show all steps. Thank you!
(b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. • The value of the test statistic is given by √n The p-value is the area under the curve to the right of the value of the test statistic. O One-tailed o Two-tailed Step 2: Enter the test statistic. (Round to 3 decimal places.) Step 3: Shade the area represented by the p- value. Step 4: Enter the p- value. (Round to 3 decimal places.) 0.3 0.2 0.1 (c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the studio. o Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90. Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90.…
A basketball analyst recently claimed that the average number of passes per possession in basketball today is less than 5. A random sample of 40 possessions had a mean number of passes of 4.1. The standard deviation of all possessions is known to be 2.4. Test the claim with 0.01 level of significance.

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