WileyPLUS C Secure | https://edugen.vvileyplus.com/ed WileyPLUS an, Introductory

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WileyPLUS C Secure | https://edugen.vvileyplus.com/ed WileyPLUS an, Introductory statistics, 9e INTRO STATS. (MQM 100) Home Read, Study & Practice Gradebook ORION Downloadable eTextbook Assionment Open Assignment BACK LLProblemoaden 11 Chapter 09, Section 9.3, Problem 047 A past study claimed that adults in America spent an average of 18 hours a week on leisure activities. A researcher wanted to test this claimm. She took a sample of 10 adults and asked them about the time they spend per weak on leisure activities. Their responses (in hours) are as follows 12.0 13.6 20.8 21.1 20.9 25.9 15 15.3 20.9 20.6 it es has changed? First Ass me that the mes spent on le sure 8cti ties by e 8du 5 are no mal y distributed. Us, g the 5% 5 g cance level. can you conch de that the averege amount of time spent by American adults on leisure ec cakculste the sample mean and the sample standerd deviation for these dats. Then make the test of hypothesis about u.) t Round the sample standard deviation to three decimal places. The claim i Question Attemptsi 0 of 2 used SAVE FOR LATER l by Reviow Result Verzion 4.2433 4:45 PM O Type here to search

Explanation / Answer

Solution:- given that values: 12,13.6,20.8,21.1,20.8,35.9,15,15.3,20.9,20.6

X-bar = sum f terms /no of terms = 196/10 = 19.6

S = 6.724

hypothesis test : H0 : mu = 18
Ha : mu not = 18

This is two tailed test.
X - bar = 19.6
s = 6.724
n = 10
df = n-1 = 10 - 1 = 9   
significance level = 0.05
  
t = (x - mu)/(s/sqrt(n))
= (19.6 - 18)/(6.724/(10))
= 2.3795
  
critical value = +/- 2/626
  
The claim is false
  

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