Question

AaBbcc Aabolcul Adbull hou LLUD HUD 1 Normal 1 No Spac... Hending 1 Heading 2 Title Subtitle HUDS Subtle Em... O M Emphasis v Rc Replace a let Paragraph Styles Editing 4. (30 points) A fitness magazine claims that the mean cost of a Yoga session is no more than $ 14. You want to test this claim. You find that a random sample of 32 yoga sessions has a mean cost of $ 15.59 and a standard deviation of $ 2.60. At a = 0.025, do you have enough evidence to reject the magazine's claim? Assume that the population is normally distributed. a) Identify the claim and state the null hypothesis and the alternative hypothesis. b) Find the standardized test statistict A ENG

Answer 1

a)

Ho :   µ <=   14                  
Ha :   µ >   14       (Right tail test)          

b)

                          
Level of Significance ,    α =    0.03                  
sample std dev ,    s =    2.6000                  
Sample Size ,   n =    32                  
Sample Mean,    x̅ =   15.5900                  
                          
degree of freedom=   DF=n-1=   31                  
                          
Standard Error , SE = s/√n =   2.6000   / √    32   =   0.4596      
t-test statistic= (x̅ - µ )/SE = (   15.590   -   14   ) /    0.4596   =   3.46
                          
critical t value, t* =        2.0395   [Excel formula =t.inv(α/no. of tails,df) ]              
                          
p-Value   =   0.0008   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       
                      

Please revert in case of any doubt.

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