Question

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20￠. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95￠ with a standard deviation of 18￠. Do the data support the claim at the 1% level? FILL IN THE BLANKS a. State null hypothesis in symbols: H0: µ = $1.00 b. State alternative hypothesis in symbols: HA: µ > .95c c. Let = The variable represents the newspaper d. Distribution: __________________________ e. z = ___________________ f. p-value = 0.357 g. Alpha = _______ h. Decision Reject or Do Not Reject the null hypothesis: _______________ i. Reason for decision: __________________________________________ j. Conclusion: ______________________________________________________________________________________ k. Confidence Interval: ____________________________

Answer 1

a)

Ho :   µ =   100 c

b)
Ha :   µ ╪   100 c

c)

variable is mean cost of a daily newspaper

d)

distribution = Z distribution

e)

Level of Significance ,    α =    0.01                  
population std dev ,    σ =    20
Sample Size ,   n =    12                  
Sample Mean,    x̅ =   95.0000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   20.0000   / √    12   =   5.7735      
Z-test statistic= (x̅ - µ )/SE = (   95.000   -   100   ) /    5.774   =   -0.8660

f)

p-Value   =   0.3865 [ Excel formula =2*NORMSDIST(z) ]

g)

α=0.01

h)

Do not reject null hypothesis   

i)

because p-value>α,

j)

Conclusion: There is not enough evidence to reject the claim at 1% level of significance

k)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   20.000   / √   12   =   5.7735
margin of error, E=Z*SE =   2.5758   *   5.774   =   14.872
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    95.00   -   14.872   =   80.1284
Interval Upper Limit = x̅ + E =    95.00   -   14.872   =   109.8716
99%   confidence interval is (   80.13   < µ <   109.87   )