Question

Given the following hypothesis:

H₀: μ = 100

H₁: μ ≠ 100

A random sample of six resulted in the following values:

114

120

119

108

113

108

Using the 0.01 significance level, can we conclude that the mean is different from 100?

a. What is the decision rule? (Negative answer should be indicated by a minus sign. Round the final answers to 3 decimal places.)

Reject H₀: μ = 100 and accept H₁: μ ≠ 100 when the test statistic is  (Click to select)  inside the interval  outside the interval  (  ,  ).

b. Compute the value of the test statistic. (Round the final answer to 2 decimal places.)

Value of the test statistic            

c. What is your decision regarding H₀?

(Click to select)  Reject  Do not reject  H₀ .

d. Estimate the p-value.

(Click to select)  Zero  Close to zero  One

e-1. Construct a 99% confidence interval.

Confidence interval is from             to  .

e-2. Use the results of the confidence interval to support your decision in part (c).

(Click to select)  Reject  Do not Reject  H₀ .

Answer 1

a)

α=0.01

degree of freedom=   DF=n-1=   5
critical t value, t* =    ±   4.032   [Excel formula =t.inv(α/no. of tails,df) ]

reject Ho when t >4.032 ot t < -4.032

Reject H₀: μ = 100 and accept H₁: μ ≠ 100 when the test statistic is  outside the interval  ( -4.032 ,4.032 )

b)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   5.1640                  
Sample Size ,   n =    6                  
Sample Mean,    x̅ = ΣX/n =    113.6667                  
                          
degree of freedom=   DF=n-1=   5                  
                          
Standard Error , SE = s/√n =   5.1640   / √    6   =   2.1082      
t-test statistic= (x̅ - µ )/SE = (   113.667   -   100   ) /    2.1082   =   6.48

c)

reject Ho

d)

p-Value   =   0.001

close to zero

e-1)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   5          
't value='   tα/2=   4.032   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.1640   / √   6   =   2.1082
margin of error , E=t*SE =   4.0321   *   2.1082   =   8.5005
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    113.67   -   8.500504   =   105.1662
Interval Upper Limit = x̅ + E =    113.67   -   8.500504   =   122.1672
99%   confidence interval is (   105.1662   < µ <   122.1672   )

e-2)

reject Ho, because 100 is outside the interval