Question

Suppose that students own an average of 4 pairs of jeans. 8 people from your class were surveyed to determine if the average for students is higher than 4.

DATA TO USE:   2, 2, 3, 4, 6, 6, 8, 9

a. Give the null and alternative hypotheses: Ho: _______________ Ha: ___________________

b. In words, CLEARLY state what your random variable or P' represents.

c. State the distribution to use for the test. If t, include the degrees of freedom. If normal, include the mean and standard deviation.

d. p-value = ______________   

e. In 1 – 2 complete sentences, explain what the p-value means for this problem.

f. Use the previous information to draw a graph of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value. The values of your sample statistic and the hypothesized value of the population parameter should be on the axis.

g. Indicate the correct decision (“reject the null hypothesis” or “do not reject the null hypothesis”) and write an appropriate conclusion, using COMPLETE SENTENCES.

   Decision:

   Conclusion:

h. Construct a 95% Confidence Interval for the true mean or proportion.   Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.

Confidence Interval: ( ___________________ , ___________________ )

i. Interpret the confidence interval in a complete sentence.

Answer 1

a)

Ho :   µ =   4      
Ha :   µ >   4       (Right tail test)

b)

random variable is number of pair of jeans

c)

t distribution

degree of freedom=   DF=n-1=   7

d)

Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.6726                  
Sample Size ,   n =    8                  
Sample Mean,    x̅ = ΣX/n =    5.0000                  
degree of freedom=   DF=n-1=   7                  
                          
Standard Error , SE = s/√n =   2.6726   / √    8   =   0.9449      
t-test statistic= (x̅ - µ )/SE = (   5.000   -   4   ) /    0.9449   =   1.06
                                
p-Value   =   0.1625   [Excel formula =t.dist(t-stat,df) ]              

e)

p value is the probability of accepting the null hypothesis true.

so, there is 0.1625 probability that Ho is true

.........

Decision:   p-value>α, Do not reject null hypothesis       
Conclusion: There is not enough evidence to say that the average for students is higher than 4.

...........

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   7          
't value='   tα/2=   2.3646   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.6726   / √   8   =   0.944911
margin of error , E=t*SE =   2.3646   *   0.94491   =   2.234360
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    5.00   -   2.234360   =   2.765640
Interval Upper Limit = x̅ + E =    5.00   -   2.234360   =   7.234360
95%   confidence interval is (   2.77   < µ <   7.23   )
  

the true mean for average number of jeans lie within range of (2.77 , 7.23)

thanks

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