Question

1.   What dessert is the most popular?  A grocery chain gave people the choice of a free sample of chocolate brownies and chocolate cake.  They were told they were only allowed to pick one of the two samples and their choices were recorded.  The statisticians ensured that the samples were plentiful to prevent people from feeling bad for taking all of one sample or the other.  Over the course of 325 customers, 200 chose the brownies.   Does this suggest that people prefer brownies over cake?

a)  (2 points) Can we perform a statistical study based on this sample?

b) (2 points) Create a Hypothesis Test for the importance of brownies over cake.

c) (2 points) Estimate the standard error for this study.  Round to three decimal places.

d) (2 points) Find the test statistic for your study.  Round to three decimal places.

e) (2 points) Find the p-value.

f) (2 points) Do we have enough evidence that people prefer brownies over cake?

2.  A local university wants to suggest its students have all scored over 500 on their SAT Math scores each year.  Gathering 60 random students from the past 10 years, the average score per year is 540 with a standard deviation of 90.

a) (2 points) Is this sample appropriate for a normalized confidence interval?

b) (1 points) Find the standard error for this sample.

c) (1 points) Find the appropriate t-distribution for the 95% confidence interval.

d) (2 points) Find the confidence interval based on the sample.

e) (2 points)  Based on your confidence interval, is the university allowed to boast about its high SAT math scores?  Why or why not?

Answer 1

1)

A)

YES, we perform a statistical study based on this sample

.

B)

Ho :   p =    0.5
H1 :   p >   0.5
.

C)

Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   200                  
Sample Size,   n =    325                  
                          
Sample Proportion ,    p̂ = x/n =    0.6154                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.028

..........

D)
Z Test Statistic = ( p̂-p)/SE = (   0.6154   -   0.5   ) /   0.0277   =   4.160
                          
3)   
   
p-Value   =   0.0000   [Excel function =NORMSDIST(-z)     

F)
Decision:   p-value<α , reject null hypothesis

we have enough evidence that people prefer brownies over cake

............................

2)

A)

YES,this sample appropriate for a normalized confidence interval

....

B)

sample std dev ,    s =    90.0000
Sample Size ,   n =    60
Sample Mean,    x̅ =   540.0000
Standard Error , SE = s/√n =   90/√60=   11.619

............

C)

Level of Significance ,    α =    0.05              
degree of freedom=   DF=n-1=   59              
't value='   tα/2=   2.001   [Excel formula =t.inv(α/2,df) ]          
..

D)

margin of error , E=t*SE =   2.0010   *   11.6190   =   23.2495
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    540.00   -   23.2495   =   516.7505
Interval Upper Limit = x̅ + E =    540.00   -   23.2495   =   563.2495
95%   confidence interval is (   516.75   < µ <   563.25   )

...............

E)

500 DO NOT LIE IN THE INTERVAL, SO students have all scored over 500

................

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