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can any body solve this question? Step 2: Consider the following study to compar

ID: 3151920 • Letter: C

Question

can any body solve this question?

Step 2:
Consider the following study to compare two popular energy drinks. Each drink was rated on a 0 to 100 scale, with 100 being the highest rating.



Is there a difference in preference? State appropriate hypotheses and carry out a matched pairs test for these data.

Determine whether the following conclusion is True or False:
The data gives evidence for the assumption that there is a difference in preference. True False step-3 Consider the following study to compare two popular energy drinks. Each drink was rated on a 0 to 100 scale, with 100 being the highest rating. For the companies producing these drinks, the real question is how much difference there is between the two preferences.


Use the data above to give a 95% confidence interval for the difference in preference between Drink A and Drink B. A 95% confidence interval for the difference in preference between the drinks is (______, _______) (Give your answers to 3 decimal places).




Answer 1
Answer 2

can any body solve this question?

Explanation / Answer

As significance level is not given, we assume it is 0.05.

Let ud = u2 - u1.              

The differences are

-1
12
17
8
22

Formulating the null and alternative hypotheses,              
              
Ho:   ud   =   0  
Ha:   ud   =/   0  

At level of significance =    0.05          

As we can see, this is a    two   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    7.986731421          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    3.571774875          
              
Calculating the mean of the differences (third column):              
              
XD =    11.6          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    3.247685088   [ANSWER, T STATISTIC]

***********************************************      
              
As df = n - 1 =    4          
              
Then the critical value of t is              
              
tcrit =    +/-   2.776445105      
              
              
Also, using p values,              
              
p =        0.031444475      
              
As |t| > 2.776, and P < 0.05, WE REJECT THE NULL HYPOTHESIS.          

Hence,

TRUE: the The data gives evidence for the assumption that there is a difference in preference. [ANSWER]

**********************************************

For the   0.95   confidence level,      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    2.776445105          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    1.683163131          
upper bound = [X1 - X2] + t(alpha/2) * sD =    21.51683687          
              
Thus, the confidence interval is              
              
(   1.683163131   ,   21.51683687   ) [ANSWER, CONFIDENCE INTERVAL]

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