Question

A student project involved collecting data to see if there was a difference in the amount of time one had to wait at the drive-thru between two fast food restaurants, A and B. She randomly selected 17 cars at fast food restaurant A and 17 cars at fast food restaurant B. For each car chosen, she recorded how much time passed from the placement of the order to receiving their food at the pick-up window. The data is given in the table below measured in Seconds. Use α=0.05.

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Entered Answer Preview Result incorrect 0.4919 0.4919 incorrect 0.4882 0.4882 incorrect not reject not reject correct At least one of the answers above is NOT correct. (1 point) A student project involved collecting data to see if there was a difference in the amount of time one had to wait at the drive-thru between two fast food restaurants, A and B. She randomly selected 17 cars at fast food restaurant A and 17 cars at fast food restaurant B. For each car chosen, she recorded how much time passed from the placement of the order to receiving their food at the pick-up window. The data is given in the table below measured in Seconds. Use a = 0.05. Fast Food A Fast Food B 201.5 150.9 177.6 181.2 214.7 182.6 228.4 138.2 187.6 206.2 214.9 206.3 203.7 242.8 181.9 195.9 191 225.4 189.4 219.2 215.9 242.6 235.9 176.6 165 213.3 202.1 202.9 128.6 193.3 189.1 184.7 207.1 232.9 Download .csv file (a) Choose the most appropriate statistical hypotheses. A. HO HA = HB HANA UB B. Ho: up = 0, Haup > 0 C. HOD=0, HAUD 70 OD. H :D = 0, HAUD <0 E. H, HA= B HALA< MB OF. HJA= UB HAHA > HB (b) Test the statistical hypotheses in (a) by carrying out the appropriate statistical test. Find the value of the test statistic for this test, use at least two decimals in your answer. Test Statistic = 0.4919 (c) Determine the P-value for this test, to at least three decimal places P = 0.4882 (d) Based on the above calculations, we should not reject the null hypothesis. Use a = 0.05.

DATA:

Fast Food A	Fast Food B
201.5	150.9
177.6	181.2
214.7	182.6
228.4	138.2
187.6	206.2
214.9	206.3
203.7	242.8
181.9	195.9
191	225.4
189.4	219.2
215.9	242.6
235.9	176.6
165	213.3
202.1	202.9
128.6	193.3
189.1	184.7
207.1	232.9

Answer 1

a) option A)

b)

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   196.141                  
standard deviation of sample 1,   s1 =    25.2762                  
size of sample 1,    n1=   17                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   199.706                  
standard deviation of sample 2,   s2 =    29.2984                  
size of sample 2,    n2=   17                  
                          
difference in sample means =    x̅1-x̅2 =    196.1412   -   199.7   =   -3.565  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    27.3613                  
std error , SE =    Sp*√(1/n1+1/n2) =    9.3849                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3.5647   -   0   ) /    9.38   =   -0.3798
                          
Degree of freedom, DF=   n1+n2-2 =    32                  
t-critical value , t* =        2.037   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho                      

c) p-value =        0.7066   (excel function: =T.DIST.2T(t stat,df) )              

d) not reject

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