A magazine assessed 2053 drive-thru visits at quick-service restaurants. One ben

Question

A magazine assessed 2053 drive-thru visits at quick-service restaurants. One benchmark that was measured was the service time. A summary of the results (in seconds) for two of the chains is shown below.

(a) Is there a difference in the average service time between these two chains? Test the null hypothesis that the chains' average service time is the same. Use a significance level of 0.05.

State the null and alternative hypotheses. (Let TB represent Taco Bell, and M represent McDonald's.)

___    H0: TB M     Ha: TB > M

___    H0: TB M      Ha: TB < M

___    H0: TB = M      Ha: TB M

___    H0: TB = M      Ha: TB < M

____ H0: TB = M      Ha: TB > M

Report the test statistic, the degrees of freedom, and the P-value. (Round your value for t to three decimal places, your degrees of freedom to the nearest whole number, and your P-value to four decimal places.)

df =

State your conclusion.

__   Reject the null hypothesis. There is significant evidence that the chains' average serice time is not the same.

__   Fail to reject the null hypothesis. There is significant evidence that the chains' average serice time is not the same.

__   Fail to reject the null hypothesis. There is not significant evidence that the chains' average serice time is not the same.

__   Reject the null hypothesis. There is not significant evidence that the chains' average serice time is not the same.

(b) Construct a 95% confidence interval for the difference in average service time. (Round your answers to two decimal places.) ,

( ________ , _________ ) sec

c) Lex plans to go to Taco Bell and Sam to McDonald's. Does the interval in part contain the difference in their service times that they're likely to encounter? Explain your answer.

___ Yes, the confidence interval contains the difference in their service times that they are likely to encounter.

___ No, the confidence interval is for the mean difference in time, not the difference for individuals.

Explanation / Answer

Given that,
mean(x)=149.69
standard deviation , s.d1=35.6
number(n1)=307
y(mean)=188.82
standard deviation, s.d2 =42.8
number(n2)=362
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.968
since our test is two-tailed
reject Ho, if to < -1.968 OR if to > 1.968
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =149.69-188.82/sqrt((1267.36/307)+(1831.84/362))
to =-12.909
| to | =12.909
critical value
the value of |t | with min (n1-1, n2-1) i.e 306 d.f is 1.968
we got |to| = 12.90882 & | t | = 1.968
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -12.9088 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -12.909
critical value: -1.968 , 1.968
decision: reject Ho
p-value: 0

b.

CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=149.69
Standard deviation( sd1 )=35.6
Sample Size(n1)=307
Mean(x2)=188.82
Standard deviation( sd2 )=42.8
Sample Size(n2)=362
CI = [ ( 149.69-188.82) ±t a/2 * Sqrt( 1267.36/307+1831.84/362)]
= [ (-39.13) ± t a/2 * Sqrt( 9.1885) ]
= [ (-39.13) ± 1.968 * Sqrt( 9.1885) ]
= [-45.0955 , -33.1645]

c.

No, the confidence interval is for the mean difference in time, not the difference for individuals.

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