Question

The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 85.2 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts (a) and (b) below. 108.9 67.8 59.8 76.9 64.3 81.5 94.9 86.9 70.2 88.8 Click the icon to view the table of correlation coefficient critical values. (a) Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be r= 0.985. Are the conditions for testing the hypothesis satisfied? the conditions satisfied. The normal probability plot linear enough, since the correlation coefficient is than the critical value. A Expected z-score 90 105 Time (sec) (b) is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of a = 0.1. First determine the appropriate hypotheses. Ho: 785.2 H: 85.2 Find the test statistic. to- (Round to two decimal places as needed.) Find the P-value. Click to select your answer(s).

Answer 1

a.
Given that,
value of r =0.985
number (n)=10
null, Ho: ρ =0
alternate, H1: ρ!=0
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.86
since our test is two-tailed
reject Ho, if to < -1.86 OR if to > 1.86
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.985/(sqrt( ( 1-0.985^2 )/(10-2) )
to =16.15
|to | =16.15
critical value
the value of |t α| at los 0.1% is 1.86
we got |to| =16.15 & | t α | =1.86
make decision
hence value of | to | > | t α| and here we reject Ho
ANSWERS
---------------
null, Ho: ρ =0
alternate, H1: ρ!=0
test statistic: 16.15
critical value: -1.86 , 1.86
decision: reject Ho
we have enough evidence to support the claim that sample correlation is equal to 0.985
b.
Given that,
population mean(u)=85.2
sample mean, x =80
standard deviation, s =15.2517
number (n)=10
null, Ho: μ=85.2
alternate, H1: μ!=85.2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.833
since our test is two-tailed
reject Ho, if to < -1.833 OR if to > 1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =80-85.2/(15.2517/sqrt(10))
to =-1.0782
| to | =1.0782
critical value
the value of |t α| with n-1 = 9 d.f is 1.833
we got |to| =1.0782 & | t α | =1.833
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.0782 ) = 0.309
hence value of p0.1 < 0.309,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=85.2
alternate, H1: μ!=85.2
test statistic: -1.0782
critical value: -1.833 , 1.833
decision: do not reject Ho
p-value: 0.309
we do not have enough evidence to support the claim that the new system is effective.