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 87.6 seconds. A manager devises a new​ drive-through system that he believes will decrease wait time. As a​ test, he initiates the new system at his 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.	107.9	82.3
	69.7	95.3
	59.1	85.9
	73.5	72.7
	64.7	88.9

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

requals=0.988

Are the conditions for testing the hypothesis​ satisfied?

▼ Yes, No, the conditions ▼ are not are satisfied. The normal probability plot ▼ is is not linear​ enough, since the correlation coefficient is ▼ less great than the critical value.

607590105-2-1012Time (sec)Expected z-score A normal probability plot has a horizontal axis labeled Time (seconds) from 50 to 115 in increments of 5 and a vertical axis labeled Expected z-score from negative 2 to 2 in increments of 0.5. Ten plotted points closely follow the pattern of a line that rises from left to right through (59, negative 1.55) and (95.5, 1). All coordinates are approximate.

​(b) Is the new system​ effective? Conduct a hypothesis test using the​ P-value approach and a level of significance of

alpha equals 0.1α=0.1.

First, determine the appropriate hypotheses.

Upper H 0H0​:

▼

sigmaσ

pp

muμ

▼

less than<

equals=

greater than>

not equals≠

87.6

Upper H 1H1​:

▼

muμ

pp

sigmaσ

▼

not equals≠

greater than>

equals=

less than<

87.6

Find the test statistic.

t 0t0equals=nothing

​(Round to two decimal places as​ needed.)

Find the​ P-value.

The​ P-value is

nothing.

​(Round to three decimal places as​ needed.)

Use the

alpha equals 0.1α=0.1

level of significance. What can be concluded from the hypothesis​ test?

A.The​ P-value is

lessless

than the level of significance so there

isnbsp sufficient

evidence to conclude the new system is effective.

B.The​ P-value is

greatergreater

than the level of significance so there

isnbsp sufficient

evidence to conclude the new system is effective.

C.The​ P-value is

greatergreater

than the level of significance so there

isnbsp not nbsp not sufficient

evidence to conclude the new system is effective.

D.The​ P-value is

lessless

than the level of significance so there

isnbsp not nbsp not sufficient

evidence to conclude the new system is effective.

Answer 1

107.9
69.7
59.1
73.5
64.7
82.3
95.3
85.9
72.7
88.9
80	mean
14.92329	sd
10	n
87.6	mu
-1.61046	TS
0.070879	p-value

formulas

107.9
69.7
59.1
73.5
64.7
82.3
95.3
85.9
72.7
88.9
=AVERAGE(A1:A10)	mean
=STDEV(A1:A10)	sd
10	n
87.6	mu
=(A12-A16)/(A13/SQRT(A14))	TS
=T.DIST(A17,A14-1,1)	p-value

yes
conditions are satisfied
the normal probability plot is linear enough
as correlation coefficient is greater than the critical value

b)
Ho: mu = 87.6
Ha: mu < 87.6

TS = -1.61
p-value = 0.071

p-value < alpha
hence there is sufficient evidence

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