A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is 5 minutes. He decides to collect data to see if the mean amount of time that college students spend in the shower is significantly different from 5 minutes. In a sample of 15 students, he found the average time was 4.29 minutes and the standard deviation was 0.75 minutes. Using this sample information, conduct the appropriate hypothesis test at the 0.01 level of significance. Assume normality.

Step-by-step explanation:

n = 15

average = bar x = 4.29

null hypothesis

h0: μ=5

alternative hypothesis=

h1 = μ≠5

we find the test statistics

t stat =

= -0.71/0.1936

= -3.6673

degree of freedom = 15 - 1 = 14

we find p value using excel's t distribution function

= T.DIST(3.664,14 2) this is a 2 tailed test

p value = 0.0025

alpha = 0.01

because 0.0025 < 0.01

conclusion

so we reject the claim that the time = 5 minutes  since p value is smaller than alpha.

answer:

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

explanation:

service only with 20% tip  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

in the opening of the question the word 'service' is associated only with the pretax value.

the closing part of the question asks: how much did she pay for the  

service

−−−−−−  

so technically the answer is  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

however, the intention of the question provider may have been that your answer is to be  

$

87.12

(

1

+

8

100

)

=

$

94.09

to 2 decimal places.

'

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

tony b

mar 21, 2017

answer:

trying to explain it another way

explanation:

service charge  

.

→

$

72.60

20% tip  

20

100

×

$

72.60

=

$

14.52

−−−−−−  

←

add

$

87.12

tax  

8

100

×

$

87.12

=

$

6.97

−−−−−  

←

add

$

94.09

related questions

more detail

step-by-step explanation: