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This is what I currently have but am stuck because there is no .0125 and .9875 o

ID: 2921324 • Letter: T

Question

This is what I currently have but am stuck because there is no .0125 and .9875 on the chi squared table. is it a 1 tailed test? and do I have the null and alternative hypothesis set up correctly?

A manager would like to know if the standard deviation in service times is greater than 4 minutes. A random sample of 14 service times is taken with the sampled times shown below. Using a level of significance of 0.025, can the manager conclude that the standard deviation of service times is greater than 4 minutes?

Sample #

Service Time (mins)

1

25.85

2

36.77

3

28.47

4

30.41

5

31.64

6

37.44

7

31.84

8

34.61

9

35.53

10

29.85

11

30.66

12

24.81

13

24.34

14

27.68

Sample #

Service Time (mins)

1

25.85

2

36.77

3

28.47

4

30.41

5

31.64

6

37.44

7

31.84

8

34.61

9

35.53

10

29.85

11

30.66

12

24.81

13

24.34

14

27.68

24 a H.Roy A''- 4.2bb She A T74 n=14 d SOLS df=13 do not 025 , OIZG Izs So 130 F @ SF

Explanation / Answer

Given that,
population standard deviation ()=4
sample standard deviation (s) =4.2665
sample size (n) = 14
we calculate,
population variance (^2) =16
sample variance (s^2)=18.20302225
null, Ho: ^2=16
alternate, H1 : ^2 >16
level of significance, = 0.025
from standard normal table,right tailed ^2 /2 =24.736
since our test is right-tailed
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(14 - 1 ) * 18.20302225 / 16 = 13*18.20302225/16 = 14.79
| ^2 cal | =14.79
critical value
the value of |^2 | at los 0.025 with d.f (n-1)=13 is 24.736
we got | ^2| =14.79 & | ^2 | =24.736
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.3206
ANSWERS
---------------
null, Ho: ^2 =16
alternate, H1 : ^2 >16
test statistic: 14.79
critical value: 24.736
p-value:0.3206
decision: do not reject Ho

manager don't have support to conclude that the standard deviation of service times is greater than 4 minutes

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