Question
In: Statistics and Probability
24. Jessica wants to find the average lifespan of cats who are kept as indoor pets....
24. Jessica wants to find the average lifespan of cats who are kept as indoor pets. She take a random sample of 13 indoor cats and find the average lifespan is 14.1 years with a standard deviation of 0.75. Assume the population distribution is approximately normal. Jessica wants to create a 96% confidence interval for the true average lifespan of all indoor cats. What is the 96% confidence interval?
| a. |
(13.65, 14.55) |
|
| b. |
(12.72, 13.68) |
|
| c. |
(13.62, 14.58) |
|
| d. |
(13.60, 14.60) |
|
| e. |
(13.67, 14.53) |
Solutions
Expert Solution
Solution :
Given that,
= 14.1
s =0.75
n =13
Degrees of freedom = df = n - 1 =13 - 1 = 12
At 96% confidence level the t is ,
t
/2,df = t0.02,12 = 2.303 ( using student t
table)
Margin of error = E = t/2,df
* (s /
= 2.303 * (0.75 /
= 0.48
The 96% confidence interval estimate of the population mean is,
- E < 
<
+ E
14.1 - 0.48<
<14.1+ 0.48
13.62 <
< 14.58
( 13.62 , 14.58)
orchestra answered 2 years agoRelated Solutions
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38
40
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36
33
36
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38.20
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