In: Statistics and Probability
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) |
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 /n)
= 2.303 * (0.75 / 13)
= 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)