In: Math
In a random sample of 24 fifth graders who took an IQ test, the average score was 101.48 with a standard deviation of 13.34. Assuming that the IQ scores are normally distributed, what will be the 98% confidence interval for the average IQ scores for all fifth graders?
Select the best answer.
96.3054 to 106.6546
95.5521 to 107.4079
96.8131 to 106.1469
94.6728 to 108.2872
Solution :
Given that,
= 101.48
s =13.34
n = 24
Degrees of freedom = df = n - 1 = 24- 1 = 23
At 98% confidence level the t is ,
= 1 - 98% = 1 - 0.98 = 0.02
/
2 = 0.02 / 2 = 0.01
t
/2,df = t0.01,23 = 2.4999
Margin of error = E = t/2,df
* (s /
n)
= 2.4999 * ( 13.34/
24)
=6.8072
The 98% confidence interval for the average IQ scores for all fifth graders
- E <
<
+ E
101.48 - 6.8072 <
< + 101.48+6.8072
94.6728 <
< 108.2872
94.6728 to 108.2872
CORRECT OPTION IS LAST
94.6728 to 108.2872