For this problem, the confidence interval is the one we are looking
for. Since the confidence level is not given, we assume that it is 95%.
The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n
Where:
<span>
</span>
α= 5%
α/2
= 2.5%
t
0.025, 19 = 2.093 (check t table)
n
= 20
df
= n – 1 = 20 – 1 = 19
So plugging in our values:
8.41 ± 2.093 * 0.77 √ 1 + 1/20
= 8.41 ± 2.093 * 0.77 (1.0247)
= 8.41 ± 2.093 * 0.789019
= 8.41 ± 1.65141676
<span>= 6.7586 < x < 10.0614</span>
Answer:
55.5
Step-by-step explanation:
find the median - in this case it's 60
then group the upper and lower halves of the data:
(11, 19, 35, 42) 60 (72, 80, 85, 88)
now we find the medians of those sets:
35 + 19/2 = 54/2 =27
80 + 85/2 = 165/2 = 82.5
then subtract those two medians to find the interquartile:
82.5 - 27 = 55.5
= 19 this is the answer I'm guessing since 16-12= 4 so 16 is 4 more than 12
