$500 dollars is the answer to this problem
It’s a little complicated but here’s how it works:
Imagine a table with the intervals
0:4 , 4:6 , 6:7 , 7:10 , 10:13 (10 year intervals)
Then we have different rows
Class width: 4 , 2 , 1 , 3 , 3
Freq density: 0.2 , 0.5 , 1.2 , 0.7 , 0.3
So now calculate frequency where freq = class width * density
Freq: 0.8 , 1 , 3.6 , 2.1 , 0.9
So to find median find cumulative frequency
(Add all freq)
Cfreq = 8.4 now divide by 2 = 4.2
So find the interval where 4.2 lies.
0.8 + 1 = 1.8 + 3.6 = 5.6
So 4.2 (median) will lie in that interval 60-70 years.
<span>It is clear to see that 0.0259 is less than 0.05. A p-value that is less than the confidence level of alpha indicates that there is sufficient evidence from the data that the null hypothesis should be rejected in favor of an alternative hypothesis.</span>
