In: Statistics and Probability
The annual incomes of a random sample of 10 individuals in a certain city are given below. Construct at 95% confidence interval for the average yearly income of individuals in this city.
65100, 52400, 32000, 33300, 72200, 39700, 39400, 42800, 54300, 68200,
the necessary calculation table :-
income (x) | x2 |
65100 | 4238010000 |
52400 | 2745760000 |
32000 | 1024000000 |
33300 | 1108890000 |
72200 | 5212840000 |
39700 | 1576090000 |
39400 | 1552360000 |
42800 | 1831840000 |
54300 | 2948490000 |
68200 | 4651240000 |
sum=499400 | sum=26889520000 |
sample size (n) = 10
[ for these values you can directly use any software or excel or your calculator ..i am just showing all calculations for your clarification]
degrees of freedom = (n-1) = (10-1) = 9
t critical value for alpha=0.05,df = 9, both tailed test be:-
[ using t distribution table ]
the 95% confidence interval for the average yearly income of individuals in this city is :-
* as nothing is mentioned i have rounded off the final answer to nearest whole number .
interpretation:-
we are 95% confident that the average yearly income of individuals in this city will be between 39412 and 60468.
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