Question
A teacher is interested in the average family income of students at his university. He randomly...
A teacher is interested in the average family income of students at his university. He randomly selected 17 students and obtained their family incomes. The distribution of family incomes is normal. The average family income of the 17 students was $110,525. Assume the population standard deviation is $15,580.
a) Find a 95% confidence interval for the average family income of students at her university.
b) Interpret your interval in context.
Homework Answers
Answer #1
Given :- 17 student are selected and
There average family income = $110525
Standard deviation =$15580
Solution:-
A)
Since significant value of Z at 95% confidence coefficient is 1.96
Therefore, 95% confidence interval is given as
=(103118.74 , 117931.262)
B)
therefore , 95% confidence interval for the average family income of students in university is lie between 103118.74 to 117931.26.
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