3,053/2,8 Alguien me ayuda porfavor? Pero con el proceso porfa
1 answer:
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Answer: a) (0.755, 0.925)
Step-by-step explanation:
Let p be the population proportion of 8th graders are involved with some type of after school activity.
As per given , we have
n= 100
sample proportion:
Significance level :
Critical z-value : (using z-value table)
Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-
i.e.
i.e.
i.e.
Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)
Solution: