Is this correct?
1 answer:
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Answer: B) 4.963±0.019.
Step-by-step explanation:
Confidence interval for population mean ( when population standard deviation is not given) is given by :-
, where = Sample mean
n= Sample size
s= sample standard deviation
t* = critical t-value.
As per given:
n= 50
Degree of freedom = n-1 =49
s= 0.067 lb
For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.
Now , substitute all values in the formula , we get
Hence, a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is .
Thus , the correct answer is B) 4.963±0.019.