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 :-
[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
n= Sample size
s= sample standard deviation
t* = critical t-value.
As per given:
n= 50
Degree of freedom = n-1 =49
[tex]\overline{x}= 4.963\ lb[/tex]
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
[tex]4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019[/tex]
Hence, a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is [tex]4.963\pm0.019[/tex].
Thus , the correct answer is B) 4.963±0.019.
