Question

A random sample of the actual weight of 5-lb bags of mulch produces a mean of 4.963 lb and a standard deviation of 0.067 lb. If n=50, which of the following will give a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company?
A) 4.963±0.016.
B) 4.963±0.019.
C) 4.963±0.067.
D) 4.963±0.009.
E) None of the above.

Answer 1

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 :-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = Sample  mean

n= Sample size

s= sample standard deviation

t* = critical t-value.

As per given:

n= 50

Degree of freedom = n-1 =49

$\overline{x}= 4.963\ lb$

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

$4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019$

Hence,  a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is $4.963\pm0.019$.

Thus , the correct answer is B) 4.963±0.019.