Question

Q: An agronomist wanted to estimate the mean weight of a certain fruit (?) in his farm. He selected a random sample of size ? = 16, and found that the sample mean ?̅ = 55 gm and a sample stranded deviation ? = 3 gm. It is assumed that the population is normal with unknown standard deviation (?).

a) Find a point estimate for (?).

b) Find a 95% confidence interval for the population mean (?).

Answer 1

Solution :

Given that,

a. Point estimate = sample mean = = 55

sample standard deviation = s = 3

sample size = n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

b. At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = 2.131

Margin of error = E = t/2,df * (s /n)

= 2.131 * ( 3 / 16 )

Margin of error = E = 1.60

The 99% confidence interval estimate of the population mean is,

- E < < + E

55 - 1.60 < < 55 + 1.60

53.4 < < 56.6