Question

A supermarket wants to estimate the population mean weight of watermelons of a certain type. A sample size n = 45 has sample mean = 20 pounds with sample standard deviation s = 2.2 pounds. Create a confidence interval at 90% confidence for the population mean μ.

Answer 1

Sol:

n=45

df=n-1=45-1=44

alpha=1-0.90=0.10

alpha/2=0.10/2=0.05

sample mean=xbar=20

sample standard deviation=s=2.2

t crit for 0.05 level of  significance and 44 df is

=T.INV(0.05;44)

=1.6802

90% confidence for the population mean μ. is

xbar-tc*s/sqrt(n),xbar+tc*s/sqrt(n)

20-(1.6802*2.2/sqrt(45)),20+(1.6802*2.2/sqrt(45))

19.44897,20.55103

lower limit= 19.44897

upper limit=20.55103

we are 90% confident that the true  population mean weight of watermelons lies in between 19.45 pounds and 20.55 pounds.