Question

An operator fill sacks. Each sack is to have 60 lbs. of sand in it. A random sample of 28 sacks was found to have the following weight statistics: sample mean = 60.71 lbs, sample standard deviation = 3.76 lbs at α=0.05, can we conclude that actual (population's) mean filled sack weight is > 60 lbs? What is the critical value of the sample mean to compare to the actual sample mean to make this conclusion? The answer may be negative to 4 decimal places.

Please show work.

Answer 1

Solution :

= 60

=60.71

S =3.76

n = 28

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :    = 60

Ha : > 60

Test statistic = t

= ( - ) / S / n

= (60.71 - 60) / 3.76 / 28

= 0.999

Test statistic = t = 0.999

The critical value =1.703

P-value =0.1633

= 0.05  

P-value >

0.1633 > 0.05

Fail to reject the null hypothesis .

There is insufficient evidence to sconclude that the population mean μ is greater than 60, at the 0.05 significance level