Question

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8​ ounces, the company wants the packages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected​ periodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 ​packages, the mean amount dispensed is 8.167 ​ounces, with a sample standard deviation of 0.054 ounce. Complete parts​ (a) and​ (b). a. Is there evidence that the population mean amount is different from 8.17 ​ounces? (Use a 0.10 level of​ significance.) State the null and alternative hypotheses.

Answer 1

Solution :

= 8.17

= 8.167

s= 0.054

n = 50

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   = 8.17

H_a :    8.17

Test statistic = t

= ( - ) / s / n

= (8.167-8.17) / 0.054 / 50

= -0.393

P (Z < -0.393) =0.6961

P-value = 0.6961

= 0.10  

p=0.6961 ≥ 0.10

Fail to reject the null hypothesis .

There is not enough evidence to claim that the population mean μ is different than 8.17, at the 0.10 significance level.