Question

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 429.0 gram setting. It is believed that the machine is underfilling the bags. A 4040 bag sample had a mean of 425.0 grams. A level of significance of 0.02 will be used. Determine the decision rule. Assume the standard deviation is known to be 11.0.

Enter the decision rule.

Answer 1

Solution :

Given that,

This is a two tailed test.

The null and alternative hypothesis is,

Ho: 429

Ha: 429

The test statistics,

Z =( - )/ (/n)

= ( 425 - 429 ) / ( 11 / 40)

= -2.30

Critical value of  the significance level is α = 0.02, and the critical value for a two-tailed test is

= +2.33 , - 2.33

Enter the decision rule.

Reject H0. if z > 2.33

P-value = 2 * P(Z < -2.30 )

= 2 * 0.01073

= 0.0214

Since it is observed that ∣z∣=2.30 <  zc=2.33 it is then concluded that the null hypothesis is fail to rejected.