Question

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449.0 gram setting. It is believed that the machine is underfilling the bags. A 37 bag sample had a mean of 441.0 grams. A level of significance of 0.05 will be used. Determine the decision rule. Assume the standard deviation is known to be 18.0.

Answer 1

the decision rule is to reject H0 if Z is < -1.645

Explanation:

this question is interested in the decision rule.

H0: u = 449

H1 : u <449

We find the test statistic

Mean bar x = 441

u = 449

Sd = 18

n = 37

Z = barX - u / sd/√n

= 441-449/18/√37

= -8/18/6.083

= -8/2.959

Z = - 2.7036

This is the value of the test statistic

The critical value at 0.05 = -1.645. this is a left tailed test

The decision rule is to reject H0 if Z < -1.645

The test statistic is less than the critical value

-2.7036 < -1.645

So we have enough evidence to reject H0. At 0.05 level we have enough evidence to support claim.