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.

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

Step-by-step 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.

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

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step-by-step explanation: