Question

A machine that is programmed to package 2.40 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 29 cereal boxes, the mean and standard deviation are calculated as 2.44 pounds and 0.17 pound, respectively. Use Table 2.

a.

Select the null and the alternative hypotheses to determine if the machine is working improperly, that is, it is either underfilling or overfilling the cereal boxes.

	H0: µ ≥ 2.40; HA: µ < 2.40
	H0: µ ≤ 2.40; HA: µ > 2.40
	H0: µ = 2.40; HA: µ ≠ 2.40

b.	Calculate the value of the test statistic. (Round your answer to 2 decimal places.)

Test statistic

c-1.

Approximate the p-value.

	0.20< p-value < 0.40
	0.01 < p-value < 0.02
	0.02 < p-value < 0.05
	p-value < 0.01
	p-value  0.4

c-2.

What is the conclusion at the 1% significance level?

	Reject H0 since the p-value is greater than α.
	Reject H0 since the p-value is smaller than α.
	Do not reject H0 since the p-value is greater than α.
	Do not reject H0 since the p-value is smaller than α.

d-1.	Calculate the critical value(s) at a 1% level of significance. (Round your answer to 3 decimal places.)

Critical value(s)

±

d-2.

Can you conclude that the machine is working improperly?

	Yes
	No

Answer 1

Solution:

a)

H₀: µ = 2.40; H_A: µ ≠ 2.40

b)

The test statistic t is

t = ( - )/[/n] = [2.44 - 2.40]/[0.17 /29] = 1.27

Test statistic : 1.27

c-1)

d.f. = n - 1 = 29 - 1 = 28

Two tailed test

t = 1.27

Using t table ,

0.20< p-value < 0.40

c -2)

Do not reject H₀ since the p-value is greater than α.

d -1)

= 1% =0.01

/2 = 0.005

Two tailed test

So , critical values are

Critical value(s) :  2.763

d -2)

No