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A machine that is programmed to package 1.90 pounds of cereal is being tested for its accuracy. In a sample of 26 ce...

A machine that is programmed to package 1.90 pounds of cereal is being tested for its accuracy. In a sample of 26 cereal boxes, the sample weight and standard deviation are calculated as 2.02 pounds and 0.24 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 ≥ 1.9; HA: µ < 1.9

H0: µ = 1.9; HA: µ ≠ 1.9

H0: µ ≤ 1.9; HA: µ > 1.9

b.

Calculate the value of the test statistic. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Test statistic   
c-1. Approximate the p-value.

0.02 < p-value < 0.05

0.05 < p-value < 0.1

0.01 < p-value < 0.02

c-2. At the 10% significance level, what is the conclusion?

Reject H0 since the p-value is greater than α.

Do not 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 smaller than α.

d-1.

Calculate the critical value(s) at a 10% 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

0 0
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Answer #1

a)

H0: µ = 1.9; HA: µ ≠ 1.9

b)

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (2.02 - 1.9)/(0.24/sqrt(26))
t = 2.55


c)

P-value = 0.0173


c2)

Reject H0 since the p-value is smaller than α.

d1)

This is two tailed test, for α = 0.1 and df = 25
Critical value of t are +/-1.708
Hence reject H0 if t < -1.708 or t > 1.708

d2)

yes, we can  conclude that the machine is working improperly

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