Question

A machine that puts corn flakes into boxes is adjusted to put an average of 15.8 ounces into each box, with standard deviation of 0.22 ounce. If a random sample of 14 boxes gave a sample standard deviation of 0.36 ounce, do these data support the claim that the variance has increased and the machine needs to be brought back into adjustment? (Use a 0.01 level of significance.)

(i) Give the value of the level of significance.


State the null and alternate hypotheses.

H₀: σ² = 0.0484; H₁: σ² > 0.0484

H₀: σ² < 0.0484; H₁: σ² = 0.0484    

H₀: σ² = 0.0484; H₁: σ² ≠ 0.0484

H₀: σ² = 0.0484; H₁: σ² < 0.0484

(ii) Find the sample test statistic. (Round your answer to two decimal places.)


(iii) Find or estimate the P-value of the sample test statistic.

P-value > 0.100

0.050 < P-value < 0.100   

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

P-value < 0.005

(iv) Conclude the test.

Since the P-value ≥ α, we fail to reject the null hypothesis.Since the P-value < α, we reject the null hypothesis.    

Since the P-value < α, we fail to reject the null hypothesis.Since the P-value ≥ α, we reject the null hypothesis.

(v) Interpret the conclusion in the context of the application.

At the 1% level of significance, there is sufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.

At the 1% level of significance, there is insufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.    

Answer 1

i)
significance = 0.01

H0: sigma^2 = 0.0484
Ha: sigma^2 > 0.0484

ii)
Test statistic,
chi-square = (n-1)*(s/sigma)^2
chi-square = 13*(0.36/0.22)^2
= 34.81

iii)
p-value = 0.0009
P-value < 0.005

iv)
Since the P-value < α, we fail to reject the null hypothesis.Since the P-value ≥ α, we reject the null hypothesis.

v)
At the 1% level of significance, there is sufficient evidence to conclude that the variance has increased and the machine needs to be adjusted.