Question

A machine that is programmed to package 3.00 pounds of cereal in each cereal box is being tested for its accuracy. In a sample of 49 cereal boxes, the mean and the standard deviation are calculated as 3.05 pounds and 0.14 pound, respectively. (You may find it useful to reference the appropriate table: z table or t table)

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 ≥ 3.00; H_A: µ < 3.00

• H₀: µ ≤ 3.00; H_A: µ > 3.00

• H₀: µ = 3.00; H_A: µ ≠ 3.00

b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

b-2. Find the p-value.

• p-value < 0.01

• p-value 0.10
• 0.05 p-value < 0.10
• 0.01 p-value < 0.02
• 0.02 p-value < 0.05

c-1. What is the conclusion at the 5% significance level?

• Do not reject H₀ since the p-value is smaller than significance level.

• Do not reject H₀ since the p-value is greater than significance level.

• Reject H₀ since the p-value is smaller than significance level.

• Reject H₀ since the p-value is greater than significance level.

c-2. Can you conclude that the machine is working improperly?

• No

• Yes

Answer 1

a) H₀: µ = 3.00; H_A: µ ≠ 3.00

Because in the question it is given that machine is programmed to pack 3 pounds of cereals, so we either it should be equal to 3 or not equal to 3. Thus it is a two sided hypothesis.

b-1)

Test statistics = (sample mean- population mean) √n /SD

Test statistic = (3.05-3.00)*7/0.14

Test statistic = 2.5

b-2) Pr (0_<Z_<2.50) = 0.4938

Pr( Z_>2.50) = 0.50- 0.4938 = 0.0062

P- value< 0.01

c-1) Reject H₀ since the p-value is smaller than significance level.

significance level = 0.05

P- value< 0.01

c-2) Yes, the machine is working improperly