A machine that is programmed to package exactly 1.20 pounds of cereal is being tested for its accuracy. Assume that the population standard

Question

A machine that is programmed to package exactly 1.20 pounds of cereal is being tested for its accuracy. Assume that the population standard deviation is 0.66 pounds. In a sample of 36 cereal boxes, the sample mean weight is calculated as 1.48 pounds.

a. State the null hypothesis and the alternate hypothesis.
b. Compute the value of test statistic.
c. State the decision rule at 10% level of significance that relates to a critical value method.
d. Conduct the test with the critical value method.

in progress 0
Katherine 3 weeks 2021-11-08T11:17:18+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-08T11:18:49+00:00

    Answer:

    Step-by-step explanation:

    Hello!

    To test the accuracy of a machine said to package exactly 1.20 pounds od cereal a random sample of 36 cereal boxes was taken.

    The study variable is:

    X: Weight of a cereal box.

    This variable has no known distribution but it’s population standard deviation is known σ= 0.66 pounds.

    The sample weight is X[bar]= 1.48 pounds.

    a.

    The claim is that the machine packages exactly 1.20 pounds per box and the parameter of interest is the population mean, so you have to test is the population mean weight packaged by the machine is 1.20 pounds or not, symbolically:

    H₀: μ = 1.20

    H₁: μ ≠ 1.20

    b.

    As I said before there is no information about the distribution of the study variable so I’ll apply the Central Limit Theorem to approximate the distribution of the sample proportion to normal.

    Remember: Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample means tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

    As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

    Then we can use the standard normal as a statistic for the test:

    Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }≈N(0;1)

    Z_{H_0}= \frac{1.48-1.20}{\frac{0.66}{\sqrt{36} } } = 2.54

    c.

    The level of significance of this test is α: 0.10

    The critical is two-tailed, wich means you will have two critical values:

    Critical value 1: Z_{\alpha /2} = Z_{0.1/2} = Z_{0.05}= -1.648

    Critical value 1: Z_{1-\alpha /2}= Z_{1-(0.1/2)}= Z_{0.95}= 1.648

    This means that you will reject the null hypothesis if  Z_{H_0} ≤ -1.648 or if Z_{H_0} ≥ 1.648.

    You will not reject the null hypothesis if -1.648 ≤ Z_{H_0} ≤ 1.648.

    d.

    Z_{H_0} is greater than the upper critical value, so the decision is to reject the null hypothesis.

    Using 10% confidence level there is enough statistical evidence to conclude that the population average weight of the cereal boxes is different from 1.2 pounds.

    I hope it helps!

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )