The joint probability mass function based on the given experiment is as seen in the attached file.
How to construct a joint probability mass function?
The whole sample space consists of 36 elements, i.e.,
Ω = {ω_ij = (i, j) : i, j = 1, ....6}
Now, with each of these 36 elements associate values of two random variables, X₁ and X₂, such that;
X₁ ≡ sum of the outcomes on the two dice.
X₂ ≡ |difference of the outcomes on the two dice|
That is;
X(ω_ij) = X₁(ω_ij) + X₂(ω_ij) = (i + j, |i - j|) i, j = 1, 2, ...., 6
Then, the bivariate rv X = (X₁, X₂) has the following joint probability mass
function (empty cells mean that the pmf is equal to zero at the relevant values of the rvs).
The joint probability mass function is seen in the attached file.
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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 426 gram setting. Based on a 13 bag sample where the mean is 433 grams and the standard deviation is 29, is there sufficient evidence at the 0.05 level that the bags are overfilled? Assume the population distribution is approximately normal.
The conclusion of the research is that we reject the null hypothesis and we conclude that the bag filling machine does not work correctly at the 426 gram setting.
How to solve hypothesis testing?We are given;
Population Mean; μ = 426
Sample mean; x' = 433
Sample size; n = 13
Standard deviation; s = 29
significance level; α = 0.05
Let us define the hypotheses;
Null Hypothesis; H₀: μ = 426 g
Alternative Hypothesis; H_a: μ < 426 g
Formula for the z-score here is;
z = (x' - μ)/(s/√n)
z = (433 - 426)/(29/√13)
z = 0.87
From online p-value from z-score calculator, we have;
p-value = 0.1922
This p-value is greater than the significance value and as such we reject the null hypothesis and we conclude that the bag filling machine does not work correctly at the 426 gram setting.
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