Answer to Question #220323 in Mechanical Engineering for J MAYAKAYAKA

Question

A consumer organisation and the manufacturer of a certain brand of margarine are in dispute. The manufacturer states that his packs of margarine contain at least 250g of margarine whereas the consumer organisation claims that the content is at most 249g. A judge has to decide the issue.The judge decides to base her judgement on a random sample of ten packs of margarine.

The consumer organisation will be adjudged to be right if the random sample has a mean weight of less than 249:5g.

The manufacturer will be adjudged to be right if this mean weight is at least 249:5g.

The starting point for the judge is that he assumes a normal distribution with mean and variance 4, that is (u; 4) to be a good model for the weight X of a pack of margarine of that brand. Below, X denotes the mean weight (in grams) of the ten packs of margarine.

a)Determine the probability distribution of X

b)Calculate the probability that the judge decides (incorrectly) that the consumer organisation is right while the true u is equal to 251.

Answer 1

Part a

null, $H_o: μ=250$

alternate, $H_1: μ<250$

the test statistic: $-0.791$

critical value: $-1.645$

decision: do not reject H_o

p-value: 0.215

we do not have enough evidence to support the claim that the mean is less than 250.

Part b

Suppose the size of the sample is $n = 10$, then the critical region

becomes,

Reject $H_o \space if \space x < 250-\frac{3.2898}{\sigma / \sqrt{10}} \space OR \space if \space x > 250+\frac{3.2898}{\sigma / \sqrt{10}}$

Reject $H_o \space if \space x < 248.9597 \space \space OR \space \space if \space \space x > 251.0403$

Implies, don't reject H_o if $248.9597> x > 251.0403$

Suppose the true mean is 249.7

Probability of Type II error,

P(Type II error) = P(Don't Reject Ho | H1 is true )

= P($248.9597> x > 251.0403$ = 249.7)

= 0.983 - 0.1209 [ Using Z Table ]

= 0.8621