A manufacturer of stone-ground deli-style mustard uses a high-speed machine to fill jars. The amount of mustard dispensed is normally distributed with a mean weight of 290 grams, and a standard deviation of 4 grams. If the actual amount dispensed is too low, then their customers will be cheated; if it’s too high, then the company could lose money. To keep the machine properly calibrated, the company periodically takes a sample of 12 jars, to see if they need to stop production temporarily and re-calibrate. A recent sample produced a mean of 292.2 grams. Should the company be concerned that µ 6= 290 grams?

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Respuesta :

This question is incomplete, the complete question is;

A manufacturer of stone-ground deli-style mustard uses a high-speed machine to fill jars. The amount of mustard dispensed is normally distributed with a mean weight of 290 grams, and a standard deviation of 4 grams. If the actual amount dispensed is too low, then their customers will be cheated; if it’s too high, then the company could lose money. To keep the machine properly calibrated, the company periodically takes a sample of 12 jars, to see if they need to stop production temporarily and re-calibrate. A recent sample produced a mean of 292.2 grams. Should the company be concerned that µ ≠ 290 grams? Use σ = 0:025. Use the p-value approach.

Answer:

P-Value = 0.08324

p-value ( 0.08324 ) is greater than significance level σ ( 0.025 );

fail to reject the null hypothesis

the company should not be concerned because, we have sufficient evidence to conclude that the mean weight is not different from 290 grams.

Step-by-step explanation:

Given the data in the question;

mean weight μ = 290 grams

sample mean x" = 292.2 grams

standard deviation s= 4 grams

sample size n = 12 jars

degree of freedom DF = n - 1 = 12 - 1 = 11

significance level σ  = 0.025

Two tailed Test

Null hypothesis             H₀ : μ = 290

Alternative hypothesis Hₐ : μ ≠ 290

Test Statistics;

t = ( x" - μ ) / ( s/√n )

we substitute our values into the equation

t = ( 292.2 - 290 ) / ( 4/√12 )

t = 2.2 / 1.1547

t = 1.905

From table; { t=1.905, df = 11, } Two tailed

P-Value = 0.08324

Hence, p-value ( 0.08324 ) is greater than significance level σ ( 0.025 );

fail to reject the null hypothesis meaning μ = 290

So the company should not be concerned because, we have sufficient evidence to conclude that the mean weight is not different from 290 grams.