Kettle Chips suspects that a machine which fills packets of 175g Sea Salt Chips may be wrongly calibrated and over-filling. A quality control manager takes a random sample of 100 packets and calculates the mean content to be 176g with a standard deviation of 5g. You want to find whether there is evidence that the machine is overfilling packets.

Draw a box model for the null hypothesis. What are the EV and SE of the mean of the 100 draws?

Question

Kettle Chips suspects that a machine which fills packets of 175g Sea Salt Chips may be wrongly calibrated and over-filling. A quality control manager takes a random sample of 100 packets and calculates the mean content to be 176g with a standard deviation of 5g. You want to find whether there is evidence that the machine is overfilling packets.

Draw a box model for the null hypothesis. What are the EV and SE of the mean of the 100 draws?

BlackTom AI · Accepted Answer

Under H0 we assume the machine is correctly calibrated so the true packet mean is 175 g. Model the population as a box of tickets whose values are packet weights with mean 175 g and (unknown) standard deviation about 5 g (we use the sample s = 5 g as an estimate). Taking 100 independent draws (the sample) and averaging them gives the sampling distribution of the sample mean.

1) Relevant concepts
- Expected value (EV) of the sample mean = population mean (under H0).
- Standard error (SE) of the sample mean = population SD / sqrt(n). For large n, the Central Limit Theorem makes the distribution approximately normal.
- Use s = 5 g as an estimate of the population SD when population SD is unknown.

2) Step-by-step reasoning
- Under the null hypothesis µ = 175 g, so the EV of the sample mean is 175 g.
- The SE = σ/√n. We substitute σ ≈ s = 5 g and n = 100.

3) Calculations (all intermediate steps)
- √n = √100 = 10
- SE = 5 / 10 = 0.5 g

4) Why this is correct
- By linearity of expectation, the expected value of the average of 100 independent draws equals the population mean (175 g) under H0. The SE formula comes from variance(average) = variance(single)/n, so SD(average) = SD(single)/√n. With s = 5 g, that gives 0.5 g.

5) Why other possible answers are incorrect
- EV = 176 would use the observed sample mean as the null mean; that is incorrect because the null hypothesis fixes the mean at 175 g. 
- SE = 5 (or any value not divided by √n) mistakes the SD of a single packet for the SE of the sample mean; the SE must shrink by √n.

6) Conclusion
Therefore the box model under H0 has EV = 175 and SE = 5/10 (which equals 0.5 g). This matches the provided answer: EV = 175; SE = 5/10.

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