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.

What are your hypotheses?

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.

What are your hypotheses?

BlackTom AI · Accepted Answer

Since the suspicion is over-filling, the alternative hypothesis must state that the true mean packet content exceeds 175 g (μ > 175). 1. Relevant knowledge points and concepts - Parameter of interest: the population mean μ (average content per packet). - Null hypothesis (H0): the assumption of no effect or no overfill; typically takes an equality form. - Alternative hypothesis (H1): expresses the suspected direction (one-sided if we suspect overfilling). - For large samples, the sampling distribution of the sample mean is approximately normal with standard error SE = s/√n. 2. Complete reasoning process - The company suspects over-filling, so we want to test whether μ is greater than 175 g. - The null should represent the status quo (μ = 175) and the alternative the claim (μ > 175). Some texts use H0: μ ≤ 175 vs H1: μ > 175; the equality formulation H0: μ = 175 is the canonical form for hypothesis testing. 3. Calculations (intermediate steps shown) - Sample mean x̄ = 176 g, sample standard deviation s = 5 g, sample size n = 100. - Standard error SE = s / √n = 5 / √100 = 5 / 10 = 0.5 g. - Test statistic (z or t approx): z = (x̄ − μ0) / SE = (176 − 175) / 0.5 = 1 / 0.5 = 2.0. (Although the question only asks for hypotheses, this shows how the sample would be used to test them.) 4. Why the correct answer is correct - Because the concern is specifically that the machine is over-filling, the alternative must be one-sided in the positive direction: μ > 175. The null is the baseline value μ = 175. 5. Why other options would be incorrect - H1: μ < 175 would assert under-filling, the opposite of the suspicion. - H1: μ ≠ 175 is two-sided and would test for any calibration error (over- or under-filling), which is not what the manager specifically suspects. - Using H0: μ ≥ 175 vs H1: μ < 175 would again contradict the over-fill concern. 6. Conclusion - The appropriate hypotheses are H0: mean = 175 versus H1: mean > 175, which matches the provided answer.

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