ENGI 4421 Probability and Statistics

Faculty of Engineering and Applied Science

Hypothesis Test Tutorial Example — Solution

An existing manufacturing process produces thin wires with a breaking load that is known to be normally distributed with a mean of 100.0 N.

A new process claims to produce thin wires with a greater breaking load.   The client will not invest in the new process unless there is clear evidence (at a level of significance of 5%) that the mean breaking load is more than 5 N greater than that of the existing wires.

A random sample of 36 wires has a mean breaking load of 106.0 N, with a sample standard deviation of 3.0 N.

1. State the appropriate null and alternative hypotheses.

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   The burden of proof is on demonstrating that the mean breaking load of the new wires is more than 5 N greater than that of the existing wires, that is, greater than 105.0 N.   This is a one-sided alternative (upper tailed test).   Therefore

   

   [Note that, as the complement of the alternative hypothesis, the null hypothesis also includes   µ < 105.0.]

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2. Conduct the appropriate hypothesis test.

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   Note that while the breaking load for wires produced by the existing process is normally distributed, the question gives no such guarantee for the distribution of the breaking load of wires produced by the new process.

   The sample size is greater than 30.
   The central limit theorem therefore ensures that the distribution of the sample mean is normal, to at least a good approximation.   Furthermore, the t distribution with 35 degrees of freedom may be replaced by the standard normal (z) distribution, provided that distances from means are rounded off to no more than two significant figures.

   Method 1:

   
   or, using the more precise t distribution with 35 degrees of freedom,
   
   

   Method 2:

   
   
   

   Method 3:

   
   Using the t Calculator spreadsheet,
   
   or, using the table of the c.d.f. of the standard normal distribution,
   
   By any of these three methods, we reject the null hypothesis in favour of the alternative hypothesis at a level of significance of 5%.

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3. Should you advise the client to invest in the new process?

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   We have sufficient evidence to reject the null hypothesis in favour of the alternative hypothesis.   Therefore

   YES

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4. What would the decision be if the sample mean had been 104 N?

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   We are seeking evidence for   µ > 105.
   which is clearly inconsistent with   µ > 105.
   The rejection region is to the right of   x = 105, but this sample mean is to the left of   x = 105.   No hypothesis test is needed here.   We would decide immediately to retain the null hypothesis and the advice would be to

   not invest in the new process

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          BONUS QUESTION:

5. If the true mean breaking load of the new wires is 106.0 N, then what is the probability that another random sample of 36 wires, with the same sample standard deviation, will lead to an incorrect decision?

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   We need Method 1 from part (b) above.
   The rejection region, calculated on the basis that   µ = 105, has its boundary at   c = 105.8448 [we need greater precision here].
   [If z is used instead of t₃₅, then c = 105.8224]
   If   µ = 106 then the null hypothesis is false.
   The incorrect decision is to retain the null hypothesis, which will happen iff the sample mean does not fall in the rejection region.   Using the more precise t distribution,
   
   Approximating by the standard normal distribution, (which allows the tables of the standard normal c.d.f. to be used),
   
   Therefore the probability of an incorrect decision in this case is

   

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6. What type of error is the event in part (e)?

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   If the population mean is 106 N, then the alternative hypothesis is true and the null hypothesis is false.
   Not rejecting the null hypothesis when it is false is a

   type II error