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Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure oilfilled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than \(.05 \mathrm{~mm}\). What hypotheses should be tested, and why? In this context, what are the type I and type II errors?

Short Answer

Expert verified
Test \( H_0: \sigma \geq 0.05 \) vs \( H_a: \sigma < 0.05 \). Type I error: Conclude \( \sigma < 0.05 \), but \( \sigma \geq 0.05 \). Type II error: Fail to conclude \( \sigma < 0.05 \), but \( \sigma < 0.05 \).

Step by step solution

01

Define the Parameter of Interest

We are interested in testing the standard deviation of sheath thickness for the polyethylene sheaths. Define \( \sigma \) as the true standard deviation of sheath thickness.
02

State the Null Hypothesis

The null hypothesis \( H_0 \) should represent the current belief that needs to be tested. Here it is whether the standard deviation is equal to or greater than \(.05 \text{ mm}\). Thus, the null hypothesis is \( H_0: \sigma \geq 0.05 \text{ mm} \).
03

State the Alternative Hypothesis

The alternative hypothesis \( H_a \) represents what we want to prove. In this case, that the standard deviation is less than \(0.05\) mm. Thus, the alternative hypothesis is \( H_a: \sigma < 0.05 \text{ mm} \).
04

Describe Type I Error

A Type I error occurs when we reject the null hypothesis \( H_0 \) when it is actually true. In this context, it means concluding that the standard deviation is less than \(0.05\) mm when in fact it is \(0.05\) mm or greater.
05

Describe Type II Error

A Type II error occurs when we fail to reject the null hypothesis \( H_0 \) when the alternative hypothesis \( H_a \) is actually true. Here, it would mean concluding that the standard deviation is not less than \(0.05\) mm when it is indeed less.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
Standard deviation is a statistical measure that provides insights into the amount of variation or dispersion of a set of values. In simpler terms, it tells us how spread out numbers are in a dataset. If the standard deviation is small, it means that the values tend to be close to the mean (average) value. Conversely, a large standard deviation indicates that the values are more spread out over a wider range.

For example, if the thickness of polyethylene sheaths is consistently around a certain measurement, say 0.045 mm, then a small standard deviation signifies that most sheath thicknesses are close to this value. The formula for standard deviation is:
\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \] Here:
  • \( \sigma \) is the standard deviation.
  • \( N \) is the number of observations.
  • \( x_i \) represents each individual observation.
  • \( \mu \) is the mean of the observations.

In the context of hypothesis testing, a company might want to ensure that variability, expressed by the standard deviation, is below a certain threshold to maintain product quality. Lower variability often implies more reliable performance.
Type I Error
In hypothesis testing, a Type I error is a critical concept that deals with the false rejection of a null hypothesis. It occurs when, based on the sample data, we wrongly conclude that there is an effect or a difference when in truth, there isn't one.

In the scenario concerning the polyethylene sheaths, a Type I error would mean the company believes the sheath thickness variability is acceptably low (below 0.05 mm), leading them to purchase the order. However, the reality is that the variability is indeed not as low as required. This error can lead to decisions based on incorrect data analysis, possibly resulting in defects or failures in the future products. The significance level, often denoted as \( \alpha \), represents the probability of making a Type I error, with common practice setting \( \alpha \) at 0.05 (5%). This implies there is a 5% risk of concluding a false positive.
Type II Error
A Type II error, practically speaking, happens when we fail to reject a false null hypothesis. In more straightforward terms, this mistake occurs when we neglect to identify an existing effect or difference.

Applying this to our exercise, if the company does not recognize the sheath thickness variability as being sufficiently low (less than 0.05 mm) when it truly is, they might pass on an opportunity to make a beneficial purchase decision based on accurate but neglected findings. The risk of making a Type II error is represented by \( \beta \). Unlike the predetermined \( \alpha \) for Type I error, \( \beta \) is generally not fixed and requires more complex considerations, including sample size and the true effect size. A low \( \beta \) value (high test power) is preferable, signifying a lower chance of missing an actual effect.

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Most popular questions from this chapter

would not conclude that more than \(50 \%\) do? 49\. A plan for an executive travelers' club has been developed by an airline on the premise that \(5 \%\) of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. a. Using this data, test at level .01 the null hypothesis that the company's premise is correct against the alternative that it is not correct. b. What is the probability that when the test of part (a) is used, the company's premise will be judged correct when in fact \(10 \%\) of all current customers qualify?

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let \(\mu\) denote the true average reflectometer reading for a new type of paint under consideration. A test of \(H_{0}: \mu=20\) versus \(H_{a}: \mu>20\) will be based on a random sample of size \(n\) from a normal population distribution. What conclusion is appropriate in each of the following situations? a. \(n=15, t=3.2, \alpha=.05\) b. \(n=9, t=1.8, \alpha=.01\) c. \(n=24, t=-.2\)

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using Minitab, resulting in the accompanying output. \(\begin{array}{lrrrrrr}\text { Variable } & N & \text { Mean } & \text { StDev } & \text { SE Mean } & \text { Z } & \text { P-Value } \\ \text { lifetime } 50 & 738.44 & 38.20 & 5.40 & -2.14 & 0.016\end{array}\) What conclusion would be appropriate for a significance level of \(.05\) ? A significance level of .01? What significance level and conclusion would you recommend?

For each of the following assertions, state whether it is a legitimate statistical hypothesis and why: a. \(H: \sigma>100\) b. \(H: \tilde{x}=45\) c. \(H: s \leq .20\) d. \(H: \sigma_{1} / \sigma_{2}<1\) e. \(H: \bar{X}-\bar{Y}=5\) f. \(H: \lambda \leq .01\), where \(\lambda\) is the parameter of an exponential distribution used to model component lifetime

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40 . If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. If a hypothesis test were to be performed on the resulting data, what null and alternative hypotheses would be of interest to the manufacturer? Describe type I and type II errors in the context of this problem situation.

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