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The print on the packages of 100-watt General Electric soft-white lightbulbs states that these lightbulbs have an average life of 750 hours. Assume that the standard deviation of the lengths of lives of these lightbulbs is 50 hours. A skeptical consumer does not think these lightbulbs last as long as the manufacturer claims, and she decides to test 64 randomly selected lightbulbs. She has set up the decision rule that if the average life of these 64 lightbulbs is less than 735 hours, then she will conclude that GE has printed too high an average length of life on the packages and will write them a letter to that effect. Approximately what significance level is the consumer using? Approximately what significance level is she using if she decides that GE has printed too high an average length of life on the packages if the average life of the 64 lightbulbs is less than 700 hours? Interpret the values you get.

Short Answer

Expert verified
Approximate significance level when average life is 735 hours is 2.28% and when it is 700 hours is 0%. The lower the average hour, higher the certainty level of the consumer in doubting the manufacturer's claim.

Step by step solution

01

Understand the problem

Given that 100-watt GE soft-white lightbulbs have an average life of 750 hours (population mean, \(\mu\)) with a standard deviation of 50 hours (\(\sigma\)). A sample of 64 bulbs are selected (n=64). The goal is to find the significance level at which the consumer is making a decision for two different average lifetimes.
02

Calculate Z-Score for 735 hours

To find the significance level is equivalent to determining the probability of getting an average life less than 735 hours. That probability is also the Z-score area in a standard normal distribution. First, the Z-score for 735 hours is calculated using the formula: \(Z = (x - \mu) / (\sigma / \sqrt{n})\), where x = 735 hours. Plugging all values, the Z-score is obtained as \(Z_1 = (735 - 750) / (50 / \sqrt{64}) = -2\).
03

Look up Value for Z-Score at -2

Now look up the area value corresponding to the Z-score of -2 using the Z-table. This value is the significance level, and is 0.0228.
04

Calculate Z-Score for 700 hours

Now, calculate the Z-score for 700 hours using the same formula. Plugging all values into the formula, the Z-score is obtained as \(Z_2 = (700 - 750) / (50 / \sqrt{64}) = -8\).
05

Look up Value for Z-Score at -8

Since -8 is off the chart of the Z-table, the corresponding probability value is incredibly tiny and essentially zero.
06

Interpret the values

At 735 hours, the significance level found is approximately 0.0228 or 2.28%, and at 700 hours, the significance level is essentially zero. This means the consumer is more confident about doubting GE's claim at an average life of 700 hours compared to 735 hours. But in both cases, the consumer is rejecting the manufacturer's claim of an average bulb life of 750 hours.

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Key Concepts

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

Z-Score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. In terms of hypothesis testing, the Z-score is crucial because it tells us how many standard deviations a data point is from the mean. For example, when the skeptical consumer checks if the lightbulbs have an average life less than 750 hours, calculating the Z-score helps identify whether this discrepancy is significant. The formula to calculate the Z-score is:
\[ Z = \frac{x - \mu}{\sigma / \sqrt{n}} \]
where:
  • \( x \) is the sample mean,
  • \( \mu \) is the population mean,
  • \( \sigma \) is the standard deviation, and
  • \( n \) is the sample size.
Using this formula, if a Z-score for an average life of 735 hours is calculated as -2, it means the sample average is 2 standard deviations below the population mean. This is important for determining the probability that such a result could occur by random chance.
Significance Level
The significance level, denoted by \( \alpha \), is the threshold at which we decide whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. A common choice for significance levels is 0.05 or 0.01, implying a 5% or 1% risk of a false positive.
In the lightbulb example, the significance level reflects the consumer's risk of incorrectly asserting that the average life is less than 750 hours. With a Z-score of -2 for 735 hours, the significance level is found to be approximately 0.0228, or 2.28%. This means there's a 2.28% chance of observing a sample mean at least this extreme due to random variation.
Importantly, the smaller the significance level, the stronger the evidence must be to reject the null hypothesis. When the test mean is 700 hours, the significance level effectively becomes zero, indicating an extremely rare occurrence under the assumption the null hypothesis is true.
Standard Deviation
Standard deviation is a measure that describes the amount of variation or dispersion of a set of values. In simpler terms, it tells us how spread out the numbers in a data set are around the mean. For hypothesis testing, it's vital because it allows us to understand the data's variability and calculate other important statistics, like the Z-score.
In the context of the lightbulbs example, a standard deviation of 50 hours means that the lifetime of the bulbs generally varies by 50 hours around the average life of 750 hours. Lower standard deviation signifies that the bulb lifetimes are closer to the mean, whereas a higher standard deviation implies more variability.
It plays a key role in determining how significant deviations from the mean really are, making it central to the hypothesis testing process. Without standard deviation, we wouldn't know the significance of our sample results.
Normal Distribution
Normal distribution, often referred to as the bell curve, is a fundamental concept in statistics representing how values in a data set are dispersed. A normal distribution is symmetric and most of the data points fall close to the mean, with fewer data points appearing as you move further from the mean.
In hypothesis testing, a normal distribution is assumed to utilize Z-scores and probabilities effectively. In the lightbulb lifespan example, it's assumed that the life of the bulbs follows a normal distribution. This assumption allows us to use the standard normal distribution table (Z-table) to evaluate probabilities associated with Z-scores.
The normal distribution is key in determining how likely or unlikely a sample mean is under the null hypothesis. It helps in setting boundaries for different significance levels and making informed decisions based on statistical evidence. Understanding and assuming normal distribution is critical to making valid inferences in hypothesis testing.

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Most popular questions from this chapter

A computer company that recently introduced a new software product claims that the mean time it takes to learn how to use this software is not more than 2 hours for people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Test at the \(1 \%\) significance level whether the company's claim is true. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

A business school claims that students who complete a 3 -month typing course can type, on average, at least 1200 words an hour. A random sample of 25 students who completed this course typed, on average, 1125 words an hour with a standard deviation of 85 words. Assume that the typing speeds for all students who complete this course have an approximately normal distribution. a. Suppose the probability of making a Type I error is selected to be zero. Can you conclude that the claim of the business school is true? Answer without performing the five steps of a test of hypothesis. b. Using the \(5 \%\) significance level, can you conclude that the claim of the business school is true? Use both approaches.

Consider \(H_{0}: \mu=80\) versus \(H_{1}: \mu \neq 80\) for a population that is normally distributed. a. A random sample of 25 observations taken from this population produced a sample mean of 77 and a standard deviation of 8 . Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 25 observations taken from the same population produced a sample mean of 86 and a standard deviation of \(6 .\) Using \(\alpha=.01\), would you reject the null hypothesis?

The Bath Heritage Days, which take place in Bath, Maine, have been popular for, among other things, an eating contest. In 2009, the contest switched from blueberry pie to a Whoopie Pie (www.timesrecord.com), which consists of two large, chocolate cake-like cookies filled with a large amount of vanilla cream. Sixty-five randomly selected adults are chosen to eat a 1 -pound Whoopie Pie, and the average time for 59 adults (out of these 65 ) is \(127.10\) seconds. Based on other Whoopie Pie-eating contests throughout the United States, suppose that the standard deviation of the times taken by all adults to consume 1-pound Whoopie pies are known to be \(23.80\) seconds. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean time to eat a 1 -pound Whoopie Pie is more than 2 minutes. Will you reject the null hypothesis at \(\alpha=.01\) ? Explain. What if \(\alpha=.02\) ? b. Test the hypothesis of part a using the critical-value approach. Will you reject the null hypothesis at \(\alpha=.01\) ? What if \(\alpha=.02\) ?

As noted in U.S. Senate Resolution \(28,9.3 \%\) of Americans speak their native language and another language fluently (Source: www.actfl.org/i4a/pages/index.cfm?pageid=3782). Suppose that in a recent sample of 880 Americans, 69 speak their native language and another language fluently. Is there significant evidence at the \(10 \%\) significance level that the percentage of all Americans who speak their native language and another language fluently is different from \(9.3 \%\) ? Use both the \(p\) -value and the critical-value approaches.

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