/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 160 A manufacturer of television set... [FREE SOLUTION] | 91Ó°ÊÓ

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A manufacturer of television sets claims that the maintenance expenditures for its product will average no more than \(110\)dollar during the first year following the expiration of the warranty. A consumer group has asked you to substantiate or discredit the claim. The results of a random sample of 50 owners of such television sets showed that the mean expenditure was \(131.60\)dollar and the standard deviation was \(42.46\)dollar At the 0.01 level of significance, should you conclude that the manufacturer's claim is true or not likely to be true?

Short Answer

Expert verified
Based on the calculations, one should conclude at the 0.01 level of significance that the manufacturer's claim is not likely to be true.

Step by step solution

01

- Determine null and alternative hypotheses

Set up the null hypothesis and the alternative hypothesis. The null hypothesis \(H_0\) would be the manufacturer's claim: the average maintenance cost is no more than $110 (\(H_0: \mu \leq 110\)). The alternative hypothesis \(H_1\) is that the average maintenance cost is more than $110 (\(H_1: \mu > 110\)).
02

- Calculate the test statistic

Calculate the test statistic using the formula for the z-score, which is \((\bar{x} - \mu) / (\sigma / \sqrt{n})\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size. Given values are \(\bar{x} = 131.60\), \(\mu = 110\), \(\sigma = 42.46\), and \(n = 50\). So, the z-score is approximately 2.55.
03

- Find the critical value

The critical value for a one-tailed test at the 0.01 significance level is about 2.33 from the z-table.
04

- Decision making

Since the z-score (2.55) is greater than the critical value (2.33), the null hypothesis should be rejected, implying that the manufacturer's claim is not likely to be true.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is crucial in hypothesis testing. These hypotheses are essentially statements about a population parameter such as the mean (.

In our television sets manufacturer's claim, the null hypothesis ( represents what the manufacturer asserts: the average maintenance cost is 110 dollars (. This is a statement of no effect or no difference, and in hypothesis testing, we assume this to be true unless the evidence suggests otherwise.

On the flip side, the alternative hypothesis ( is a statement that opposes the null hypothesis. It's what the consumer group might believe: that the average maintenance expense is greater than 110 (. This is what we aim to support through our data and analysis. When conducting a hypothesis test, only one of these hypotheses can be supported by the evidence. It's like a court trial where the null represents the 'innocent until proven guilty' stance and the alternative represents 'guilty as charged'.
Test Statistic Calculation
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's vital as it helps determine how far the sample mean deviates from the null hypothesis's asserted mean, and whether this deviation is significant enough to warrant a rejection of the null hypothesis.

To find our test statistic, we use the z-score formula which involves subtracting the claimed population mean ( from the sample mean (, and then dividing this difference by the standard deviation ( divided by the square root of the sample size (. In our example, using the given data, we computed a z-score of approximately 2.55. This numerical value is pivotal as it will be used in comparison with a critical value to make our conclusive decision on the hypothesis.
Significance Level
The significance level, often denoted by , is the probability of rejecting the null hypothesis when it is actually true. This is also known as a Type I error. It reflects the level of risk we are willing to take to draw a conclusion about the population based on our sample.

Choosing a significance level is a subjective decision that depends on the consequences of making an error. In many scientific studies, a 0.05 level is common, but in our television set example, we've used a more stringent level of 0.01. This means we have less than a 1% chance of incorrectly rejecting the manufacturer's claim if it's true. A lower significance level implies we demand stronger evidence against the null hypothesis before we are willing to reject it. This stringent threshold ensures that our conclusion to discredit the manufacturer's claim is not made lightly.
Z-Score
A z-score, in the context of hypothesis testing, is a measure of how many standard deviations a data point (in our case, the sample mean) is from the population mean stated in the null hypothesis. Not only does it gauge the distance, but it also indicates the direction (above or below the mean).

For our exercise, the calculation yielded a z-score of approximately 2.55, which means the sample mean is 2.55 standard deviations above the manufacturer's claimed average cost of $110. The z-score is compared against a critical value that corresponds to the significance level. If our z-score exceeds the critical value (2.33 at the 0.01 significance level for a one-tailed test), we have statistically significant evidence to reject the null hypothesis and accept that the alternative hypothesis may be true, indicating the manufacturer's claim isn't likely to be accurate.

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Most popular questions from this chapter

To test the hypothesis that the standard deviation on a standard test is \(12,\) a sample of 40 randomly selected students' exams was tested. The sample variance was found to be \(155 .\) Does this sample provide sufficient evidence to show that the standard deviation differs from 12 at the 0.05 level of significance?

Use a computer or calculator to find the area (a) to the left, and (b) to the right of \(\chi^{2} \star=20.2\) with df \(=15\).

Length is not very important in evaluating the quality of corks because it has little to do with the effectiveness of a cork in preserving wine. Winemakers have several lengths to choose from and order the length of cork they prefer (long corks tend to make a louder pop when the bottle is uncorked). Length is monitored very closely, though, because it is a specified quality of the cork. The lengths of no. 9 natural corks \((24 \mathrm{mm}\) diameter by \(45 \mathrm{mm}\) length) have a normal distribution. Twelve randomly selected corks were measured to the nearest hundredth of a millimeter. $$\begin{array}{llllll}\hline 44.95 & 44.95 & 44.80 & 44.93 & 45.22 & 44.82 \\\45.12 & 44.62 & 45.17 & 44.60 &44.60 & 44.75 \\\\\hline\end{array}$$ a. Does the preceding sample give sufficient reason to show that the mean length is different from \(45.0 \mathrm{mm}\) at the 0.02 level of significance? A different random sample of 18 corks is taken from the same batch. $$\begin{array}{lllllllll}\hline 45.17 & 45.02 & 45.30 & 45.14 & 45.35 & 45.50 & 45.26 & 44.88 & 44.71 \\\44.07 & 45.10 & 45.01 & 44.83 & 45.13 & 44.69 & 44.89 & 45.15 & 45.13 \\\\\hline\end{array}$$ b. Does the preceding sample give sufficient reason to show that the mean length is different from \(45.0 \mathrm{mm}\) at the 0.02 level of significance? c. What effect did the two different sample means have on the calculated test statistic in parts a and b? Explain. d. What effect did the two different sample sizes have on the calculated test statistic in parts a and b? Explain. e. What effect did the two different sample standard deviations have on the calculated test statistic in parts a and b? Explain.

a. The central \(90 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? b. The central \(95 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? c. The central \(99 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values?

It has been suggested that abnormal male children tend to be born to older- than-average parents. Case histories of 20 abnormal males were obtained, and the ages of the 20 mothers were as follows: $$\begin{array}{lllllllllll}31 & 21 & 29 & 28 & 34 & 45 & 21 & 41 & 27 & 31 \\\43 & 21 & 39 & 38 & 32 & 28 & 37 & 28 & 16 & 39\end{array}$$ The mean age at which mothers in the general population give birth is 28.0 years. a. Calculate the sample mean and standard deviation. b. Does the sample give sufficient evidence to support the claim that abnormal male children have olderthan-average mothers? Use \(\alpha=0.05 .\) Assume ages have a normal distribution.

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