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The true average diameter of ball bearings of a certain type is supposed to be \(0.5\) in. What conclusion is appropriate when testing \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5 \mathrm{in}\) each of the following situations: a. \(n=13, t=1.6, \alpha=.05\) b. \(n=13, t=-1.6, \alpha=.05\) c. \(n=25, t=-2.6, \alpha=.01\) d. \(n=25, t=-3.6\)

Short Answer

Expert verified
In case a and b we fail to reject the null hypothesis while in case c we reject the null hypothesis. In case d, there is no definitive answer without a significance level but the t-value indicates it might reject the null hypothesis.

Step by step solution

01

Determine the critical t-value

For situation a. Using a t-table or t-distribution calculator, with \(n = 13\) thus degrees of freedom equals \(n-1 = 12\) and \(\alpha = .05\), we get a critical t-value of approximately ±2.18; For situation b. Like in case a, we get the same critical t-value which is approximately ±2.18; For situation c. In this case, with \(n=25\), thus degrees of freedom equals \(n-1 = 24\) and \(\alpha = .01\), we get a critical t-value of approximately ±2.80; For situation d. In absence of a significance level, we can only infer a conclusion based on the t-value alone.
02

Compare the t-values

For situation a. Given t-value is \(1.6\), which is less than the critical t-value \(|2.18|\). Therefore, we fail to reject the null hypothesis; For situation b. Given t-value is \(-1.6\), which again is within the range of \(-2.18\) and \(2.18\). Therefore, we fail to reject the null hypothesis; For situation c. Given t-value is \(-2.6\), which is less than the lower limit of the critical range \(-2.8\). Therefore, we reject the null hypothesis; For situation d. We cannot make definitive conclusion but the provided value \(-3.6\) suggests that it might reject the null hypothesis if we had a significance level.

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Key Concepts

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

Understanding t-distribution
The t-distribution is a critical component in hypothesis testing, especially useful when dealing with smaller sample sizes. It represents the distribution of sample means drawn from a normally distributed population when the population standard deviation is unknown. This distribution is symmetrical and more spread out compared to the standard normal distribution, which makes it better suited for small datasets.

A key aspect of the t-distribution involves "degrees of freedom," calculated as the sample size minus one. This number helps determine the shape of the t-distribution. The fewer the degrees of freedom, the wider and more variegated the distribution becomes, leading to larger critical values. As the sample size increases, the t-distribution approaches a normal distribution. Understanding this behavior is crucial for analyzing small sample data effectively.
The role of the critical t-value
The critical t-value is a threshold that helps determine whether an observed t-statistic suggests significant evidence against the null hypothesis. This value is derived from the t-distribution chart or calculator, based on the degrees of freedom and the desired significance level.

For example, in a hypothesis test where the null hypothesis states the population mean is 0.5 inches, the critical t-value will dictate the range within which we would fail to reject the null hypothesis. If your calculated t-value from the sample data falls outside of this range, it signifies that the sample provides enough evidence to reject the null hypothesis. It acts as a cutoff point that marks the significance boundary for making decisions in hypothesis testing.
Understanding the null hypothesis
The null hypothesis (\(H_{0}\)) is a fundamental part of hypothesis testing. It is a statement that asserts no effect or no difference exists. In our example, the null hypothesis states that the average diameter of ball bearings is exactly 0.5 inches.

When conducting a hypothesis test, the goal is to gather evidence to decide whether to reject this null hypothesis in favor of an alternative hypothesis (\(H_{a}\)), which in this case is that the average diameter is not equal to 0.5 inches.
  • The process involves computing a sample statistic (such as a t-value)
  • Comparing it to a critical value derived from a relevant distribution
Failing to reject the null hypothesis doesn't prove it true; rather, it suggests that the sample data isn’t sufficient to demonstrate a significant difference.
The significance level in hypothesis testing
The significance level, often denoted as \(\alpha\), is a threshold set by the researcher to determine how extreme the sample data must be in order to reject the null hypothesis. Common significance levels are 0.05, 0.01, or 0.10. These numbers represent the risk of committing a Type I error, which is rejecting a true null hypothesis.

Choosing a significance level is essential in hypothesis testing as it balances the risk of false positives. For instance, an \(\alpha\) of 0.05 implies a 5% risk that the null hypothesis will be incorrectly rejected. It provides a boundary for how unlikely your results must be (assuming the null hypothesis is true) to consider them statistically significant.

So, in a hypothesis test examination such as the ball bearing example, the significance level helps quantify the uncertainty in drawing conclusions about the population.

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Most popular questions from this chapter

The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let \(\pi\) be the true proportion of apartments that prohibit children. If \(\pi\) exceeds \(.75\), the city council will consider appropriate legislation. a. If 102 of the 125 . sampled apartments exclude renters with children, would a level 05 test lead you to the conclusion that more than \(75 \%\) of all apartments exclude children? b. What is the power of the test when \(\pi=.8\) and \(\alpha=.05\) ?

According to a survey of 1000 adult Americans conducted by Opinion Research Corporation, 210 of those surveyed said playing the lottery would be the most practical way for them to accumulate \(\$ 200,000\) in net wealth in their lifetime ("One in Five Believe Path to Riches Is the Lottery," San Luis Obispo Tribune, January 11,2006 ). Although the article does not describe how the sample was selected, for purposes of this exercise, assume that the sample can be regarded as a random sample of adult Americans. Is there convincing evidence that more than \(20 \%\) of adult Americans believe that playing the lottery is the best strategy for accumulating \(\$ 200,000\) in net wealth?

White remains the most popular car color in the United States, but its popularity appears to be slipping. According to an annual survey by DuPont (Los Angeles Times, February 22,1994 ), white was the color of \(20 \%\) of the vehicles purchased during 1993 , a decline of \(4 \%\) from the previous year. (According to a DuPont spokesperson, white represents "innocence, purity, honesty, and cleanliness.") A random sample of 400 cars purchased during this period in a certain metropolitan area resulted in 100 cars that were white. Does the proportion of all cars purchased in this area that are white appear to differ from the national percentage? Test the relevant hypotheses using \(\alpha=.05\). Does your conclusion change if \(\alpha=.01\) is used?

Much concern has been expressed in recent years regarding the practice of using nitrates as meat preservatives. In one study involving possible effects of these chemicals, bacteria cultures were grown in a medium containing nitrates. The rate of uptake of radio-labeled amino acid was then determined for each culture, yielding the following observations: \(\begin{array}{llllllll}7251 & 6871 & 9632 & 6866 & 9094 & 5849 & 8957 & 7978 \\\ 7064 & 7494 & 7883 & 8178 & 7523 & 8724 & 7468 & \end{array}\) Suppose that it is known that the true average uptake for cultures without nitrates is 8000 . Do the data suggest that the addition of nitrates results in a decrease in the true average uptake? Test the appropriate hypotheses using a significance level of \(.10\).

Pizza Hut, after test-marketing a new product called the Bigfoot Pizza, concluded that introduction of the Bigfoot nationwide would increase its sales by more than \(14 \%\) (USA Today, April 2, 1993). This conclusion was based on recording sales information for a random sample of Pizza Hut restaurants selected for the marketing trial. With \(\mu\) denoting the mean percentage increase in sales for all Pizza Hut restaurants, consider using the sample data to decide between \(H_{0}: \mu=14\) and \(H_{a}: \mu>14\). a. Is Pizza Hut's conclusion consistent with a decision to reject \(H_{0}\) or to fail to reject \(H_{0}\) ? b. If Pizza Hut is incorrect in its conclusion, is the company making a Type I or a Type II error?

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