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A lathe is adjusted so that the mean of a certain dimension of the parts is \(20 \mathrm{~cm}\). A random sample of 10 of the parts produced a mean of \(20.3 \mathrm{~cm}\) and a standard deviation of \(.2 \mathrm{~cm}\). Do the results indicate that the machine is out of adjustment? Test at the \(.05\) level of significance.

Short Answer

Expert verified
We reject the null hypothesis H₀ since the test statistic (4.74) falls in the rejection region (± 2.262). The results indicate that the machine is out of adjustment at the 0.05 level of significance.

Step by step solution

01

Compute the test statistic t

To calculate the test statistic, we can use the formula: \(t = \frac{x̄ - µ}{\frac{s}{\sqrt{n}}}\) Plugging in the given values, we have: \(t = \frac{20.3 - 20}{\frac{0.2}{\sqrt{10}}}\) When you calculate the test statistic, you get: \(t \approx 4.74\)
02

Determine the critical value and rejection region

Since we are performing a two-tailed test at a 0.05 level of significance, we need to find the critical value t(α/2) with degrees of freedom (df) = n - 1 = 9. Using the t-distribution table or a calculator, we find the critical value to be: \(t(0.025, 9) \approx 2.262\) This provides us with the rejection region for our test: We will reject H₀ if the test statistic is either less than -2.262 or greater than 2.262.
03

Compare the test statistic to the rejection region

Now, compare our calculated test statistic t (4.74) to the critical values (± 2.262). Since 4.74 > 2.262, our test statistic falls in the rejection region.
04

Make a decision based on the comparison

As our test statistic (4.74) falls in the rejection region, we reject the null hypothesis Hâ‚€. Therefore, the results indicate that the machine is out of adjustment at the 0.05 level of significance.

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Key Concepts

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

t-distribution
The t-distribution is similar to the normal distribution but has thicker tails, which makes it more appropriate for smaller sample sizes. When working with smaller samples (usually less than 30), the t-distribution provides a more accurate reflection of variability in the data compared to a normal distribution.

Characteristics of the t-distribution include:
  • It is symmetrical around zero.
  • The shape of the distribution depends on the degrees of freedom. As the degrees of freedom increase, the t-distribution approaches the shape of a normal distribution.
  • Thicker tails than the normal distribution, providing a higher probability for extreme values.
In hypothesis testing, the t-distribution is used to determine the probability of the calculated test statistic when the sample size is small. This allows us to make more reliable decisions in determining if there is enough evidence to support a claim, such as whether a machine is out of adjustment.
level of significance
The level of significance, denoted by α, represents the threshold at which you are prepared to reject the null hypothesis. It reflects the probability of making a Type I error, which is rejecting a true null hypothesis.

In many scientific experiments, a 5% level of significance (α = 0.05) is standard, though this can vary depending on the context.
  • A smaller α (like 0.01) means you need stronger evidence to reject the null hypothesis, reducing the chance of a Type I error.
  • A larger α (like 0.10) allows for easier rejection of the null hypothesis, increasing the chance of a Type I error.
In the exercise, a 0.05 level of significance was used to test whether the machine is out of adjustment. Determining the level of significance helps set up the critical region, which provides the values that the test statistic must exceed to indicate a result significant enough for hypothesis rejection.
critical value
The critical value is a point on the test statistic distribution that defines the threshold for rejecting the null hypothesis. When conducting a two-tailed test, you have two critical values: one positive and one negative.

Here is how the critical value is used:
  • The critical values are determined from the t-distribution table based on the desired level of significance (e.g., 0.05) and the degrees of freedom in the sample.
  • For a 0.05 level of significance with 9 degrees of freedom, the critical values in the example were ±2.262.
  • If the calculated test statistic falls beyond these critical values (in the rejection region), the null hypothesis is rejected.
Understanding critical values is essential because they draw the line between results that occur by random chance and those that are statistically significant enough to question the status quo, such as whether a machine setting needs adjustment.
standard deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It indicates how much individual data points deviate from the mean of the dataset.
  • A low standard deviation means that the data points tend to be close to the mean.
  • A high standard deviation indicates that the data points are spread out over a larger range of values.
In hypothesis testing, standard deviation helps quantify the uncertainty or variability in the data. It's used in calculating the test statistic, which is crucial for determining whether to accept or reject the null hypothesis.

In the given exercise, the standard deviation of the part dimensions was 0.2 cm, which suggested how much each part's measurement differed from the sample mean of 20.3 cm. This figure was important for computing the test statistic that helped decide if the lathe machine is properly adjusted or not.

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Most popular questions from this chapter

Suppose that for the preceding problem, we have taken a sample of 15 pairs of specimens. The differences in the average maximum pits for each coating, \(\underline{d}\) is 8 with a standard deviation for the differences of \(11.03\). Test whether there is a significant difference between the corrosion preventing abilities of coatings \(\mathrm{A}\) and \(\mathrm{B}\) at a \(.05\) level of significance.

In a recent science fiction novel, aliens invented a death ray that killed 3 Earthman. The lethal doses were 10,11 , and 12 units of the ray. The mean lethal dose for Moonmen is \(13.30\) units. For this sample of 3 Earthmen \(\underline{\mathrm{X}}=11\) and \(\mathrm{S}=1.0\) and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=3.98\), not significant at the \(\alpha=.05\) level of significance. Then 6 more Earthmen were killed with lethal doses of \(9,10,11,12,13\) and \(14 .\) For this sample, \(\underline{\mathrm{X}}=11.5\), and \(\mathrm{S}=1.87\), and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=2.36\), also not significant at \(\alpha=.05\). What can be said if we combine the results of these two samples?

For the following samples of data, compute \(\mathrm{t}\) and determine whether \(\mu_{1}\) is significantly less than \(\mu_{2}\). For your test use a level of significance of \(10 .\) Sample \(1: \mathrm{n}=10, \underline{\mathrm{X}}=10.0\) \(\mathrm{S}=5.2\); sample \(2: \mathrm{n}=10, \underline{\mathrm{X}}=13.3, \mathrm{~S}=5.7\)

A college dean wants to determine if the students entering the college in a given year have higher IQ's than the students entering the same college in the previous year. The IQ's of the college entrants in the two years are known to be normally distributed and to have equal variances. A random sample of four students from this year's entering class gives IQ scores of \(110,113,116\), and 117. Another random sample of four students from last year's entering class gives IQ scores of \(109,111,112\), and 112 . At the \(5 \%\) level of significance, can we conclude that this year's students have a higher IQ than last years?

In investigating several complaints concerning the weight of the "NET WT. 12 OZ." jar of a local brand of peanut butter, the Better Business Bureau selected a sample of 36 jars. The sample showed an average net weight of \(11.92\) ounces and a standard deviation of \(.3\) ounce. Using a \(.01\) level of significance, what would the Bureau conclude about the operation of the local firm?

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