/*! 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 64 A random sample of \(n=12\) obse... [FREE SOLUTION] | 91Ó°ÊÓ

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A random sample of \(n=12\) observations from a normal population produced \(\bar{x}=47.1\) and \(s^{2}=4.7 .\) Test the hypothesis \(H_{0}: \mu=48\) against \(H_{\mathrm{a}}: \mu \neq 48 .\) Use the Small-Sample Test of a Population Mean applet and a \(5 \%\) significance level.

Short Answer

Expert verified
Based on the small-sample t-test with a 5% significance level, we fail to reject the null hypothesis, meaning there is not enough evidence to support the claim that the population mean is different from 48.

Step by step solution

01

State the null and alternative hypotheses

Null hypothesis (\(H_0\)): \(\mu = 48\). Alternative hypothesis (\(H_a\)): \(\mu \neq 48\).
02

Calculate the t-statistic

To calculate the t-statistic, use the formula: \(t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\) where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, we have: \(\bar{x} = 47.1\) \(\mu_0 = 48\) \(s^2 = 4.7\) (so \(s = \sqrt{4.7} \approx 2.1679\)) \(n = 12\) Plug these values into the formula: \(t = \frac{47.1 - 48}{\frac{2.1679}{\sqrt{12}}} \approx -1.2551\)
03

Determine the critical t-value

Since we are given a 5% significance level, and we are performing a two-tailed test, we will use a significance level of \(\frac{5\%}{2} = 2.5\%\) for each tail. Using a t-distribution table or an online calculator with degrees of freedom (\(df = n-1 = 11\)) and a significance level of \(2.5\%\), we find the critical t-values to be \(\pm 2.20098\).
04

Compare t-statistic to critical t-value and decide

Now we compare our t-statistic and the critical t-value: \(-2.20098 < -1.2551 < 2.20098\) Our t-statistic (\(-1.2551\)) is not in the critical region (outside the range \(-2.20098\) to \(2.20098\)). Therefore, we fail to reject the null hypothesis.
05

Conclusion

We fail to reject the null hypothesis at the 5% significance level. There is not enough evidence to support the alternative hypothesis that the population mean is different from \(48\).

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Key Concepts

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

Normal Distribution
The concept of normal distribution is foundational in statistics and is often depicted as the bell-shaped curve known for its symmetry and unimodality.

In a normal distribution, the mean, median, and mode are all located at the center of the distribution, and the curve extends infinitely in both directions, getting closer to the horizontal axis but never touching it.
  • The total area under the curve is equal to 1, which represents the probability of all possible outcomes.
  • Approximately 68% of the data falls within one standard deviation from the mean, 95% within two standard deviations, and nearly all (99.7%) within three standard deviations.
Understanding the properties of the normal distribution is essential, especially when conducting hypothesis tests, as many statistical methods assume normality.

It allows us to relate probabilities directly to distances from the mean measured in standard deviations.
T-Statistic
The t-statistic is a critical component in many hypothesis tests, particularly when dealing with small sample sizes or unknown population standard deviations.

It quantifies the difference between the observed sample mean and the hypothesized population mean, scaled by the sample standard deviation and size. The formula for the t-statistic is: \( t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \) Where:
  • \(\bar{x}\) is the sample mean
  • \(\mu_0\) is the hypothesized population mean
  • \(s\) is the sample standard deviation
  • \(n\) is the sample size
The t-statistic tells us how many standard errors the sample mean is from the hypothesized population mean.

It is used to determine whether to reject the null hypothesis, based on how far the observed value is from expected under the null.
Significance Level
A significance level, often denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true.

It represents the tolerance for error, or the risk we are willing to accept for making a Type I error (a false positive). Common significance levels used in hypothesis testing are 0.05, 0.01, or 0.10.
  • For a 0.05 significance level, there is a 5% chance of incorrectly rejecting the null hypothesis.
  • The significance level affects the critical value, which determines the threshold for rejecting or not rejecting the null hypothesis.
A smaller significance level implies a stricter criterion for rejecting the null hypothesis, thus less risk of a Type I error but more risk of a Type II error (failing to reject a false null).

Choosing the significance level depends on the context of the test and the consequences of making an error.
Null and Alternative Hypotheses
In the context of hypothesis testing, forming null and alternative hypotheses is a crucial step that sets the stage for the analysis.

The null hypothesis, \(H_0\), usually represents a statement of no effect or no difference. It is a default or starting assumption for the test.
  • For example, \(H_0: \mu = 48\) means the true population mean is equal to 48.
The alternative hypothesis, \(H_a\), represents what we want to test for - it is the statement that there is an effect or a difference.
  • In our example, \(H_a: \mu eq 48\) suggests the population mean differs from 48 in either direction.
The decision to accept or reject these hypotheses is based on the evidence provided by the sample data and the calculated test statistic relative to the critical value.

Understanding these hypotheses helps in interpreting results and deciding the course of action based on statistical analysis.

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Most popular questions from this chapter

A manufacturer of industrial light bulbs likes its bulbs to have a mean life that is acceptable to its customers and a variation in life that is relatively small. If some bulbs fail too early in their life, customers become annoyed and shift to competitive products. Large variations above the mean reduce replacement sales, and variation in general disrupts customers' replacement schedules. A sample of 20 bulbs tested produced the following lengths of life (in hours): \(\begin{array}{lllllllll}2100 & 2302 & 1951 & 2067 & 2415 & 1883 & 2101 & 2146 & 2278 & 2019\end{array}\) $$ \begin{array}{llllllll} 1924 & 2183 & 2077 & 2392 & 2286 & 2501 & 1946 & 2161 & 2253 & 1827 \end{array} $$ The manufacturer wishes to control the variability in length of life so that \(\sigma\) is less than 150 hours. Do the data provide sufficient evidence to indicate that the manufacturer is achieving this goal? Test using \(\alpha=.01\).

Industrial wastes and sewage dumped into our rivers and streams absorb oxygen and thereby reduce the amount of dissolved oxygen available for fish and other forms of aquatic life. One state agency requires a minimum of 5 parts per million (ppm) of dissolved oxygen in order for the oxygen content to be sufficient to support aquatic life. Six water specimens taken from a river at a specific location during the low-water season (July) gave readings of \(4.9,5.1,4.9,5.0,5.0,\) and \(4.7 \mathrm{ppm}\) of dissolved oxygen. Do the data provide sufficient evidence to indicate that the dissolved oxygen content is less than 5 ppm? Test using \(\alpha=.05 .\)

Refer to Exercise \(10.7,\) in which we measured the dissolved oxygen content in river water to determine whether a stream had sufficient oxygen to support aquatic life. A pollution control inspector suspected that a river community was releasing amounts of semi treated sewage into a river. To check his theory, he drew five randomly selected specimens of river water at a location above the town, and another five below. The dissolved oxygen readings (in parts per million) are as follows: \begin{tabular}{l|lllll} Above Town & 4.8 & 5.2 & 5.0 & 4.9 & 5.1 \\ \hline Below Town & 5.0 & 4.7 & 4.9 & 4.8 & 4.9 \end{tabular} a. Do the data provide sufficient evidence to indicate that the mean oxygen content below the town is less than the mean oxygen content above? Test using \(\alpha=05 .\) b. Suppose you prefer estimation as a method of inference. Estimate the difference in the mean dissolved oxygen contents for locations above and below the town. Use a \(95 \%\) confidence interval.

Organic chemists often purify organic compounds by a method known as fractional crystallization. An experimenter wanted to prepare and purify 4.85 grams (g) of aniline. Ten 4.85 -g quantities of aniline were individually prepared and purified to acetanilide. The following dry vields were recorded: $$ \begin{array}{lllll} 3.85 & 3.80 & 3.88 & 3.85 & 3.90 \\ 3.36 & 3.62 & 4.01 & 3.72 & 3.82 \end{array} $$ Estimate the mean grams of acetanilide that can be recovered from an initial amount of \(4.85 \mathrm{~g}\) of aniline. Use a \(95 \%\) confidence interval.

Exercise 10.24 describes a dental experiment conducted to investigate the effectiveness of an oral rinse used to inhibit the growth of plaque on teeth. Subjects were divided into two groups: One group used a rinse with an anti plaque ingredient, and the control group used a rinse containing inactive ingredients. Suppose that the plaque growth on each person's teeth was measured after using the rinse after 4 hours and then again after 8 hours. If you wish to estimate the difference in plaque growth from 4 to 8 hours, should you use a confidence interval based on a paired or an unpaired analysis? Explain.

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