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Chapter 9: Problem 14

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation of \(s=.29 .\) Your research objective is to show that the population mean \(\mu\) exceeds 2.3. Use this information to answer the questions. Give the null and alternative hypotheses for the test.

Short Answer

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= (0.1) / (0.29 / 鈭�35) = (0.1) / (0.29 / 5.916) = (0.1) / 0.049 鈮� 2.04 #tag_title#Step 3: Determine the critical value and p-value#tag_content# To determine the critical value, first find the degrees of freedom (df), which is the sample size (n) minus 1. df = n - 1 df = 35 - 1 df = 34 Using a t-distribution table or software, find the critical t-value for a one-tailed test at a 0.05 significance level (伪) with 34 degrees of freedom. Here, we find that the critical t-value is approximately 1.69. Now, by comparing the t-value (2.04) with the critical t-value (1.69), we see that the t-value is greater than the critical t-value. This means that we reject the null hypothesis in favor of the alternative hypothesis. To determine the p-value, we can use a calculator or software for the t-distribution. For a t-value of 2.04 and 34 degrees of freedom, the p-value is approximately 0.025. Since the p-value (0.025) is less than the significance level (0.05), we reject the null hypothesis and accept the alternative hypothesis. #tag_title#Conclusion#tag_content# The hypothesis test results show that there is enough evidence to support the claim that the population mean is greater than 2.3, at a 0.05 significance level.

Step by step solution

01

Define the null and alternative hypotheses

The null hypothesis (H鈧�) states that there is no difference or effect, and in this case, it assumes that the population mean 碌 is equal to 2.3. The alternative hypothesis (H鈧�) is the opposite of the null hypothesis and, in this case, suggests that the population mean 碌 is greater than 2.3. Thus, we have: H鈧�: 碌 = 2.3 H鈧�: 碌 > 2.3
02

Calculate the test statistic

To perform the hypothesis test, we will use one-sample t-test. First, calculate the t-value as follows: t = (x虅 - 碌鈧�) / (s / 鈭歯) Where: - x虅 is the sample mean - 碌鈧� is the value of the population mean under the null hypothesis - s is the sample standard deviation - n is the sample size In this case: t = (2.4 - 2.3) / (0.29 / 鈭�35)

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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 null and alternative hypotheses is crucial when embarking on hypothesis testing. The null hypothesis, typically denoted as H鈧�, is a statement suggesting that there is no effect or no difference, and it serves as the starting assumption for the testing procedure. In contrast, the alternative hypothesis, symbolized as H鈧� or Ha, asserts that there is a significant effect or a difference.

For instance, assume that you're tasked with establishing whether a new teaching method improves student grades. Your null hypothesis would state that the mean student grades with the new method are equal to the grades without it (no improvement). Conversely, the alternative hypothesis would suggest that the grades have indeed improved with the new teaching method.

During hypothesis testing, one attempts to accumulate evidence through data to reject the null hypothesis. If sufficient evidence is found, the null hypothesis is rejected in favor of the alternative. If not, the null hypothesis remains in place, though it is not necessarily proven to be true.
One-Sample t-Test
The one-sample t-test is a statistical procedure used when comparing the mean of a single sample to a known population mean, especially when the population standard deviation is unknown. Under this method, the t-statistic is calculated 鈥� which measures the size of the difference relative to the variation in your sample data.

Using the formula:
\[ t = \frac{{\bar{x} - \mu_0}}{{s / \sqrt{n}}} \]
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. The calculated t-value is then compared against a critical value from the t-distribution table based on the degrees of freedom and the level of significance desired. This comparison will determine if the results are statistically significant, allowing us to confidently accept or reject the null hypothesis.
Population Mean
The population mean, denoted as \( \mu \), is a parameter that represents the average of all possible observations in the group you're studying. It's a critical value in statistics because it provides a central point or the expected value for a population's data distribution.

However, since it's often impractical to measure every member of a population, we typically estimate the population mean using a representative sample. The sample mean, \( \bar{x} \), serves as a close approximation of the population mean. The accuracy of this estimate increases with the size of the sample and the randomness of sample selection. Understanding the population mean is vital to making inferences about the population based on sample data, as is commonly the case in hypothesis testing.
Standard Deviation
The standard deviation (denoted as \( s \) in samples and \( \sigma \) in populations) quantifies how much variation or dispersion exists from the average (mean) in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation points to more spread out values.

In the context of a one-sample t-test, the standard deviation is used to help measure the variability of your sample. It plays a critical role as part of the denominator in the t-test formula, where it is adjusted for the sample size using the square root of \( n \). This adjustment, known as the standard error, standardizes the deviation and is pivotal in calculating the t-value for hypothesis testing.

Understanding and accurately calculating standard deviation is important not only for hypothesis testing but also for interpreting data dispersion and the reliability of mean as a measure of central tendency.

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Most popular questions from this chapter

An experiment was conducted to test the effect of a new drug on a viral infection. After the infection was induced in 100 mice, the mice were randomly split into two groups of \(50 .\) The control group received no treatment for the infection, while the other group received the drug. After a 30 -day period, the proportions of survivors, \(\hat{p}_{1}\) and \(\hat{p}_{2},\) in the two groups were found to be .36 and \(.60,\) respectively. a. Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use \(\alpha=.05\). b. Use a \(95 \%\) confidence interval to estimate the actual difference in the survival rates for the treated versus the control groups.

Has the consumption of red meat decreased over the last 10 years? A researcher selected hospital nutrition records for 400 subjects surveyed 10 years ago and compared the average amount of beef consumed per year to amounts consumed by an equal number of subjects interviewed this year. The data are given in the table. $$ \begin{array}{lcc} \hline & \text { Ten Years Ago } & \text { This Year } \\ \hline \text { Sample Mean } & 73 & 63 \\ \text { Sample Standard Deviation } & 25 & 28 \\ \hline \end{array} $$ a. Do the data present sufficient evidence to indicate that per-capita beef consumption has decreased over the last 10 years? Test at the \(1 \%\) level of significance. b. Find a \(99 \%\) lower confidence bound for the difference in the average per-capita beef consumptions for the two groups. Does the confidence bound confirm your conclusions in part a? Explain. What additional information does the confidence bound give you?

Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a screening program for the early detection of breast cancer was started in order to increase the survival rate \(p\) of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were screened by the program and who were diagnosed to have the disease. Let \(x\) represent the number of those in the sample who survive the disease. a. If you wish to determine whether the screening program has been effective, state the alternative hypothesis that should be tested. b. State the null hypothesis. c. If 164 women in the sample of 200 survive the disease, can you conclude that the screening program was effective? Test using \(\alpha=.05\) and explain the practical conclusions from your test. d. Find the \(p\) -value for the test and interpret it.

Although there is a big difference in costs between a Tesla \(S\) and a Chevrolet Bolt, is there a difference in the range of miles for these vehicles between charges? Suppose that the results of driving tests are as follows: \({ }^{-10}\) $$ \begin{array}{lccc} \hline & & \text { Standard } & \text { Sample } \\ \text { Car } & \text { Mean } & \text { Deviation } & \text { Size } \\ \hline \text { Tesla S } & 230.2 & 14.3 & 40 \\ \text { Bolt } & 236.8 & 18.8 & 40 \\ \hline \end{array} $$ Is there sufficient evidence to indicate a difference in the average driving ranges for these two vehicles? Test using \(\alpha=.01\)

In a head-to-head taste test of storebrand foods versus national brands, Consumer Reports found that it was hard to find a taste difference in the two. \(^{16}\) If the national brand is indeed better than the store brand, it should be judged as better more than \(50 \%\) of the time. a. State the null and alternative hypothesis to be tested. Is this a one- or a two-tailed test? b. Suppose that, of the 35 food categories used for the taste test, the national brand was found to be better than the store brand in eight categories. Use this information to test the hypothesis in part a with \(\alpha=.01 .\) What practical conclusions can you draw from the results?
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