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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. Repeat the calculation of \(\beta\) for \(\mu=2.3,2.5,\) and \(2.6 .\)

Short Answer

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Question: Calculate the probability of a Type II error (尾) for population means 渭 = 2.3, 2.5, and 2.6 using the given random sample of size n = 35, mean 饾懃虆 = 2.4, and standard deviation s = 0.29. Answer: The probability of a Type II error (尾) for 渭 = 2.3, 2.5, and 2.6 are approximately 0.0017, 0.9983, and 1.0000, respectively.

Step by step solution

01

Determine the critical value

Students need to know the level of significance (伪) to determine the critical value. Without the level of significance, we cannot proceed with the problem. In most cases, 伪 = 0.05 as a default value. Assuming 伪 = 0.05, the critical value for a one-tailed t-test with 34 degrees of freedom (n-1) is approximately 1.6909 (you will typically obtain this value from a t-distribution table).
02

Calculate the standardized critical value

Divide the critical value by the standard error of the mean to find the standardized critical value: Standardized critical value = critical value / (s / sqrt(n)) Standardized critical value = 1.6909 / (0.29 / sqrt(35)) = 9.8178
03

Calculate the non-centrality parameter

For each population mean (2.3, 2.5, and 2.6), calculate the non-centrality parameter (位) as follows: 位 = (饾懃虆 - 渭) / (s / sqrt(n)) For 渭 = 2.3, 位 = (2.4 - 2.3) / (0.29 / sqrt(35)) = 3.2738 For 渭 = 2.5, 位 = (2.4 - 2.5) / (0.29 / sqrt(35)) = -3.2738 For 渭 = 2.6, 位 = (2.4 - 2.6) / (0.29 / sqrt(35)) = -6.5476
04

Calculate 尾

Using the non-central t-distribution table or calculator and the value of the standardized critical value from step 2, calculate 尾. For 渭 = 2.3, 尾(位 = 3.2738, df = 34) = 0.0017 (approx.) For 渭 = 2.5, 尾(位 = -3.2738, df = 34) = 0.9983 (approx.) For 渭 = 2.6, 尾(位 = -6.5476, df = 34) = 1.0000 (approx.) The probability of a Type II error (尾) for 渭 = 2.3, 2.5, and 2.6 are approximately 0.0017, 0.9983, and 1.0000, respectively.

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Key Concepts

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

Probability and Statistics
At the heart of any statistical analysis lies the dual concepts of probability and statistics. They form the cornerstone for understanding data, making predictions, and making informed decisions.

Probability refers to the likelihood of an event occurring, often quantified as a number between 0 and 1. For example, the flip of a fair coin has a probability of 0.5 for landing heads. Statistics, on the other hand, is the science of collecting, analyzing, interpreting, and presenting data. It allows us to make sense of the randomness and uncertainty in the data and draw conclusions about the larger population from which a sample is drawn.

Understanding Type II error, or \(\beta\), involves both of these fields. A Type II error occurs when a statistical test fails to reject a false null hypothesis - in simpler terms, it's missing the detection of an effect or difference when one actually exists. The calculation of \(\beta\) is a key component of hypothesis testing that provides important information about the power of the test 鈥 the probability that the test correctly rejects the false null hypothesis.
T-Distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but with heavier tails. It arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown.

The t-distribution is vital for conducting hypothesis tests for the population mean when these conditions are met. It is parameterized by degrees of freedom, which relates to the number of independent values in a calculation - in practice, this is typically the sample size minus one. As the sample size increases, the t-distribution approaches the normal distribution.

Understanding the t-distribution is crucial because a Type II error calculation often involves the t-statistic - a value computed from sample data that falls under the t-distribution. This is especially relevant when we're dealing with small samples from our population where the standard deviation isn't known and must be estimated.
Non-Centrality Parameter
The non-centrality parameter (NCP) or \(\lambda\), in statistical analysis, is a measure used in non-central distributions, which arise in the context of power calculations for hypothesis tests, such as the Type II error (\(\beta\)) calculations.

In a non-central t-distribution, which is used when the null hypothesis is false, the NCP quantifies the degree to which the distribution is shifted from the center. It is a function of the true mean value, the hypothesized mean value under the null hypothesis, the sample size, and the sample standard deviation. Calculating the non-centrality parameter for different true means provides an understanding of how likely it is to observe a test statistic as extreme as the calculated value, given that the null hypothesis is false.

Knowing the NCP enables us to explore the effect size and how it changes the probability of making a Type II error, allowing researchers to evaluate the sensitivity of their statistical tests effectively.
Standard Error of the Mean
The standard error of the mean (SEM) is a statistical term that tells us the precision with which we can estimate the population mean based on a sample mean. It represents the variability of sample means around the actual population mean if we were to take multiple samples from the same population.

SEM is calculated by dividing the sample standard deviation \(s\) by the square root of the sample size \(n\). A small SEM suggests that the sample mean is likely to be close to the population mean, whereas a larger SEM indicates less precision and greater variability between the sample mean and the population mean.

In the context of Type II error calculation, the SEM contributes to determining the standardized test statistic and subsequently calculating the probability of a Type II error. A crucial step in hypothesis testing, the SEM affects the critical values of the test, and understanding its role can aid in interpreting statistical results with greater accuracy.

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Most popular questions from this chapter

The weights of 3 -month-old baby girls are known to have a mean of 5.86 kilograms. \(^{2}\) Doctors at an inner city pediatric facility suspect that the average weight of 3 -month-old baby girls at their facility may be less than 5.86 kilograms. They select a random sample of 403 -month-old baby girls and find \(\bar{x}=5.56\) and \(s=0.70\) kilogram. a. Does the data indicate that the average weight of 3 -month-old baby girls at their facility is less than 5.86 kilograms? Test using \(\alpha=.05 .\) b. What is the \(p\) -value associated with the test in part a? Can you reject \(H_{0}\) at the \(5 \%\) level of significance using the \(p\) -value?

In a Pew Research report concerning the rise of automation in the United States, \(56 \%\) of the participants indicated that they would not ride in a driverless car, and \(87 \%\) favor a requirement of having a human in the driver's seat in case of an emergency. \({ }^{14}\) Suppose that the number of participants was \(n=500\). Is there sufficient evidence to conclude that more than a simple majority of Americans would not ride in a driverless car? a. Use a formal test of hypothesis with \(\alpha=.05\) to determine whether more than \(50 \%\) of Americans would not ride in a driverless car. b. Use the \(p\) -value approach. Do the two approaches lead to the same conclusion?

To determine whether there is a significant difference in the weights of boys and girls beginning kindergarten, random samples of 50 boys and 50 girls aged 5 years produced the following information: \({ }^{12}\) $$ \begin{array}{lccc} \hline & & \text { Standard } & \text { Sample } \\ & \text { Mean } & \text { Deviation } & \text { Size } \\ \hline \text { Boys } & 19.4 \mathrm{~kg} & 2.4 & 50 \\ \text { Girls } & 17.0 \mathrm{~kg} & 1.9 & 50 \end{array} $$ a. Do you have a preconceived idea of what to expect when examining the average weights of 5 -year-old boys and girls? Based on your answer, state the null and alternative hypotheses to be tested. b. Test the hypothesis in part a using \(\alpha=.05\).

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.

Does Mars, Inc. use the same proportion of red M\&M'S in its plain and peanut varieties? Random samples of plain and peanut M\&M'S provide the following sample data: $$ \begin{array}{lcc} \hline & \text { Plain } & \text { Peanut } \\ \hline \text { Sample Size } & 56 & 32 \\ \text { Number of Red M\&M'S } & 12 & 8 \end{array} $$ a. Use a test of hypothesis to determine whether there is a significant difference in the proportions of red candies for the two types of M\&M'S. Let \(\alpha=.05 .\) b. Calculate a \(95 \%\) confidence interval for the difference in the proportion of red candies for the two types of M\&M'S. Does this interval confirm your results in part a?

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