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Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu=10\) vs \(H_{a}: \mu \neq 10\) when the sample has \(n=75, \bar{x}=11.3,\) and \(s=0.85,\) with \(S E=0.10\).

Short Answer

Expert verified
The value of the standardized z-test statistic in this case is 13.0.

Step by step solution

01

Identifying the Given Values

We start by identifying the given values from the problem statement, which include: \n\nSample size (\( n \)) = 75\nSample mean (\( \bar{x} \)) = 11.3\nStandard deviation (\( s \)) = 0.85\nStandard error (\( SE \)) = 0.10\nHypothesized mean (\( \mu_0 \)) = 10
02

Formula for Z-Score Calculation

The formula to calculate the z-score in this case is defined as \[ Z = \frac{\bar{x}-\mu_0}{SE} \] where, \n\n\( \bar{x} \) = Sample Mean\n\( \mu_0 \) = Hypothesized Mean\n\( SE \) = Standard Error
03

Calculating the Z-Score

Insert the given values into the z-score formula \[ Z = \frac{\bar{x}-\mu_0}{SE} = \frac{11.3 – 10.0}{0.10} = 13.0\] Thus, z-score is 13.00

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions based on data analysis. It is widely used in various fields to determine if there is enough evidence to support a specific hypothesis about a population parameter, such as the mean.

When conducting hypothesis testing, we set up two opposing hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)). The null hypothesis usually states that there is no effect or difference, while the alternative hypothesis proposes that there is some effect or difference. In the context of this problem, the null hypothesis is \( H_0: \mu = 10 \), indicating that the population mean equals 10. The alternative hypothesis is \( H_a: \mu eq 10 \), suggesting the population mean is not 10.

The aim is to gather evidence in favor of either \( H_0 \) or \( H_a \), using statistical data from a sample. Calculating the z-test statistic, as in this case, helps determine how far our sample mean is from the hypothesized population mean, measured in terms of standard error. If the \( z \)-score is beyond a certain threshold, we can reject the null hypothesis.
Standard Error
The standard error (SE) measures the variability or dispersion of a sample statistic, like the sample mean, from an experiment or study. It essentially tells us how much the sample mean would vary if we were to take multiple samples from the population.

Standard error is calculated as the sample standard deviation (\( s \)) divided by the square root of the sample size (\( n \)):\[ SE = \frac{s}{\sqrt{n}} \]

In practical terms, a smaller standard error signifies that the sample mean is likely to be closer to the population mean, providing more confidence in our sample estimate. In hypothesis testing, standard error is crucial as it is part of the calculation for the z-test statistic, allowing us to understand the reliability of our results. In this exercise, the given standard error of 0.10 indicates the variability of the sample mean when estimating the population mean.
Sample Mean
The sample mean (\( \bar{x} \)) is a key statistic computed from a dataset and represents the average value of the observed data. It is a point estimate of the population mean, providing us with a snapshot of what the typical value might be within the population.

For example, when you have a dataset of 75 observations (as in this exercise), and the average value calculated is 11.3, this result is the sample mean. The sample mean can fluctuate from sample to sample, depending on randomly drawn observations.

By calculating the sample mean, statisticians can make inferences about the population mean, especially when combined with the standard error to perform hypothesis testing. The closer the sample mean is to the hypothesized population mean, relative to the standard error, the more likely the null hypothesis might be true. However, in our case, a sample mean of 11.3 considerably differs from the hypothesized mean of 10, which results in a high z-score when performing hypothesis testing.

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Most popular questions from this chapter

How Do You Get Your News? A study by the Pew Research Center \(^{9}\) reports that in \(2010,\) for the first time, more adults aged 18 to 29 got their news from the Internet than from television. In a random sample of 1500 adults of all ages in the US, \(66 \%\) said television was one of their main sources of news. Does this provide evidence that more than \(65 \%\) of all adults in the US used television as one of their main sources for news in \(2010 ?\) A randomization distribution for this test shows \(S E=0.013\). Find a standardized test statistic and compare to the standard normal to find the p-value. Show all details of the test.

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(90 \%\) confidence interval for a mean \(\mu\) if the sample has \(n=30\) with \(\bar{x}=23.1\) and \(s=5.7\), and th standard error is \(S E=1.04\)

Smoke-Free Legislation and Asthma Hospital admissions for asthma in children younger than 15 years was studied \(^{22}\) in Scotland both before and after comprehensive smoke-free legislation was passed in March \(2006 .\) Monthly records were kept of the annualized percent change in asthma admissions. For the sample studied, before the legislation, admissions for asthma were increasing at a mean rate of \(5.2 \%\) per year. The standard error for this estimate is \(0.7 \%\) per year. After the legislation, admissions were decreasing at a mean rate of \(18.2 \%\) per year, with a standard error for this mean of \(1.79 \% .\) In both cases, the sample size is large enough to use a normal distribution. (a) Find and interpret a \(95 \%\) confidence interval for the mean annual percent rate of change in childhood asthma hospital admissions in Scotland before the smoke-free legislation. (b) Find a \(95 \%\) confidence interval for the same quantity after the legislation. (c) Is this an experiment or an observational study? (d) The evidence is quite compelling. Can we conclude cause and effect?

In Exercises 5.15 to 5.17 , find the specified areas for a normal distribution. For a \(N(160,25)\) distribution (a) The area to the right of 140 (b) The area to the left of 200

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