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Chapter 5: Problem 53

Exercises 5.50 to 5.55 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 z-test statistic is 13.

Step by step solution

01

Identifying the z-test statistic formula

The first step involves identifying the z-test statistic formula. The formula for the z-test statistic is given by, \[ z = \frac{\bar{x} - \mu}{SE} \], where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under null hypothesis and SE is the standard error.
02

Substituting the given values into the z-test statistic formula

Next, we substitute the given values into the z-test statistic equation. Here, \(\bar{x} = 11.3\), \(\mu = 10\), and \(SE = 0.10\). So substituting we get, \[ z = \frac{11.3 - 10}{0.10} \]
03

Solve the equation

After substituting the values into the formula, solve the equation. Doing the subtraction in the numerator first and then dividing by the denominator gives the value of the z-statistic.

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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 method used in statistics to determine whether there is enough evidence in a sample to infer a particular property about a population. It involves two opposing hypotheses: the null hypothesis (
  • Null hypothesis ( H_{0} ): This is a statement that there is no effect or difference; it is the hypothesis that the parameter takes a specific value, in this case, \(\mu=10\) .
  • Alternative hypothesis ( H_{a} ): This is the statement that contradicts the null hypothesis. It's what we suspect might be true instead, shown here as \(\mu eq 10\)
The test helps us decide if the observed data are sufficiently different from what the null hypothesis predicts. We calculate a test statistic (in our case, a z-statistic) to determine this difference. If the test statistic falls outside a pre-determined threshold (critical value), we reject the null hypothesis.
Standard Error
Standard Error (SE) is a measure of how much the sample mean is expected to vary from the true population mean. It's crucial in quantifying the precision of a sample statistic as an estimate of the population parameter. For a sample mean, SE is calculated using the formula: \[SE = \frac{s}{\sqrt{n}}\],where:
  • \(s\) represents the sample standard deviation.
  • \(n\) is the size of the sample.
The smaller the standard error, the closer the sample mean is expected to be to the population mean. In our example, an SE of 0.10 indicates a relatively small expected variation of the sample mean from the population mean if the null hypothesis were true.
Sample Mean
The sample mean, denoted as \(\bar{x}\), is the average value of all observations in a sample. It is a critical statistic used to estimate the population mean, which is a key aspect of hypothesis testing. When we compare the sample mean to the hypothesized population mean (\(\mu\)), it helps us establish whether there is substantial evidence against the null hypothesis. For instance, in our exercise, the given sample mean is 11.3, which is compared to the hypothesized population mean of 10. Calculating the standard deviation and then the standard error are essential steps in assessing whether the difference between 11.3 and 10 is statistically significant. The larger the difference between \(\bar{x}\) and \(\mu\), the stronger the evidence against the null hypothesis.

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Most popular questions from this chapter

Smoke-Free Legislation and Asthma Hospital admissions for asthma in children younger than 15 years was studied \(^{11}\) 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, both before and after the legislation was passed. 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 \(1.79 \% . \mathrm{In}\) 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?

(a) The area below 80 on a \(N(75,10)\) distribution (b) The area above 25 on a \(N(20,6)\) distribution (c) The area between 11 and 14 on a \(N(12.2,1.6)\) distribution

Exercises 5.50 to 5.55 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 \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) when the sample has \(n=50\) and \(\hat{p}=0.41,\) with \(S E=0.07\).

Penalty Shots in World Cup Soccer A study \(^{13}\) of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only \(41 \%\) of the time. The article notes that this is "slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a \(5 \%\) significance level to see whether there is evidence that the percent guessed correctly is less than \(50 \%\). The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is \(S E=0.043 .\)

In Exercises 5.44 to 5.49 , find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a difference in proportions \(p_{1}-p_{2}\) if the samples have \(n_{1}=50\) with \(\hat{p}_{1}=0.68\) and \(n_{2}=80\) with \(\hat{p}_{2}=0.61,\) and the standard error is \(S E=0.085\)
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