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The following information is available. $$\begin{array}{l}H_{0}: \mu=50 \\\H_{1}: \mu \neq 50\end{array}$$ The sample mean is \(49,\) and the sample size is \(36 .\) The population follows the normal distribution and the standard deviation is \(5 .\) Use the .05 significance level.

Short Answer

Expert verified
Fail to reject the null hypothesis; the mean is not significantly different from 50.

Step by step solution

01

State the Hypotheses

The null hypothesis is stated as \( H_0: \mu = 50 \), which means the population mean is hypothesized to be 50. The alternative hypothesis is \( H_1: \mu eq 50 \), indicating the mean is not equal to 50. This is a two-tailed test since we are testing for inequality.
02

Identify the Significance Level and Critical Value

The significance level is given as 0.05. For a two-tailed test at \( \alpha = 0.05 \), the critical values for a Z-test are approximately -1.96 and 1.96, since 0.025 in each tail corresponds to these Z-values.
03

Calculate the Test Statistic

To calculate the test statistic \( z \), use the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where \( \bar{x} = 49 \), \( \mu = 50 \), \( \sigma = 5 \), and \( n = 36 \). Plugging in the values, \[ z = \frac{49 - 50}{\frac{5}{\sqrt{36}}} = \frac{-1}{\frac{5}{6}} = \frac{-1}{0.8333} \approx -1.2 \].
04

Make a Decision

Compare the calculated test statistic \( z = -1.2 \) with the critical values. Since \(-1.96 \leq -1.2 \leq 1.96\), the test statistic falls within the range between the critical values. Thus, we fail to reject the null hypothesis \( H_0 \).
05

Conclusion

Since we failed to reject the null hypothesis, there is not enough statistical evidence at the 0.05 significance level to say that the population mean is different from 50.

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Key Concepts

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

Null Hypothesis
The null hypothesis is like a starting assumption in hypothesis testing. It is a statement that suggests there is no effect or no difference in a situation. Think of it as the "default" position that researchers assume to be true until proven otherwise. In the example provided, the null hypothesis is stated as \( H_0: \mu = 50 \). This means we assume that the population mean, \( \mu \), is 50. This hypothesis is usually tested to see if there is enough evidence to support concluding that it is false.When we say "we fail to reject the null hypothesis," it's important to note that we are not saying the null hypothesis is true. Instead, we are saying that there's not enough evidence to conclude it is false. This is why researchers say "fail to reject" rather than "accept." Ultimately, the null hypothesis serves as a baseline or a claim that our sample data attempts to refute through statistical testing.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold that researchers set to decide how much evidence is needed to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. Common levels of significance are 0.01, 0.05, and 0.10, depending on how rigorous the test needs to be.For the given exercise, the significance level is set at 0.05. This means there is a 5% risk of concluding that the population mean is different from the hypothesized mean when it really isn't. The choice of significance level depends on the context of the test and the potential consequences of making a type I error (i.e., incorrectly rejecting the null hypothesis).At a significance level of 0.05, we use critical values in standard normal (Z) distribution to determine the rejection regions. By establishing this threshold, researchers can control the rate of false positives in their work.
Test Statistic
The test statistic is a standardized value that comes from the sample data during hypothesis testing. It measures how far the sample statistic (such as the sample mean) is from the hypothesized population parameter, given the standard deviation of the data.In the provided example, the test statistic \( z \) is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size. The test statistic helps us decide in which "tail" of the distribution our sample mean lies.With a calculated \( z \) value of approximately -1.2, we evaluate whether this number falls within the critical value range. Since it lies between \(-1.96\) and \(1.96\) for a two-tailed test, this indicates that the observed sample mean isn't significantly different from the hypothesized mean at the 0.05 significance level.
Two-tailed Test
A two-tailed test in hypothesis testing checks for the possibility of the relationship in both directions—greater than or less than a certain value. It's used when we want to detect any difference from the hypothesized value, regardless of direction.In the exercise provided, since the alternative hypothesis is \( H_1: \mu eq 50 \), it indicates that we're interested in finding if the mean is either less than or greater than 50. This suggests a two-tailed test because we're looking at deviations in both directions from the hypothesized mean of 50.For two-tailed tests, the significance level \( \alpha \) is split between both tails of the distribution. At a significance level of 0.05, each tail has 0.025, leading us to the critical values of approximately \(-1.96\) and \(1.96\) when using a Z-test. The decision criteria to accept or reject the null hypothesis considers both negative and positive extremes, ensuring a comprehensive examination of potential differences.

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Most popular questions from this chapter

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