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Chapter 17: Problem 16

A test of \(H_{0}: \mu=0\) against \(H_{a}: \mu>0\) has test statistic \(z=1.65\). Is this test statistically significant at the \(5 \%\) level \((\alpha=0.05)\) ? Is it statistically significant at the \(1 \%\) level \((\alpha=0.01)\) ?

Short Answer

Expert verified
Yes, significant at 5% but not at 1% level.

Step by step solution

01

Understand the Test Context

We are testing a hypothesis where the null hypothesis is \( H_0: \mu = 0 \) and the alternative hypothesis is \( H_a: \mu > 0 \). We are given a test statistic \( z = 1.65 \) and need to determine if the results are statistically significant at the \( 5\% \) and \( 1\% \) significance levels.
02

Significance Level and Critical Value (5% level)

At a significance level of \( \alpha = 0.05 \), we need to find the critical value for a one-tailed test (right-tailed test). Using the standard normal distribution table, the critical value is \( z_{0.05} = 1.645 \).
03

Evaluate 5% Significance

Compare the test statistic \( z = 1.65 \) with the critical value at 5% level (\( z_{0.05} = 1.645 \)). Since \( 1.65 > 1.645 \), the test statistic exceeds the critical value, indicating statistical significance at the \( 5\% \) level.
04

Significance Level and Critical Value (1% level)

At a significance level of \( \alpha = 0.01 \), find the critical value for a one-tailed test. From the standard normal distribution table, the critical value is \( z_{0.01} = 2.33 \).
05

Evaluate 1% Significance

Compare the test statistic \( z = 1.65 \) with the critical value at 1% level (\( z_{0.01} = 2.33 \)). Since \( 1.65 < 2.33 \), the test statistic does not exceed the critical value, indicating no statistical significance at the \( 1\% \) level.

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Key Concepts

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

Significance Level
When conducting a hypothesis test, the significance level, denoted by \( \alpha \), is of great importance. The significance level helps us determine the threshold for rejecting the null hypothesis. This threshold acts as a boundary, representing the probability of rejecting the null hypothesis when it is actually true. This is also known as the Type I error probability.For example, a 5% significance level, or \( \alpha = 0.05 \), means we are accepting a 5% chance of committing a Type I error. Here鈥檚 how it works in practice:
  • We set \( \alpha = 0.05 \), and then identify the critical values marking the boundary of rejection for a hypothesis test.
  • We compare the test statistic to this critical value to determine statistical significance.
By adjusting the significance level, researchers can control their levels of certainty and potentially strengthen or weaken the conclusions drawn from the test. A smaller \( \alpha \) means a stricter threshold for statistical significance, while a larger \( \alpha \) implies a more lenient one. This is essential when validating the outcomes from experimental data.
Critical Value
Critical values are pivotal in hypothesis testing as they define the boundary between the acceptance region and the rejection region. These values are derived from the chosen significance level \( \alpha \) and the distribution model used for the test.In the case of standard normal distribution, we use a Z-table:
  • For a 5% significance level in a one-tailed test, the critical value is \( z_{0.05} = 1.645 \). This means any test statistic greater than 1.645 leads to rejection of the null hypothesis.
  • For a 1% significance level, the critical value changes to \( z_{0.01} = 2.33 \). The test statistic must exceed this value to reject the null hypothesis, reflecting a stricter criterion.
By comparing these critical values to the test statistic, we can infer whether the evidence against the null hypothesis is strong enough. Critical values serve as the yardstick in hypothesis testing, decisively influencing the interpretation of the test outcomes.
Standard Normal Distribution
Understanding the standard normal distribution is essential in hypothesis testing. This distribution forms the basis for determining critical values and assessing test statistics. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.This symmetric bell-shaped curve is central to many statistical tests, including Z-tests.
  • The values along this curve represent Z-scores, which are the number of standard deviations a data point is from the mean.
  • A Z-score allows us to calculate the probability of observing a data point within a certain range, and further helps derive critical values by considering area under the curve.
  • The 0.05 significance level corresponds to the critical z-value of \( z = 1.645 \) in the standard normal distribution for a one-tailed test.
  • The 0.01 significance level, a stricter criterion, corresponds to \( z = 2.33 \).
By using the standard normal distribution, statisticians ensure that hypothesis testing is grounded in uniform probability concepts, making the outcomes more reliable and understandable, regardless of sample sizes or means. It's truly a foundational tool in statistical analysis.

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Most popular questions from this chapter

You are testing \(H_{0}: \mu=0\) against \(H_{\mathrm{a}}: \mu \neq 0\) based on an SRS of 20 observations from a Normal population. What values of the \(z\) statistic are statistically significant at the \(\alpha=0.005\) level? (a) All values for which \(z>2.576\) (b) All values for which \(z>2.807\) (c) All values for which \(|z|>2.807\)

The gas mileage for a particular model SUV varies, but is known to have a standard deviation of \(\sigma=1.0\) mile per gallon in repeated tests in a controlled laboratory environment at a fixed speed of 65 miles per hour. For a fixed speed of 65 miles per hour, gas mileages in repeated tests are Normally distributed. Tests on three SUVs of this model at 65 miles per hour give gas mileages of \(19.3,19.9\), and \(19.8\) miles per gallon. The \(z\) statistic for testing \(H_{0}: \mu=20\) miles per gallon based on these three measurements is (a) \(z=-0.333\). (b) \(z=-0.577\). (c) \(z=0.577 .\)

The average income of American women who work fulltime and have only a high school degree is \(\$ 35,713\). You wonder whether the mean income of female graduates from your local high school who work full-time but have only a high school degree is different from the national average. You obtain income information from an SRS of 62 female graduates who work fulltime and have only a high school degree and find that \(x^{-} \bar{x}=\$ 35,053\). What are your null and alternative hypotheses?

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a noise as stimulus. The sample mean is \(\mathrm{x}^{-} \bar{x}=16.5\) seconds. The null hypothesis for the test of significance is (a) \(H_{0}: \mu=18\) (b) \(H_{0}: \mu=16.5\). (c) \(H_{0}: \mu<18\).

A confidence interval for the population mean \(\mu\) tells us which values of \(\mu\) are plausible (those inside the interval) and which values are not plausible (those outside the interval) at the chosen level of confidence. You can use this idea to carry out a test of any null hypothesis \(H_{0}: \mu=\mu_{0}\) starting with a confidence interval: reject \(H_{0}\) if \(\mu_{0}\) is outside the interval and fail to reject if \(\mu_{0}\) is inside the interval. The alternative hypothesis is always two-sided, \(H_{a}: \mu \neq \mu_{0}\) because the confidence interval extends in both directions from \(\mathrm{x}^{-} \bar{x}\). A \(95 \%\) confidence interval leads to a test at the \(5 \%\) significance level because the interval is wrong \(5 \%\) of the time. In general, confidence level \(C\) leads to a test at significance level \(\alpha=1-C\). (a) In Example 17.7, a medical director found mean blood pressure \(\mathrm{x}^{-} \bar{x}=\) \(128.07\) for an SRS of 72 male executives between the ages of 50 and 59 . The standard deviation of the blood pressures of all males \(50-59\) years of age is \(\sigma=15\). Give a \(90 \%\) confidence interval for the mean blood pressure \(\mu\) of all executives in this age group, assuming the standard deviation is the same as for all males \(50-59\) years of age. (b) The hypothesized value \(\mu_{0}=130\) falls inside this confidence interval. Carry out the \(z\) test for \(H_{0}: \mu=130\) against the two-sided alternative. Show that the test is not statistically significant at the \(10 \%\) level. (c) The hypothesized value \(\mu_{0}=131\) falls outside this confidence interval. Carry out the \(z\) test for \(H_{0}: \mu=131\) against the two-sided alternative. Show that the test is statistically significant at the \(10 \%\) level.
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