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91Ó°ÊÓ

The \(z\) statistic for a one-sided test is \(z=2.29\). This test is (a) not statistically significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) statistically significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) statistically significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

Short Answer

Expert verified
(b) The test is statistically significant at \(\alpha=0.05\) but not at \(\alpha=0.01\).

Step by step solution

01

Understanding the Z Statistic

The Z statistic helps determine whether to reject the null hypothesis in hypothesis testing. It compares the sample data to the null hypothesis, allowing us to calculate the p-value to measure statistical significance.
02

Determine the Critical Z-Values

For a one-sided test at a significance level of \(\alpha=0.05\), the critical Z-value is 1.645. At \(\alpha=0.01\), the critical Z-value is 2.33. These values represent the minimum threshold needed to reject the null hypothesis.
03

Compare the Given Z-Statistic to Critical Values

Given \(z=2.29\), we compare it against the critical Z-values. Since 2.29 is greater than 1.645 but less than 2.33, the test is significant at \(\alpha=0.05\) but not at \(\alpha=0.01\).
04

Identify the Correct Statistical Significance

Since the calculated Z-value is not greater than 2.33, the result does not reach significance at \(\alpha=0.01\), but it does exceed 1.645, making it significant at \(\alpha=0.05\).

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Key Concepts

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

Z-statistic
The Z-statistic is a measure used in hypothesis testing when dealing with population samples. It tells us how far, in standard deviations, our sample mean is from the population mean under the null hypothesis. This statistic is particularly useful when the sample size is large (typically over 30), as the Central Limit Theorem assures us the distribution of the sample mean will be approximately normal.

The formula for calculating the Z-statistic is given by:\[Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}\]where:
  • \(\bar{x}\) is the sample mean,
  • \(\mu\) is the population mean,
  • \(\sigma\) is the population standard deviation,
  • \(n\) is the sample size.
Once the Z-statistic is calculated, it is compared against critical values from the standard normal distribution to determine statistically significant differences between the sample and population means.
Significance Level
The significance level, denoted by \(\alpha\), is the threshold we set to decide whether our statistical test results are meaningful. Common significance levels used in hypothesis testing are 0.05 and 0.01. Essentially, a significance level is the probability of rejecting the null hypothesis when it is actually true, also known as the "Type I error rate."

Choosing a lower \(\alpha\) (e.g., 0.01 instead of 0.05) means we have stricter criteria for claiming a significant result, reducing the risk of a Type I error. However, this also increases the chance of a Type II error, where the test fails to detect a true effect. In the context of the given exercise, the test was significant at \(\alpha=0.05\) because the Z-statistic exceeded the critical value of 1.645. However, it was not significant at \(\alpha=0.01\), since the Z-statistic did not exceed 2.33, the higher threshold for significance.
Null Hypothesis
The null hypothesis, abbreviated as \(H_0\), forms a critical part of hypothesis testing. It is the starting assumption that there is no effect or difference, and any observed variation is due to random chance. In statistical terms, the null hypothesis often posits that there is no association between variables, or no difference between groups.

The main goal of hypothesis testing is to determine whether there is enough evidence in the sample data to reject \(H_0\). If the evidence is strong enough, indicated by the test statistic and significance level, we reject the null hypothesis. Otherwise, we fail to reject it. Importantly, failing to reject \(H_0\) does not prove it is true; it merely indicates insufficient evidence against it. In our example, the preset threshold for the test was exceeded at \(\alpha=0.05\), suggesting enough evidence to reject \(H_0\) at this level.
P-Value
The p-value is a measure of the probability that the observed data would occur by random chance under the null hypothesis. It is calculated during hypothesis testing and serves as a decision tool for determining statistical significance.

If the p-value is less than the chosen significance level, \(\alpha\), we have strong evidence to reject the null hypothesis. This indicates the observed data are unlikely due to random variation alone. A small p-value suggests that the null hypothesis is less likely to be true.

For the exercise provided, the p-value associated with the Z-statistic of 2.29 would be calculated and compared against the significance levels (\(\alpha=0.05\) and \(\alpha=0.01\)). Since the Z-statistic was significant at 0.05 but not at 0.01, the corresponding p-value would be smaller than 0.05 but larger than 0.01, indicating a result that is statistically significant at the 0.05 level.

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Most popular questions from this chapter

Successful hotel managers must have personality characteristics traditionally stereotyped as feminine (such as "compassionate") as well as those traditionally stereotyped as masculine (such as "forceful"). The Bem Sex-Role Inventory (BSRI) is a personality test that gives separate ratings for "female" and "male" stereotypes, both on a scale of 1 to 7 . Although the BSRI was developed at a time when these stereotypes were more pronounced, it is still widely used to assess personality types. Unfortunately, the ratings are often referred to as femininity and masculinity scores. A sample of 148 male general managers of three-star and four-star hotels had mean BSRI femininity score \(\mathrm{x}^{-} \bar{x}=5.29 .{ }^{6}\) The mean score for the general male population is \(\mu=5.19\). Do hotel managers, on the average, differ statistically significantly in femininity score from men in general? Assume that the standard deviation of scores in the population of all male hotel managers is the same as the \(\sigma=0.78\) for the adult male population. (a) State null and alternative hypotheses in terms of the mean femininity score \(\mu\) for male hotel managers. (b) Find the \(z\) test statistic. (c) What is the \(P\)-value for your \(z\) ? What do you conclude about male hotel managers?

The examinations in a large multisection statistics class are scaled after grading so that the mean score is 75 . The professor thinks that students in the 8:00 A.M. class have trouble paying attention because they are sleepy and suspects that these students have a lower mean score than the class as a whole. The students in the 8:00 A.M. class this semester can be considered a sample from the population of all students in the course, so the professor compares their mean score with 75 . State the hypotheses \(H_{0}\) and \(H_{a}\).

A test of \(H_{0}: \mu=0\) against \(H_{a}: \mu \neq 0\) 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 \((\alpha=0.01)\) ?

A randomized comparative experiment examined the effect of the attractiveness of an instructor on the performance of students on a quiz given by the instructor. The researchers found a statistically significant difference in quiz scores between students in a class with an instructor rated as attractive and students in a class with an instructor rated as unattractive \((P=\) \(0.005) \cdot{ }^{7}\) When asked to explain the meaning of " \(P=0.005\)," a student says, "This means there is only probability of \(0.005\) that the null hypothesis is true." Explain what \(P=0.005\) really means in a way that makes it clear that the student's explanation is wrong.

A random number generator is supposed to produce random numbers that are uniformly distributed on the interval from 0 to 1 . If this is true, the numbers generated come from a population with \(\mu=0.5\) and \(\sigma=0.2887\). A command to generate 100 random numbers gives outcomes with mean \(x^{-} \bar{x}=0.4365\). Assume that the population \(\sigma\) remains fixed. We want to test $$ \begin{aligned} &H_{0}: \mu=0.5 \\ &H_{a}: \mu \neq 0.5 \end{aligned} $$ (a) Calculate the value of the \(z\) test statistic. (b) Use Table \(\mathrm{C}\) : is \(z\) statistically significant at the \(5 \%\) level \((\alpha=0.05)\) ? (c) Use Table \(\mathrm{C}\) : is \(z\) statistically significant at the \(1 \%\) level \((\alpha=0.01)\) ? (d) Between which two Normal critical values \(z *\) in the bottom row of Table \(C\) does \(z\) lie? Between what two numbers does the \(P\)-value lie? Does the test give good evidence against the null hypothesis?

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