91Ó°ÊÓ

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?

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis \( H_0 \) states that there is no difference between the mean femininity score of male hotel managers and the general male population, that is \( H_0: \mu = 5.19 \). The alternative hypothesis \( H_a \) states that there is a difference, that is \( H_a: \mu eq 5.19 \).

02

Calculate the Z Test Statistic

To find the \( z \) test statistic, use the formula:\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]Substitute \( \bar{x} = 5.29 \), \( \mu = 5.19 \), \( \sigma = 0.78 \), and \( n = 148 \):\[ z = \frac{5.29 - 5.19}{\frac{0.78}{\sqrt{148}}} \]\[ z \approx \frac{0.10}{0.0641} \approx 1.56 \]

03

Determine the P-Value

For \( z = 1.56 \), the P-value can be found using a standard normal distribution table. Since this is a two-tailed test, we look for twice the tail area for \( |z| > 1.56 \). The P-value is approximately 0.1199.

04

Make a Conclusion

The P-value of 0.1199 is greater than the common significance level of 0.05. Therefore, we do not reject the null hypothesis. This indicates that there is not enough statistical evidence to support a claim that male hotel managers have a significantly different mean femininity score compared to the general male population.

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Key Concepts

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

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement we aim to test. It represents a condition of "no effect" or "no difference." In this scenario, our null hypothesis (\( H_0 \)) states that the mean femininity score of male hotel managers is equal to that of the general male population.

• \( H_0: \mu = 5.19 \)

Here, \( \mu \) denotes the mean femininity score for male hotel managers. By setting up this hypothesis, our objective is to see if statistical evidence suggests a deviation from this assumed average.
This assumption helps in understanding whether any observed difference in means is significant or merely due to random chance.

Alternative Hypothesis

The alternative hypothesis is the opposite of the null hypothesis, and it provides a contrast for testing statistical significance. In this case, the alternative hypothesis (\( H_a \)) suggests that there is a difference between the average femininity score of male hotel managers and that of the general male population.

• \( H_a: \mu eq 5.19 \)

This is a two-tailed hypothesis, meaning that the test checks for a difference in either direction (greater or less) from the mean score. By testing \( H_a \), analysts can determine if the sample provides enough evidence to conclude the existence of a meaningful difference.

Z Test Statistic

The Z Test Statistic helps determine whether to reject the null hypothesis. It calculates the deviation of the sample mean from the population mean, adjusted by the standard deviation. Its formula is:\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]In this exercise, we substitute \( \bar{x} = 5.29 \), \( \mu = 5.19 \), \( \sigma = 0.78 \), and \( n = 148 \):\[ z = \frac{5.29 - 5.19}{\frac{0.78}{\sqrt{148}}} \]This simplifies to:\[ z \approx 1.56 \]The Z value helps understand how far the sample mean is from the population average in terms of the number of standard deviations. A larger Z value indicates a larger deviation from the mean.

P-Value

The P-Value quantifies the evidence against the null hypothesis. It's a probability measure of obtaining a test statistic equal to or more extreme than the observed one, under the assumption that the null hypothesis is true.For a Z value of 1.56, one can consult a standard normal distribution table to find the tail area associated with this Z. In a two-tailed test, like this one, we multiply the single-tail area by 2.Here, the approximate P-value is 0.1199. This result helps in decision-making:- If the P-value is less than the chosen significance level (\( \alpha \), say 0.05), it indicates strong evidence against the null hypothesis.- Since 0.1199 > 0.05, we fail to reject \( H_0 \), suggesting insufficient evidence to claim a significant difference in scores.

Standard Deviation

Standard deviation measures the amount of variability or dispersion in a set of data points. In the context of this hypothesis test, it reflects how much femininity scores vary among the population.With \( \sigma = 0.78 \), this standard deviation means the femininity scores of the general male population vary on average by 0.78 units from the mean. It's crucial for calculating the Z test statistic as it contextualizes sample variation:- A lower standard deviation indicates more clustered data around the mean.- Larger standard deviations imply a wider spread of scores.Understanding standard deviation is pivotal in assessing whether sample findings are typical or an indication of possible differences from population characteristics.