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Multiple choice: Select the best answer for Exercises 29 to 32. A sample survey interviews SRSs of 500 female college students and 550 male college students. Each student is asked whether he or she worked for pay last summer. In all, 410 of the women and 484 of the men say 鈥淵es.鈥� Exercises 29 to 31 are based on this survey. In an experiment to learn whether Substance M can help restore memory, the brains of 20 rats were treated to damage their memories. The rats were trained to run a maze. After a day, 10 rats (determined at random) were given M and 7 of them succeeded in the maze. Only 2 of the 10 control rats were successful. The two-sample z test for 鈥渘o difference鈥� against 鈥渁 significantly higher proportion of the M group succeeds鈥� (a) gives \(z=2.25, P<0.02\) (b) gives \(z=2.60, P<0.005\) (c) gives \(z=2.25, P<0.04\) but not \(<0.02\) (d) should not be used because the Random condition is violated. (e) should not be used because the Normal condition is violated.

Step by step solution

01

Define the Hypotheses

First, define the null and alternative hypotheses for the two-sample z test. The null hypothesis \( H_0 \) is that there is no difference in the proportions of successful maze-running rats between the treatment and control groups. The alternative hypothesis \( H_a \) is that the proportion in the M group is significantly higher.

02

Identify Sample Proportions

Calculate the sample proportions for both groups. For the M group: \( \hat{p}_1 = \frac{7}{10} = 0.7 \). For the control group: \( \hat{p}_2 = \frac{2}{10} = 0.2 \).

03

Calculate the Pooled Proportion

The pooled proportion \( \hat{p} \) is found using the formula \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{7 + 2}{10 + 10} = 0.45 \].

04

Calculate the Standard Error

Compute the standard error of the difference in proportions. \[ SE = \sqrt{\hat{p}(1-\hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.45(1-0.45)\left(\frac{1}{10} + \frac{1}{10}\right)} = 0.213 \]

05

Compute the Z-statistic

The z-statistic is computed by \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.7 - 0.2}{0.213} = 2.35 \].

06

Decide on the Best Answer

Based on the z statistic, a z value of 2.35 does not precisely match any option, but closest realistic comparisons are options (a) \( z=2.25, P<0.02\) and (c) \( z=2.25, P<0.04\) but not \(<0.02\). The latter matches the level of certainty around the computed z.

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Key Concepts

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

Sample Proportions

In statistics, sample proportions help us to understand the characteristics of a smaller, more manageable group of data, which is representative of a larger group or population. In the given exercise, the sample proportions are vital as they summarize how many rats were successful within each group: treatment and control.

• The proportion for the M group is calculated by dividing the number of successes (7 rats) by the total number of M-treated rats (10). So, the sample proportion for this group is 0.7.
• Similarly, the proportion for the control group is calculated by dividing the number of successes (2 rats) by the total number in the control group (10), giving a sample proportion of 0.2.

These proportions are the foundation for conducting a two-sample z test, as they provide the numerical summary needed to explore differences in successes between the groups.

Null Hypothesis

The null hypothesis (\( H_0 \)) is a crucial concept in hypothesis testing, representing a statement that there is no effect or difference. It is the default assumption that any observed difference is due to sampling or experimental error rather than a real effect.
In this exercise, the null hypothesis posits that there is no difference in the success proportions between the rats treated with Substance M and the control rats. Mathematically, this can be stated as: \( H_0: p_1 = p_2 \), where \( p_1 \) and \( p_2 \) are the population proportions of successful maze runs by M-treated and control rats, respectively.Understanding this hypothesis helps in determining if any calculated differences can be viewed as statistically significant or just a result of chance.

Alternative Hypothesis

Unlike the null hypothesis, the alternative hypothesis (\( H_a \)) suggests a specific difference between groups. It is what researchers aim to support - the change or difference they want to prove.
In the context of the exercise, the alternative hypothesis is that the success proportion for the M-treated rats is higher than that of the control group. This can be written as: \( H_a: p_1 > p_2 \). Here, \( p_1 \) and \( p_2 \) represent the population proportions of rats that successfully ran the maze in M-treated and control groups, respectively.The alternative hypothesis guides the direction of the test, indicating that we're particularly interested in finding evidence that supports a higher success rate in the M group.

Pooled Proportion

The pooled proportion is a combined estimate of the success observations from both groups, calculated to provide a weighted average of the proportions.
It offers a single proportion under the assumption that there is no difference between groups, and it's utilized in calculating the standard error required for the z-test.
The formula to find the pooled proportion \( \hat{p} \) in the exercise is:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2}\]where \( x_1 \) and \( x_2 \) are the number of successes in each group, and \( n_1 \) and \( n_2 \) are the total number of observations in each. For this case, \( \hat{p} = 0.45 \), indicating that 45% of all rats combined were successful.This pooled proportion is then used to understand the expected standard error if there were no true difference.

Standard Error

Standard error provides insight into how much the sample proportion can be expected to fluctuate due to random sampling. It's essential in determining the reliability of our test results.
In this exercise, the standard error of the difference between two proportions (\( SE \)) is calculated using the formula:\[ SE = \sqrt{\hat{p}(1-\hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\]where \( \hat{p} \) is the pooled proportion, and \( n_1 \) and \( n_2 \) are sample sizes of the two groups.
Calculating standard error helps assess if the observed differences between sample proportions can be attributed to actual differences in population proportions, considering sampling variability.

Statistical Significance

Statistical significance is a term that reflects the confidence with which we can reject the null hypothesis. A result is statistically significant if observed differences are unlikely due to chance alone.
In the exercise, once the z-statistic is calculated (here it is 2.35), it is compared against a standard normal distribution to determine the associated p-value.
The p-value indicates the probability of observing the test results under the null hypothesis. If this value is below a predetermined significance level (commonly 0.05), the null hypothesis is rejected in favor of the alternative.
A small p-value in this case suggests the treatment group performs better than the control, indicating Memory-enhancing benefits of Substance M. Statistical significance provides a structured way to support or refute the initial hypothesis with calculated confidence.