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A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them.

Short Answer

Expert verified
There is insufficient evidence to conclude that local use is higher than national use.

Step by step solution

01

Define the Hypothesis

Define the null and alternative hypotheses. Null Hypothesis ( H_0 ): The proportion of local seniors using drugs or alcohol is less than or equal to the national proportion. Alternative Hypothesis ( H_1 ): The proportion of local seniors using drugs or alcohol is greater than the national proportion.
02

Identify the Given Values

Identify the values given in the problem. Locally, 65 out of 100 seniors use drugs or alcohol. Nationally, 60 out of 100 seniors do the same. This gives us \( p_{local} = 0.65 \) and \( p_{national} = 0.60 \), with \( n_{local} = n_{national} = 100 \).
03

Calculate the Test Statistic

Use the formula for the test statistic for two proportions: \[ z = \frac{(p_{local} - p_{national})}{\sqrt{p(1-p)(\frac{1}{n_{local}} + \frac{1}{n_{national}})}} \]where \( p = \frac{x_{local} + x_{national}}{n_{local} + n_{national}} \). Calculate \( p \), then substitute in the values to find \( z \).
04

Calculate the Pooled Proportion

Calculate the pooled proportion \( p \): \[ p = \frac{65 + 60}{100 + 100} = \frac{125}{200} = 0.625 \]
05

Compute the Standard Error

Compute the standard error using the pooled proportion:\[ SE = \sqrt{0.625 \times (1 - 0.625) \times \left(\frac{1}{100} + \frac{1}{100}\right)} = \sqrt{0.625 \times 0.375 \times 0.02} \approx 0.0493 \]
06

Calculate the Z-Score

Calculate the z-score:\[ z = \frac{0.65 - 0.60}{0.0493} \approx \frac{0.05}{0.0493} \approx 1.014 \]
07

Compare to Critical Value

Determine the critical value for a one-sided test at a standard significance level (e.g., 0.05). The critical z-value is approximately 1.645. Since 1.014 < 1.645, we do not reject the null hypothesis.
08

Draw Conclusion

Based on the z-score being less than the critical value, there is insufficient evidence at the 0.05 significance level to prove that the proportion of drug and alcohol use is higher locally than nationally.

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Key Concepts

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

Z-test
The z-test is a statistical tool used to determine whether there are significant differences between two population proportions in hypothesis testing. It is particularly useful when comparing the proportions of two different groups, like in our exercise with local and national seniors' drug and alcohol use.
The z-test is applicable when the sample size is large enough, typically more than 30. It involves calculating a z-score, which tells us how many standard deviations a result is from the expected mean.
  • A positive z-score suggests that the sample proportion is greater than the mean or expected proportion.
  • A negative z-score indicates the opposite.
For the test, we calculate a test statistic using the observed differences and the standard error, then compare this statistic with a critical value from the z-distribution table to decide whether to reject the null hypothesis. This is done to assess whether the observed difference is statistically significant or just due to random variation.
Pooled Proportion
The pooled proportion is a key component when conducting a z-test for comparing two proportions. This involves combining the data from two samples to compute a single overall proportion. This is under the assumption that the samples are from the same population.
The pooled proportion, denoted as \( p \), is calculated by:\[ p = \frac{x_{local} + x_{national}}{n_{local} + n_{national}} \]Where \( x_{local} \) and \( x_{national} \) are the counts of successes (e.g., seniors using drugs or alcohol) in each group, and \( n_{local} \) and \( n_{national} \) are the sample sizes.
The purpose of the pooled proportion is to calculate a more accurate measure of the standard error when testing the hypothesis. It's essential in determining the combined variance for both groups, which influences the reliability of the z-score calculated.
Alternative Hypothesis
The alternative hypothesis is a key idea in hypothesis testing, representing the statement that we are seeking evidence for. It is denoted as \( H_1 \) and suggests that there is a significant effect or difference present, opposing the null hypothesis.
In the context of our drug and alcohol study, the alternative hypothesis states that the proportion of drug and alcohol use among local seniors is greater than that among national seniors (\( p_{local} > p_{national} \)).It's essential because it represents the hypothesis that we want to test and potentially prove through our statistical analysis. If the evidence proves strong enough against the null hypothesis upon calculating the z-score and comparing to a critical value, we consider the alternative hypothesis to be true.
Null Hypothesis
The null hypothesis is a foundational concept in statistics, denoted as \( H_0 \). It represents a baseline assumption that there is no effect or difference. In our case, it states there is no higher proportion of drug and alcohol use among local seniors compared to national seniors (\( p_{local} \leq p_{national} \)).
It's a crucial part of hypothesis testing because it offers a standard against which the alternative hypothesis is tested.
  • The null hypothesis is assumed true until evidence suggests otherwise.
  • If the calculated z-score falls within a calculated critical range (based on a predetermined significance level, e.g., 0.05), we fail to reject the null hypothesis.
  • Conversely, falling outside this range would lead us to reject it in favor of the alternative hypothesis.
Understanding the null hypothesis helps to manage expectations and emphasizes the importance of evidence-based conclusions in statistical testing.

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Most popular questions from this chapter

Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The 鈥渄ay鈥 subscript refers to the statistics day students. The 鈥渘ight鈥 subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is: a. \(\mu_{\text { day }}>\mu_{\text { night }}\) b. \(\mu_{\text { day }}<\mu_{\text { night }}\) c. \(\mu\) day \(=\mu_{\text { night }}\) d. \(\mu_{\text { day }} \neq \mu_{\text { night }}\)

Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. $$\begin{array}{|l|l|l|}\hline & {\text { Sample Mean Melting Temperatures }\left(^{\circ} F\right)} & {\text { Population Standard Deviation }} \\\ \hline \text { Alloy Gamma } & {800}&{95} \\ \hline \text { Alloy zeta } & {900} &{105} \\ \hline\end{array}$$ What is the p-value?

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. Is this a right-tailed, left-tailed, or two-tailed test?

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.

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