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A random sample of \(n=1000\) observations from a binomial population produced \(x=279 .\) a. If your research hypothesis is that \(p\) is less than .3 , what should you choose for your alternative hypothesis? Your null hypothesis? b. What is the critical value that determines the rejection region for your test with \(\alpha=.05 ?\) c. Do the data provide sufficient evidence to indicate that \(p\) is less than .3 ? Use a \(5 \%\) significance level.

Short Answer

Expert verified
Answer: The data does not provide sufficient evidence to support the claim that the population proportion (\(p\)) is less than \(0.3\) at the \(5\%\) significance level.

Step by step solution

01

Set up Hypotheses

Since the research hypothesis is that \(p<0.3\), we can set up our null hypothesis (\(H_0\)) and alternative hypothesis (\(H_1\)) as follows: \(H_0: p = 0.3\) \(H_1: p < 0.3\)
02

Calculate Test Statistic

Let's calculate the test statistic using the sample proportions. Using the sample proportion \(\hat{p} = \frac{x}{n} = \frac{279}{1000} = 0.279\). The test statistic for a single sample proportion is given by: \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\) Where: \(z\) - Test statistic \(\hat{p}\) - Sample proportion \(p_0\) - Null hypothesis proportion (here, \(0.3\)) \(n\) - Sample size By substituting our values, we get: \(z = \frac{0.279 - 0.3}{\sqrt{\frac{0.3(1-0.3)}{1000}}}\) \(z \approx -1.08\)
03

Find Critical Value

We are given a significance level of \(\alpha=0.05\). As we have a left-tailed test, we need to find the critical value \(z_\alpha\) such that the probability to the left of this critical value is \(0.05\). Using the standard normal distribution table or a calculator, the critical value is: \(z_{0.05} \approx -1.645\)
04

Compare Test Statistic and Critical Value

We now compare the calculated test statistic \(z\) with the critical value \(z_\alpha\): In our case, \(z=-1.08\) and \(z_\alpha=-1.645\). Since \(z > z_\alpha\) (-1.08 > -1.645), we do not reject the null hypothesis.
05

Conclusion

As we do not reject the null hypothesis, the data does not provide sufficient evidence to support the claim that the population proportion \(p\) is less than \(0.3\). At the \(5\%\) significance level, we cannot conclude that \(p<0.3\).

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Key Concepts

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

Binomial Distribution
The binomial distribution is a discrete probability distribution. It describes the number of successes in a fixed number of independent trials, each with the same probability of success. In the context of this problem, which involves hypothesis testing with a sample size of 1000, the aim is to determine how likely an observed outcome is if this particular distribution model is assumed. Consider a scenario where we have a population with a true proportion, denoted as \( p \). If we assume \( p = 0.3 \), each individual trial of this binomial experiment has a 30% chance of "success." The outcome we observe, the "successes," can then be tested to see if they significantly deviate from this assumption. This can be quantified using a test statistic, which is computed based on the sample proportion (in this case, \( \hat{p} = 0.279 \)).Our task when dealing with such a distribution in hypothesis testing is typically to determine if this observed proportion both aligns with a hypothesized proportion \( p_0 \) and occurs due to random chance.
Significance Level
A significance level, expressed as \( \alpha \), determines the threshold at which we reject a null hypothesis. It reflects how willing we are to make a Type I error - rejecting a true null hypothesis. In this example, the significance level is set at 0.05, or 5%.This choice of \( \alpha = 0.05 \) signifies that there is a 5% risk of incorrectly rejecting the null hypothesis. This 5% threshold is often chosen because it balances caution with power - namely, not rejecting true hypotheses too frequently, while still allowing sensitivity enough to detect false hypotheses.The significance level also directly influences the critical value used in the hypothesis test. It defines the rejection region: for a left-tailed test as in this example, this is the area in the left tail of the distribution where, if the test statistic falls, we reject the null hypothesis.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement suggesting that there is no effect or no difference. In the context of this exercise, it proposes that the population proportion \( p = 0.3 \).Why do we set up a null hypothesis this way? It's initially assumed true to understand any deviations observed in our sample. We then test if these deviations are statistically significant. If not, we continue adhering to \( H_0 \).It's crucial because it shapes the entire hypothesis test. When the null hypothesis is constructed, calculations such as the test statistic depend directly on its parameters. In this test, \( H_0: p = 0.3 \), implies that the hypothesized population proportion is 30%. All calculations, including establishing the critical value and determining whether to reject \( H_0 \), rely on this hypothesis assumption.
Alternative Hypothesis
The alternative hypothesis, represented as \( H_1 \), offers a competing claim against the null hypothesis. Here, it states that the population proportion \( p \) is less than 0.3, denoted as \( H_1: p < 0.3 \).Unlike the null hypothesis, which is a statement of no effect, the alternative is typically the research hypothesis - the statement you aim to investigate. In this scenario, stating \( p < 0.3 \) means the researcher suspects the true proportion is lower than initially assumed.Deciding on \( H_1 \) guides the entire direction of the test. We choose the test type (in this case, a left-tailed test) based on it, fundamentally shaping the statistical procedure to either support or refute this alternative claim. The aim of hypothesis testing is to gather enough evidence to consider this alternative plausible, often leading to significant practical or theoretical conclusions.

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Most popular questions from this chapter

Sweet Potato Whitefly Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected, and 25 are found to be infested with whitefly. a. Assuming that the experiment satisfies the conditions of the binomial experiment, do the data indicate that the proportion of infested fields is greater than expected? Use the \(p\) -value approach, and test using a \(5 \%\) significance level. b. If the proportion of infested fields is found to be significantly greater than \(.10,\) why is this of practical significance to the agronomist? What practical conclusions might she draw from the results?

Female Models In a study to assess various effects of using a female model in automobile advertising, 100 men were shown photographs of two automobiles matched for price, color, and size, but of different makes. One of the automobiles was shown with a female model to 50 of the men (group \(\mathrm{A}\) ), and both automobiles were shown without the model to th other 50 men (group B). In group A, the automobile shown with the model was judged as more expensive by 37 men; in group \(B\), the same automobile was judged as the more expensive by 23 men. Do these results indicate that using a female model influences the perceived cost of an automobile? Use a one-tailec test with \(\alpha=.05 .\)

PCBs, continued Refer to Exercise \(9.75 .\) a. Calculate \(\beta\) and \(1-\beta\) if the true mean ppm of PCBs is 6 ppm. b. Calculate \(\beta\) and \(1-\beta\) if the true mean ppm of PCBs is 7 ppm. c. Find the power, \(1-\beta,\) when \(\mu=8,9,10,\) and 12\. Use these values to construct a power curve for the test in Exercise \(9.75 .\) d. For what values of \(\mu\) does this test have power greater than or equal to \(90 ?\)

Adolescents and Social Stress In a study to compare ethnic differences in adolescents' social stress, researchers recruited subjects from three middle schools in Houston, Texas. \({ }^{21}\) Social stress among four ethnic groups was measured using the Social Attitudinal Familial and Environment Scale for Children (SAFE-C). In addition, demographic information abou the 316 students was collected using self-administered questionnaires. A tabulation of student responses to a question regarding their socioeconomic status (SES) compared with other families in which the students chose one of five responses (much worse off, somewhat worse off, about the same, better off, or much better off) resulted in the tabulation that follows. a. Do these data support the hypothesis that the proportion of adolescent African Americans who state that their SES is "about the same" exceeds that for adolescent Hispanic Americans? b. Find the \(p\) -value for the test. c. If you plan to test using \(\alpha=.05,\) what is your conclusion?

Breaking Strengths of Cables A test of the breaking strengths of two different types of cables was conducted using samples of \(n_{1}=n_{2}=100\) pieces of each type of cable. Do the data provide sufficient evidence to indicate a difference between the mean breaking strengths of the two cables? Use \(\alpha=.05 .\)

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