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Basic Computation: Testing \(p\) A random sample of 30 binomials trials resulted in 12 successes. Test the claim that the population proportion of successes does not equal \(0.50 .\) Use a level of significance of 0.05 (a) Check Requirements Can a normal distribution be used for the \(\hat{p}\) distribution? Explain. (b) State the hypotheses. (c) Compute \(\hat{p}\) and the corresponding standardized sample test statistic. (d) Find the \(P\) -value of the test statistic. (e) Do you reject or fail to reject \(H_{0}\) ? Explain. (f) Interpretation What do the results tell you?

Short Answer

Expert verified
Fail to reject \( H_0 \); insufficient evidence to claim \( p \neq 0.50 \).

Step by step solution

01

Check Requirements

For using a normal distribution to approximate the sampling distribution of the sample proportion \( \hat{p} \), the sample size \( n \) should be large enough such that both \( n \times p \) and \( n \times (1 - p) \) are greater than 5. Here, we have a sample size \( n = 30 \) and assumed population proportion \( p = 0.50 \). Calculate:\[ n \times p = 30 \times 0.5 = 15 \]\[ n \times (1 - p) = 30 \times 0.5 = 15 \]Both values are greater than 5, so the normal distribution can be used.
02

State the Hypotheses

The null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_1 \)) are as follows:- \( H_0: p = 0.50 \) (the population proportion is 0.50)- \( H_1: p eq 0.50 \) (the population proportion is not 0.50)This is a two-tailed test because we are checking for any difference from 0.50.
03

Compute Sample Proportion and Test Statistic

The sample proportion \( \hat{p} \) of successes is calculated as:\[ \hat{p} = \frac{x}{n} = \frac{12}{30} = 0.40 \]The standardized test statistic \( z \) is computed using:\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}} = \frac{0.40 - 0.50}{\sqrt{\frac{0.50 \times 0.50}{30}}} = \frac{-0.10}{0.09129} \approx -1.095 \]
04

Find the P-value

Since this is a two-tailed test, we need to find the P-value for \( z = -1.095 \).Using a standard normal distribution table or calculator, we find:- The P-value for \( z = -1.095 \) (either tail) is roughly 0.1368.The two-tailed P-value is therefore:\[ 2 \times 0.1368 = 0.2736 \]
05

Make a Decision

At a significance level \( \alpha = 0.05 \), compare the P-value to \( \alpha \):\[ 0.2736 > 0.05 \]Since the P-value is greater than 0.05, we fail to reject the null hypothesis \( H_0 \).
06

Interpretation of Results

There is insufficient evidence to reject the claim that the population proportion of successes is 0.50. Therefore, we do not have enough statistical evidence to support the alternative hypothesis that the proportion of successes is different from 0.50.

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Key Concepts

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

population proportion
The population proportion represents the fraction of the entire population that holds a particular characteristic. In probability and statistics, it's symbolized as \( p \). For example, if we're assessing the proportion of people who prefer chocolate ice cream in a town, and we assume the population proportion is 0.50, it means that 50% of the population prefer chocolate ice cream. This parameter is usually unknown and is estimated from a sample. The value \( p = 0.50 \) is central to the hypothesis tested in the exercise.
- **Importance:** Understanding the population proportion helps in making predictions about the entire population based on a sample.
- From the exercise, the hypothesis test is set to determine if the population proportion \( p \) is equal to 0.50 or not. This is typically done by comparing sample data to the assumed population standard.
By using statistical methods, researchers can infer whether the observed data significantly deviates from the expected data, suggesting a different population proportion.
normal distribution
Normal distribution is a crucial concept in statistics that allows approximation of various datasets. In the context of hypothesis testing for population proportions, the sampling distribution of the sample proportion \( \hat{p} \) is used. When certain conditions are met, this distribution resembles a normal distribution (bell-shaped curve).
- **Conditions:** The sample size \( n \) must be large enough so that \( n \times p \) and \( n \times (1-p) \) are both greater than 5, ensuring the sample mean distribution appears bell-shaped.
- **Utility:** A normal distribution is beneficial because it simplifies the computation of probabilities and critical values using well-established statistical techniques.
The exercise shows when a normal distribution can approximate \( \hat{p} \). With a sample size of 30 and a population proportion of 0.50, we find both critical conditions are satisfied, permitting this approximation.
sample proportion
The sample proportion, noted as \( \hat{p} \), reflects the observed proportion of successes in a sample. It is calculated using the formula: \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of successes in the sample and \( n \) is the total sample size.
- **Example:** In the exercise, 12 successes out of 30 trials yield a sample proportion \( \hat{p} = 0.40 \). This proportion is then compared to the hypothesized population proportion, \( p = 0.50 \).
- **Significance:** It provides a point estimate of the population proportion and is used in hypothesis testing to determine if there is substantial evidence to claim that the population proportion differs from the hypothesized value. By comparing \( \hat{p} \) with \( p \), we can assess how far our sample result is from what was expected, under the null hypothesis.
P-value
The P-value quantifies the probability of observing a test statistic as extreme as, or more so, than the observed results, under the assumption that the null hypothesis is true. It is a central concept in hypothesis testing, providing evidence against or in support of the null hypothesis.
- **Calculation:** By using the standard normal distribution, one determines the P-value by considering the area under the curve to the left of the observed test statistic and the symmetrically equivalent area to the right in a two-tailed test.
- **Interpretation:** A smaller P-value indicates stronger evidence against the null hypothesis. In the exercise, a P-value of 0.2736 indicates a relatively high probability of observing the given sample proportion under the assumption that the null hypothesis is true. Since the P-value exceeds the 0.05 significance level, evidence is insufficient to reject \( H_0 \).
This probabilistic approach helps researchers decide on the null hypothesis based on calculated likelihoods.
significance level
In hypothesis testing, the significance level \( \alpha \) is a threshold used to determine whether the P-value indicates enough evidence to reject the null hypothesis. Common levels are 0.05, 0.01, and 0.10, with 0.05 being widely used. It represents the risk of a Type I error—that is, rejecting a true null hypothesis.
- **Role:** The significance level sets the critical value boundaries against which the P-value is compared. If the P-value is less than or equal to \( \alpha \), the null hypothesis is rejected.
- **Example from exercise:** Here, with \( \alpha = 0.05 \), the decision rule is to reject \( H_0 \) if the P-value \( \leq 0.05 \). Since the P-value of 0.2736 exceeds 0.05, the sample does not provide sufficient evidence to reject \( H_0 \). The results suggest the observed sample proportion could easily occur with a true population proportion of 0.5.
The significance level helps strike a balance between being cautious about false positives and the need for statistical evidence.

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Most popular questions from this chapter

Plato's Dialogues: Prose Rhythm Symposium is part of a larger work referred to as Plato's Dialogues. Wishart and Leach (see source in Problem 15 ) found that about \(21.4 \%\) of five-syllable sequences in Symposium are of the type in which four are short and one is long. Suppose an antiquities store in Athens has a very old manuscript that the owner claims is part of Plato's Dialogues. A random sample of 493 five-syllable sequences from this manuscript showed that 136 were of the type four short and one long. Do the data indicate that the population proportion of this type of five-syllable sequence is higher than that found in Plato's Symposium? Use \(\alpha=0.01.\)

To use the normal distribution to test a proportion \(p,\) the conditions \(n p>5\) and \(n q>5\) must be satisfied. Does the value of \(p\) come from \(H_{0}\) or is it estimated by using \(\hat{p}\) from the sample?

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the \(z\) value of the sample test statistic. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in \(\mathrm{mg} / 100 \mathrm{ml}\) ). The sample mean is \(\bar{x} \approx 93.8 .\) Let \(x\) be a random variable representing glucose readings taken from Gentle Ben. We may assume that \(x\) has a normal distribution, and we know from past experience that \(\sigma=12.5 .\) The mean glucose level for horses should be \(\mu=85 \mathrm{mg} / 100 \mathrm{ml}\) (Reference: Merck Veterinary Manual). Do these data indicate that Gentle Ben has an overall average glucose level higher than \(85 ?\) Use \(\alpha=0.05\)

Consider a hypothesis test of difference of means for two independent populations \(x_{1}\) and \(x_{2} .\) What are two ways of expressing the null hypothesis?

Myers-Briggs: Extroverts Are most student government leaders extroverts? According to Myers-Briggs estimates, about \(82 \%\) of college student government leaders are extroverts (Source: Myers -Briggs Type Indicator Atlas of Type Tables). Suppose that a Myers-Briggs personality preference test was given to a random sample of 73 student government leaders attending a large national leadership conference and that 56 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from \(82 \% ?\) Use \(\alpha=0.01.\)

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