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Determine whether it is appropriate to use the normal distribution to estimate the p-value. If it is appropriate, use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from a random sample and use a \(5 \%\) significance level. Test \(H_{0}: p=0.8\) vs \(H_{a}: p>0.8\) using the sample results \(\hat{p}=0.88\) with \(n=50\)

Short Answer

Expert verified
We can use the normal distribution to calculate the p-value given the large enough sample size. After calculating the test statistic (z) and the p-value, we can compare the p-value to the 0.05 significance level to decide whether to reject or accept the null hypothesis.

Step by step solution

01

Verify the assumptions

First, check if the sample size is large enough to assume normal distribution for \(p\). The conditions for the normal approximation are \(np ≥ 10\) and \(n(1-p) ≥ 10\). Here, \(n = 50\) and under the null hypothesis, \(p = 0.8\). Check if both \(50 * 0.8 = 40\) and \(50 * (1 - 0.8) = 10\) are greater than 10. Since they are, we can use the normal distribution.
02

Calculate the test statistic

The test statistic for a proportion is a z-score (z). It can be calculated using the formula: \[ z = \frac{(\hat{p} - p_0)}{\sqrt{ \frac{(p_0 * (1-p_0))}{n}}} \]where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion under the null hypothesis, and \(n\) is the sample size. Substituting the given values we obtain \[ z = \frac{(0.88 - 0.8)}{\sqrt{(0.8 * (1 - 0.8))/50}}\]
03

Find the p-value using the test statistic

Next, use the calculated z-score to find the p-value. The p-value is the probability that, assuming the null hypothesis is true, we would get a sample as extreme or more extreme than the one we have. As our alternate hypothesis is \(p > 0.8\), we are interested in the area to the right of our test statistic on a standard normal distribution.It means we want to find the probability \(P(Z > z)\) on a standard normal table or using a calculator's functionality. Given our Z score, we can find the p-value.
04

Make a decision

Compare the p-value to the given significance level of 0.05. If the p-value is less than \(0.05\), we reject the null hypothesis in favor of the alternative hypothesis. If not, we fail to reject the null hypothesis.

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Key Concepts

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

Normal Approximation to Binomial
When we work with probabilities related to the number of successes in a series of independent and identical trials (binomial scenarios), the calculations can become complex. However, under certain conditions, we can simplify the situation by approximating the binomial distribution with a normal distribution. This method is known as the normal approximation to the binomial distribution. It's most appropriate to use when the sample size is large and both the conditions, namely, np and n(1-p), are equal to or greater than 10.

This normal approximation is particularly useful because it allows us to use the well-known properties of the normal distribution to calculate probabilities and critical values without complex binomial calculations. In the context of hypothesis testing, if the sample size is adequate, as suggested by these conditions, the test statistic can be calculated using the z-score formula associated with the normal distribution.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to decide whether to reject the null hypothesis, \(H_{0}\), or not. The test statistic is calculated assuming the null hypothesis is true and indicates how far the sample statistic, such as the sample proportion, is from the expected population parameter according to \(H_{0}\).

In our case, the test statistic is a z-score that reflects the number of standard deviations the observed sample proportion, \(\hat{p}\), is from the hypothesized population proportion, \(p_0\). Calculating the z-score involves using the standard error of the proportion, which accounts for the sample size and the expected variability under the null hypothesis. This z-score is then used to ascertain the p-value, which in turn informs the decision to accept or reject the null hypothesis in the context of the given significance level.
P-value
The p-value is a critical component in hypothesis testing. It represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true. Essentially, it quantifies how surprising the sample data are, given the null hypothesis.

A small p-value indicates that the observed data are unlikely under the null hypothesis, suggesting that the null hypothesis may not adequately explain the data. In hypothesis testing, we compare the p-value to a pre-established significance level, \(\alpha\), to make a decision regarding the null hypothesis. If the p-value is less than the significance level, it suggests that the sample data are inconsistent with the null hypothesis, and we may reject \(H_{0}\) in favor of the alternative hypothesis, \(H_{a}\).
Significance Level
The significance level, denoted by \(\alpha\), is a threshold chosen by the researcher to evaluate the strength of the evidence against the null hypothesis. It defines the risk we are willing to take of incorrectly rejecting the null hypothesis, which is also known as Type I error. Commonly used significance levels are 0.05, 0.01, or 0.10.

In the exercise at hand, a significance level of 0.05 means that there is a 5% chance of concluding that a difference exists when there is no actual difference. In the realm of hypothesis testing, this level is weighed against the p-value. When the p-value falls below the chosen significance level, the null hypothesis is rejected, suggesting that the observed result is statistically significant. Choosing an appropriate significance level is crucial, as it balances the need for evidence against the risk of making an incorrect inference about the population.
Sample Proportion
The sample proportion, denoted as \(\hat{p}\), is a statistic that estimates the proportion of successes in a sample. It's calculated by dividing the number of successes observed in the sample by the total number of observations. The sample proportion is a point estimate of the population proportion, which we aim to infer about.

In hypothesis testing, we often use the sample proportion to assess a claim about the population proportion. The sample proportion plays a vital role in determining the test statistic and, subsequently, the p-value. The reliability of conclusions drawn from hypothesis tests greatly depends on the accuracy of the sample proportion, which in turn depends on the random nature and size of the sample. If the sample is representative and sufficiently large, we can be more confident about the inferences made regarding the population proportion based on the sample proportion.

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Most popular questions from this chapter

Football Air Pressure During the NFL's 2014 AFC championship game, officials measured the air pressure on game balls following a tip that one team's balls were under-inflated. In exercise 6.124 we found that the 11 balls measured for the New England Patriots had a mean psi of 11.10 (well below the legal limit) and a standard deviation of 0.40. Patriot supporters could argue that the under-inflated balls were due to the elements and other outside effects. To test this the officials also measured 4 balls from the opposing team (Indianapolis Colts) to be used in comparison and found a mean psi of \(12.63,\) with a standard deviation of 0.12. There is no significant skewness or outliers in the data. Use the t-distribution to determine if the average air pressure in the New England Patriot's balls was significantly less than the average air pressure in the Indianapolis Colt's balls.

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 25 from Population 1 with mean 6.2 and standard deviation 3.7 and samples of size 40 from Population 2 with mean 8.1 and standard deviation 7.6

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of lie detector trials in which the technology misses a lie, with \(n=48\) and \(\hat{p}=0.354\)

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=556.9, s_{d}=\) \(143.6, n_{d}=100\)

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of women's tears on men, levels of testosterone are measured in 50 men after they sniff women's tears and after they sniff a salt solution. The order of the two treatments was randomized and the study was double-blind.

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