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The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) using \(\hat{p}=0.58\) with each of the following sample sizes: (a) \(\hat{p}=29 / 50=0.58\) (b) \(\hat{p}=290 / 500=0.58\)

Short Answer

Expert verified
Using the z-statistics method, calculate the p-value for case (a) and case (b) and then compare them. The test with the smaller p-value provides the strongest evidence for the alternative hypothesis.

Step by step solution

01

Calculation of p-value for case (a)

Using the z-statistics method, the z-score is defined as \(Z = \frac\{(\hat{p}-p_0)}{ \sqrt\{(p_0*(1-p_0)) / n} } \), where \(p_0\) is the assumed true population proportion in the null hypothesis, and \(n\) is the sample size. For case (a), we can substitute \(\hat{p} = 0.58, p_0 = 0.5\) and \(n = 50 \) in the z-score formula. After we estimated the z-score, we calculate the p-value , which is the probability under the standard normal curve to the right of the calculated z-score. If the p-value \(< 0.05\), the result is considered significant at the \(5 \%\) level.
02

Calculation of p-value for case (b)

The same method applied for case (b), but this time the sample size \(n=500\). All another values stay the same: \(\hat{p} = 0.58, p_0 = 0.5\). After we estimated the z-score, calculate the p-value. If the p-value \(< 0.05\), the result is considered significant at the \(5 \%\) level.
03

Comparing the statistical impact of two sample sizes

After having the p-values for both sample sizes, compare them. The smaller the p-value, the stronger the evidence against the null hypothesis. Hence, the sample size providing the smaller p-value would be providing the strongest evidence for the alternative hypothesis.

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Key Concepts

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

P-Value Significance
When conducting hypothesis testing, understanding the p-value can seem complex, but it's crucial in determining statistical significance. The p-value indicates the probability of obtaining a result at least as extreme as the one in your sample data, assuming that the null hypothesis is true. In simpler terms, it measures how compatible your data is with the null hypothesis.

Within hypothesis testing, if the p-value is low, it suggests that the observed data is unusual under the assumption of the null hypothesis. Traditionally, researchers use a significance level, often set at 0.05 or 5%, to decide whether or not to reject the null hypothesis. If the p-value falls below this cutoff, the results are deemed 'statistically significant,' implying there's evidence to support the alternative hypothesis. This doesn't prove the alternative hypothesis, but rather suggests it's a more plausible explanation for the data compared to the null hypothesis.

Imagine we're tossing a coin and want to check if it's fair. If we find a significant p-value after many tosses, it might indicate the coin is biased. Remember to consider other factors like sample size, which can affect the p-value, as we'll discuss later.
Alternative Hypothesis
In the realm of hypothesis testing, the alternative hypothesis (\( H_a \) or \( H_1 \)) represents what we aim to support, based on sample data. It's defined in opposition to the null hypothesis (\( H_0 \)), which typically suggests no effect or no difference. An alternative hypothesis is formulated based on what we suspect might be true.

For example, let's go back to our coin tossing: the null hypothesis might assert that the coin is fair (\( H_0: p = 0.5 \)), and the alternative hypothesis might claim that the coin is biased towards heads (\( H_a: p > 0.5 \)). We use evidence from the sample to support the alternative hypothesis, looking for a small p-value to reject the null. In our original exercise, the alternative hypothesis suggests that the true proportion is greater than 0.5, inspired by the sample statistic (\( \bar{p} = 0.58 \)).

When we have a significant result, it strengthens the claim of our alternative hypothesis; however, it's essential to avoid overstating the conclusion. A result aligning with \( H_a \) simply diminishes the credibility of \( H_0 \) but doesn't prove \( H_a \) is correct definitively.
Sample Size Effect
Sample size plays a pivotal role in statistical inference. The larger the sample size, the more confident we are about our estimates of the population's characteristics. This confidence comes from the Law of Large Numbers, which states that as a sample size grows, the sample mean gets closer to the population mean.

When dealing with hypothesis testing and p-values, sample size can dramatically impact results. Larger samples reduce the margin of error and therefore can detect smaller differences from the null hypothesis value. This can lead to a smaller p-value and stronger evidence against the null hypothesis. Conversely, small samples may be more susceptible to random chance, potentially leading to less reliable results.

Referring to our original exercise, when we compare the p-values obtained from different sample sizes, a larger sample size providing the same sample statistic (\( \bar{p} \)) will usually give us more robust evidence for the alternative hypothesis. So, when using sample sizes of 50 and 500, with the same observed proportion (\( \bar{p} = 0.58 \)), the larger sample size would generate a p-value that more convincingly argues against the null hypothesis.

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Most popular questions from this chapter

Figure 4.25 shows a scatterplot of the acidity (pH) for a sample of \(n=53\) Florida lakes vs the average mercury level (ppm) found in fish taken from each lake. The full dataset is introduced in Data 2.4 on page 71 and is available in FloridaLakes. There appears to be a negative trend in the scatterplot, and we wish to test whether there is significant evidence of a negative association between \(\mathrm{pH}\) and mercury levels. (a) What are the null and alternative hypotheses? (b) For these data, a statistical software package produces the following output: $$ r=-0.575 \quad p \text { -value }=0.000017 $$ Use the p-value to give the conclusion of the test. Include an assessment of the strength of the evidence and state your result in terms of rejecting or failing to reject \(H_{0}\) and in terms of \(\mathrm{pH}\) and mercury. (c) Is this convincing evidence that low \(\mathrm{pH}\) causes the average mercury level in fish to increase? Why or why not?

Testing 50 people in a driving simulator to find the average reaction time to hit the brakes when an object is seen in the view ahead.

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) using \(\hat{p}=0.55\) with each of the following sample sizes: (a) \(\hat{p}=55 / 100=0.55\) (b) \(\hat{p}=275 / 500=0.55\) (c) \(\hat{p}=550 / 1000=0.55\)

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Testing to see whether taking a vitamin supplement each day has significant health benefits. There are no (known) harmful side effects of the supplement.

Match the four \(\mathrm{p}\) -values with the appropriate conclusion: (a) The evidence against the null hypothesis is significant, but only at the \(10 \%\) level. (b) The evidence against the null and in favor of the alternative is very strong. (c) There is not enough evidence to reject the null hypothesis, even at the \(10 \%\) level. (d) The result is significant at a \(5 \%\) level but not at a \(1 \%\) level. I. 0.00008 II. 0.0571 III. 0.0368 IV. \(\quad 0.1753\)

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