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Chapter 10: Problem 80

In 100 tosses of a coin, 63 heads and 37 tails are ohserved. Is this a balanced coin? Use a 0.05 level of significance.

Short Answer

Expert verified
Based on the Z-test, we reject the null hypothesis at the 0.05 level of significance. Therefore, it is concluded that the coin is not balanced.

Step by step solution

01

- Formulate null and alternative hypotheses

The first step is to set the null hypothesis and the alternative hypothesis. The null hypothesis (\(H_0\)) is that the coin is balanced, meaning that the probability of getting heads (\(p\)) is 0.5. The alternative hypothesis (\(H_a\)): is that \(p \neq 0.5\). This is a two-tailed test because we are looking for a difference in either direction from 0.5.
02

- Calculate sample proportion and standard error

Next, we calculate our sample proportion (\(\hat{p}\)) which is the observed number of successes (heads) divided by the total number of trials. In this case, \(\hat{p} = 63/100 = 0.63\). The standard error (SE) under the null hypothesis is given by \(\sqrt{p(1 - p)/n}\), where \(p\) is the assumed probability of success under \(H_0\), and \(n\) is the number of trials. So, SE = \(\sqrt{0.5*(1 - 0.5)/100} = 0.05\).
03

- Compute Z-Score

Now, calculate the z-score using the formula \(z = (\hat{p} - p) / SE\). The z-score indicates how many standard errors away our observed sample proportion is from the hypothesized population proportion. Here, \(z = (0.63 - 0.5) / 0.05 = 2.6\).
04

- Determine critical value and make conclusion

For a 0.05 level of significance in a two-tailed test, the critical z-values are -1.96 and +1.96. If our calculated z-score lies outside this range, then we reject the null hypothesis. As our calculated z-value (2.6) is greater than the critical z-value (1.96), we reject the null hypothesis. So, we conclude that the coin is not balanced.

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Key Concepts

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

Null and Alternative Hypotheses
When we analyze a statistical problem like determining if a coin is balanced, we start with stating hypotheses. The null hypothesis, usually denoted as \( H_0 \), is a statement of no effect or no difference; in this case that the coin is fair (\(p = 0.5\)). The alternative hypothesis, represented as \( H_a \) or \( H_1 \) in some texts, stands for the opposite and asserts that there is an effect or a difference, suggesting that the coin is not fair (\(p eq 0.5\)).

This construction sets the stage for a statistical test. If observed data provide enough evidence against the null hypothesis, it is rejected in favor of the alternative. This decision-making process lays the foundation for many statistical analyses.
Standard Error
Standard error (SE) measures the variability or precision of an estimated population parameter. To be more precise, for our coin-tossing exercise, it quantifies how much the sample proportion (\(\hat{p}\)) might differ from the actual population proportion if the null hypothesis were true.

Mathematically, SE is calculated using the formula \( SE = \sqrt{p(1 - p)/n} \), where \(p\) is the probability of success, and \(n\) is the sample size. A smaller SE suggests our estimate is more precise, whereas a larger SE indicates greater variability. Understanding the standard error is crucial as it directly impacts the subsequent calculation of the z-score.
Z-Score
The z-score is a statistical metric that tells us how many standard errors a data point (such as a sample mean or proportion) is from the population mean under the null hypothesis. It's calculated as \( z = (\hat{p} - p) / SE \).

In the context of our coin toss example, the z-score shows how far the observed proportion of heads (\(0.63\)) is from the expected proportion (\(0.5\)) if the coin were actually balanced. A high absolute value of the z-score indicates that the observed result is far from what we would expect if the null hypothesis were true, leading us to consider the possibility that the null hypothesis may not hold.
Statistical Significance
Statistical significance is a key concept that helps us decide whether to reject the null hypothesis or not. It's determined by comparing the calculated test statistic, in this case the z-score, against a critical value that corresponds to a pre-set significance level (\(\alpha\)).

The significance level, usually set at 0.05 (5%), reflects the probability of rejecting the null hypothesis when it is actually true (Type I error). If the absolute z-score is greater than the critical value (in absolute terms), we can declare the results statistically significant and reject the null hypothesis. In our coin example, since the absolute z-score of 2.6 is greater than the critical z-value of 1.96, we have significant evidence to suggest that the coin is not balanced.

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Most popular questions from this chapter

Pnst data indicate that the amount of money contributed by the working residents of a large city to a voluntect rescue squad is a normal random variable with a standard deviation of 81.40 . It has been suggested that the contributions to the rescue squad from iust the employees of the sanitation department are much more variable. If the contributions of a random sample of 12 employees from the sanitation department had a standard deviation of \(\$ 1.75,\) can we conclude at the 0.01 level of significance that the standard deviation of the contributions of all sanitation workers is greater than that of all workers living in this city?

10.52 On testing $$\begin{array}{l}\text { Ho: } p=14 \\\H_{1}: \quad \mu \neq 14\end{array}$$ an \(\alpha=0.05\) level \(t\) -test is being considered. What sample size is necessary in order that the probability is 0.1 of falsely failing to reject \(H o\) when the true population mean differs from 14 by \(0.5 ?\) From a preliminary sample we estimate \(\sigma\) to be \(1.25 .\)

The hydrocarbon emissions are known to have decreased dramatically during the 1980 s. A study was conducted to compare the hydrocarbon emissions at idling speed, in parts per million (ppm), for automobiles of 1980 and \(1990 .\) Twenty cars of each year model were randomly selected and their hydrocarbon emission leycls were recorded. The data are as follows 1980 models: \(\begin{array}{llllllllll}141 & 359 & 247 & 940 & 882 & 494 & 306 & 210 & 105 & 880 \\ 200 & 223 & 188 & 940 & 241 & 190 & 300 & 435 & 241 & 380\end{array}\) 1990 models: \(\begin{array}{rrrrrrrrrr}140 & 160 & 20 & 20 & 223 & 60 & 20 & 95 & 360 & 70 \\\ 220 & 400 & 217 & 58 & 235 & 380 & 200 & 175 & 85 & 65\end{array}\) Test the hypothesis that \(\sigma_{1}=\sigma_{2}\) against the alternative that \(\sigma_{1} \neq \sigma_{2}\). Assume both populations are normal. Use a P-value.

In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that 63 of 100 urban residents favor the construction while only 59 of 125 suburban residents are in favor. Is there a significant difference between the proportion of urban and suburban residents who favor construction of the nuclear plant? Make use of a P-value.

10.2 A sociologist is concerned about the effectiveness of a training course designed to get more drivers to use seat, belts in automobiles. (a) What hypothesis is she testing if she commits a type I error by erroneously concluding that the training course is ineffective? b) What hypothesis is she testing if she commits a type II error by erroneously concluding that the training course is effective?
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