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Soon after the euro was introduced as currency in Europe, it was widely reported that someone had spun a euro coin 250 times and gotten heads 140 times. We wish to test a hypothesis about the fairness of spinning the coin. a. Estimate the true proportion of heads. Use a \(95 \%\) confidence interval. Don't forget to check the conditions. b. Does your confidence interval provide evidence that the coin is unfair when spun? Explain. c. What is the significance level of this test? Explain.

Short Answer

Expert verified
a) The estimated true proportion of heads is 0.56. The 95% confidence interval is [0.4982, 0.6218]. b) The confidence interval suggests that the coin could be unfair, since 0.5 lies outside of it. c) The significance level of the test is 5%, indicating a 5% risk of concluding that a difference exists when there is actually none.

Step by step solution

01

Estimate the true proportion of heads

The true proportion of heads (\(p\)) equals the number of heads (\(x\)) divided by the total number of spins (\(n\)). In this case, \(x = 140\) and \(n = 250\), so the estimated proportion \(\hat{p} = \frac{x}{n} = \frac{140}{250} = 0.56\). This indicates that a head appeared 56% of the time.
02

Construct a 95% confidence interval

Constructing a 95% confidence interval involves calculating the standard error and using the z-score corresponding to a 95% confidence level. The standard error (SE) is calculated using the formula: \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). Substituting the values: \(SE = \sqrt{\frac{0.56(1-0.56)}{250}} \approx 0.0315\). To find the 95% confidence interval, use the formula: \(\hat{p} \pm z*SE\), where \(z\) is approximately 1.96 for a 95% confidence interval. Thus, the interval is \(0.56 \pm 1.96*0.0315 = [0.4982, 0.6218]\).
03

Evaluate if the confidence interval suggests unfairness

An unbiased, 'fair' coin would have a true proportion of heads equal to 0.5. If 0.5 lies within the 95% confidence interval calculated in step 2, we do not have enough evidence to suggest the coin is unfair. In this case however, 0.5 lies outside the confidence interval [0.4982, 0.6218], indicating we have evidence that the coin is not fair when spun.
04

Determine the significance level

The significance level is the probability of rejecting the null hypothesis when it is true. For a 95% confidence interval, the significance level is 5% or 0.05. This means that there is a 5% risk of concluding that a difference exists when there is no actual difference.

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Key Concepts

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

Confidence Interval
When we're working with statistics, particularly when estimating unknown parameters, a confidence interval (CI) is an invaluable tool. It provides a range in which we believe the true population parameter lies, with a specified degree of certainty. For instance, if we calculate a 95% confidence interval, we’re saying that we’re 95% confident that the true parameter resides within this range.

To figure out the 95% confidence interval for the proportion of heads from the euro coin example, we must first determine the sample proportion \(\hat{p}\) from the experiment. Here, 140 heads in 250 spins suggest a proportion of 0.56. When calculating the confidence interval, we include a margin of error which accounts for the sampling variability, and is influenced by both the standard error (SE) of our estimate and the desired confidence level. In this context, using the normal distribution, we calculate the standard error and then the z-score, which for a 95% confidence level is approximately 1.96.

Ultimately, the confidence interval allows us to gauge the reliability of our sample statistic to estimate the true population parameter and is mathematically represented as \( \hat{p} \pm z*SE \) which gives us the span [0.4982, 0.6218] in this particular case. Notice that due to the confidence interval not containing the hypothesized population proportion of 0.5 (for a fair coin), we suspect that the coin might not be fair.
Significance Level
Linked closely with the concept of a confidence interval is the significance level, denoted by \( \alpha \). This is a threshold we set to decide whether to accept or reject the null hypothesis in hypothesis testing. In layman's terms, it represents our tolerance for making a mistake by incorrectly rejecting a true null hypothesis—known as a Type I error.

For our euro coin, when we speak about a 95% confidence interval, inversely we have a 5% significance level (\( \alpha = 0.05 \)). This translates to a 5% probability of concluding that the coin is not fair when it actually is. It's a balance; a lower significance level means being more cautious about claiming a finding, reducing our risk of a false positive. However, it also increases the chances of missing a real effect (Type II error), showing the delicate balance in choosing an appropriate \( \alpha \).

Setting the significance level before performing the test helps to protect against random fluctuation that might lead one to erroneously believe there's an effect or difference when there's not.
Proportion Test
When examining the fairness of a coin, we're inherently conducting a type of hypothesis test known as a proportion test. In this specific case, it's a test about the proportion of heads in flips of a coin. The null hypothesis \(H_0\) commonly posits that the coin is fair, therefore the proportion of heads should be 0.5. Conversely, the alternative hypothesis \(H_A\) suggests a proportion different from 0.5, indicating potential unfairness.

The procedure generally involves computing the sample proportion—like the 0.56 found from the 250 euro coin spins—and assessing whether this sample statistic is significantly different from the hypothesized proportion, under the assumption that the null hypothesis is true. This process involves calculating a test statistic and comparing it to a theoretical distribution, such as the normal or t-distribution, depending on the sample size and conditions met.

The outcome, positioned against the predetermined significance level, will lead us to either reject the null hypothesis in favor of the alternative, or to fail to reject it, which is not equivalent to accepting it but rather withholding judgment due to insufficient evidence.
Fairness of Coin
Determining the fairness of a coin involves approaching it with a statistical mindset. A fair coin should have an equal chance of landing heads or tails, setting the expected proportion of either at 0.5. To statistically test the fairness, we employ hypothesis testing procedure—using the test of proportion we discussed.

In the reported problem, we found a 95% confidence interval not containing the value 0.5. This seems to suggest that the coin might not be fair. But remember, this interval estimation is a method to make an inference about the population parameter, not a definitive proof. We must be cautious to understand that while statistics can guide us to evidence of unfairness, it's not absolute. Real-world factors, such as imperfections in the coin or biases in the spinning method, should also be considered.

Overall, claiming a coin's fairness or unfairness involves interpreting the confidence interval within the context of hypothesis testing and significance levels, making sure to account for sample variability and to set appropriate levels of statistical certainty.

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Most popular questions from this chapter

A new reading program may reduce the number of elementary school students who read below grade level. The company that developed this program supplied materials and teacher training for a large-scale test involving nearly 8500 children in several different school districts. Statistical analysis of the results showed that the percentage of students who did not meet the grade- level goal was reduced from \(15.9 \%\) to \(15.1 \%\). The hypothesis that the new reading program produced no improvement was rejected with a P-value of 0.023 . a. Explain what the P-value means in this context. b. Even though this reading method has been shown to be significantly better, why might you not recommend that your local school adopt it?

Which of the following statements are true? If false, explain briefly. a. Using an alpha level of \(0.05,\) a P-value of 0.04 results in rejecting the null hypothesis. b. The alpha level depends on the sample size. C. With an alpha level of \(0.01,\) a \(P\) -value of 0.10 results in rejecting the null hypothesis. d. Using an alpha level of \(0.05,\) a P-value of 0.06 means the null hypothesis is true.

Have harsher penalties and ad campaigns increased seat-belt use among drivers and passengers? Observations of commuter traffic failed to find evidence of a significant change compared with three years ago. Explain what the study's P-value of 0.17 means in this context.

For each of the following situations, state whether a Type I, a Type II, or neither error has been made. a. A test of \(\mathrm{H}_{0}: \mu=25\) vs. \(\mathrm{H}_{\mathrm{A}}: \mu>25\) rejects the null hypothesis. Later it is discovered that \(\mu=24.9\). b. A test of \(\mathrm{H}_{0}: p=0.8\) vs. \(\mathrm{H}_{\mathrm{A}}: p<0.8\) fails to reject the null hypothesis. Later it is discovered that \(p=0.9\). c. A test of \(\mathrm{H}_{0}: p=0.5\) vs. \(\mathrm{H}_{\mathrm{A}}: p \neq 0.5\) rejects the null hypothesis. Later it is discovered that \(p=0.65\). d. A test of \(\mathrm{H}_{0}: p=0.7\) vs. \(\mathrm{H}_{\mathrm{A}}: p<0.7\) fails to reject the null hypothesis. Later it is discovered that \(p=0.6\).

Spam filters try to sort your e-mails, deciding which are real messages and which are unwanted. One method used is a point system. The filter reads each incoming e-mail and assigns points to the sender, the subject, key words in the message, and so on. The higher the point total, the more likely it is that the message is unwanted. The filter has a cutoff value for the point total; any message rated lower than that cutoff passes through to your inbox, and the rest, suspected to be spam, are diverted to the junk mailbox. We can think of the filter's decision as a hypothesis test. The null hypothesis is that the e-mail is a real message and should go to your inbox. A higher point total provides evidence that the message may be spam; when there's sufficient evidence, the filter rejects the null, classifying the message as junk. This usually works pretty well, but, of course, sometimes the filter makes a mistake. a. When the filter allows spam to slip through into your inbox, which kind of error is that? b. Which kind of error is it when a real message gets classified as junk? c. Some filters allow the user (that's you) to adjust the cutoff. Suppose your filter has a default cutoff of 50 points, but you reset it to 60 . Is that analogous to choosing a higher or lower value of \(\alpha\) for a hypothesis test? Explain. d. What impact does this change in the cutoff value have on the chance of each type of error?

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