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Chapter 9: Problem 105

An insurance company states that \(90 \%\) of its claims are settled within 30 days. A consumer group selected a random sample of 75 of the company's claims to test this statement. If the consumer group found that 55 of the claims were settled within 30 days, does it have sufficient reason to support the contention that less than \(90 \%\) of the claims are settled within 30 days? Use \(\alpha=0.05\). a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Short Answer

Expert verified
Using both the p-value approach and the classical approach, there is sufficient evidence to reject the null hypothesis, which suggests that the proportion of claims settled within 30 days is less than 90%.

Step by step solution

01

Set the Hypotheses

Set up the null hypothesis and the alternative hypothesis. The null hypothesis (\(H_0\)) is the insurance company's claim, which is that 90% of their claims are settled within 30 days. That is, \(p=0.90\). The alternative hypothesis (\(H_a\)) contends that less than 90% of the claims are settled within 30 days. That is, \(p<0.90\).
02

Calculate the Test Statistic

After forming the hypotheses, calculate the test statistic using the formula: \[Z = \frac{\bar{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\]. Here, \(\bar{p}\) is the sample proportion, which is \(\frac{55}{75} = 0.7333\), \(p\) is the population proportion, which is 0.90, and \(n\) is the sample size, which is 75. After calculation, the test statistic, Z is -2.497.
03

P-value Approach

Under the p-value approach, compute the p-value, which is the probability that the test statistic is as extreme as the current observation, assuming the null hypothesis is true. Using a standard normal table, the p-value for Z = -2.497 is found to be 0.0063. If the p-value is less than the significance level (\(\alpha = 0.05\)), reject the null hypothesis. Here, the p-value is less than the significance level, so reject the null hypothesis. This suggests that less than 90% of the claims are settled within 30 days.
04

Classical Approach

In the classical approach, find the critical value for the given level of significance. This is a left-tailed test because the alternative hypothesis suggests that the actual proportion is less than the hypothesized proportion. The critical Z value for a left-tailed test at the 0.05 level of significance is -1.645. If the calculated Z score is to the left of the critical Z value, reject the null hypothesis. Here, -2.497 is to the left of -1.645 on the number line. So, reject the null hypothesis. This also indicates that less than 90% of the claims are settled within 30 days.

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Key Concepts

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

P-Value Approach
The p-value approach in hypothesis testing is a method used to determine the significance of the results from a statistical test. It is the probability of obtaining a result at least as extreme as the one observed, assuming that the null hypothesis is true.

In the context of the given exercise, the null hypothesis (\(H_0\)) states that 90% of insurance claims are settled within 30 days, while the alternative hypothesis (\(H_a\) contends that less than 90% are settled in that time frame. After calculating the test statistic, which is a z-score of -2.497 in this case, the corresponding p-value is found using statistical tables or software.

By comparing the p-value to the significance level (\(\alpha = 0.05\)), we decide whether to reject or fail to reject the null hypothesis. A p-value of 0.0063, which is less than 0.05, provides sufficient evidence to reject the null hypothesis, indicating that the actual proportion of claims settled within 30 days is likely less than 90%.
Test Statistic
A test statistic is a standardized value calculated from sample data during a hypothesis test. It's essentially a tool used to measure how far the sample data deviates from the null hypothesis.

In this problem, we use the formula for a one-sample z-test for proportions: \[Z = \frac{\bar{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\] where \(\bar{p}\) is the sample proportion, \(p\) is the population proportion as stated in the null hypothesis, and \(n\) is the sample size. With the numbers provided (55 out of 75 claims), we find that the sample proportion \(\bar{p}\) is about 0.7333, and the z-test statistic is -2.497. A negative test statistic signals that the sample proportion is less than the population proportion hypothesized.
Null Hypothesis
The null hypothesis (\(H_0\) is the default assumption that there is no effect or no difference; it's what we assume to be true before collecting any data. In hypothesis testing, it represents the idea that any observed effect is due merely to sampling error or chance.

For the insurance claim example, the null hypothesis posits that 90% of claims are settled within 30 days (\(p = 0.90\)). The test's goal is to gather evidence to either support or refute this claim. If the evidence suggests that the actual proportion is significantly lower, we can reject the null hypothesis in favor of the alternative hypothesis.
Alternative Hypothesis
Contrasting the null hypothesis is the alternative hypothesis (\(H_a\) which posits that there is an effect or a difference. It's what we suspect might be true and are trying to gather evidence to support.

In our insurance claim case, the alternative hypothesis suggests that less than 90% of claims are settled within 30 days (\(p < 0.90\)). We want to test this claim rigorously. If the data leads us to reject the null hypothesis, it is implicit that we are accepting the alternative hypothesis鈥攑roviding evidence that indeed, the proportion of claims settled within the specified period is statistically lower than 90%.

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Most popular questions from this chapter

Acetaminophen is an active ingredient found in more than 600 over-the-counter and prescription medicines, such as pain relievers, cough suppressants, and cold medications. It is safe and effective when used correctly, but taking too much can lead to liver damage. A researcher believes the mean amount of acetaminophen per tablet in a particular brand of cold tablets is different from the 600 mg claimed by the manufacturer. A random sample of 30 tablets had a mean acetaminophen content of \(596.3 \mathrm{mg}\) with a standard deviation of \(4.7 \mathrm{mg}\). a. Is the assumption of normality reasonable? Explain. b. Construct a \(99 \%\) confidence interval for the estimate of the mean acetaminophen content. c. What does the confidence interval found in part b suggest about the mean acetaminophen content of one pill? Do you believe there is 600 mg per tablet? Explain.

A random sample of 51 observations was selected from a normally distributed population. The sample mean was \(\bar{x}=98.2,\) and the sample variance was \(s^{2}=37.5 .\) Does this sample show sufficient reason to conclude that the population standard deviation is not equal to 8 at the 0.05 level of significance? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Karl Pearson once tossed a coin 24,000 times and recorded 12,012 heads. a. Calculate the point estimate for \(p=P(\) head ) based on Pearson's results. b. Determine the standard error of proportion. c. Determine the \(95 \%\) confidence interval estimate for \(p=P(\text { head })\). d. It must have taken Mr. Pearson many hours to toss a coin 24,000 times. You can simulate 24,000 coin tosses using the computer and calculator commands that follow. (Note: A Bernoulli experiment is like a "single" trial binomial experiment. That is, one toss of a coin is one Bernoulli experiment with \(p=0.5;\) and 24,000 tosses of a coin either is a binomial experiment with \(n=24,000\) or is 24,000 Bernoulli experiments. Code: \(0=\) tail, \(1=\) head. The sum of the 1 s will be the number of heads in the 24,000 tosses.) e. How do your simulated results compare with Pearson's? f. Use the commands (part d) and generate another set of 24,000 coin tosses. Compare these results to those obtained by Pearson. Also, compare the two simulated samples to each other. Explain what you can conclude from these results.

Oranges are selected at random from a large shipment that just arrived. The sample is taken to estimate the size (circumference, in inches) of the oranges. The sample data are summarized as follows: \(n=100\) \(\Sigma x=878.2,\) and \(\Sigma(x-\bar{x})^{2}=49.91\). a. Determine the sample mean and standard deviation. b. What is the point estimate for \(\mu,\) the mean circumference of all oranges in the shipment? c. Find the \(95 \%\) confidence interval for \(\mu\)

Use a computer or calculator to find the area (a) to the left, and (b) to the right of \(\chi^{2} \star=14.7\) with \(\mathrm{df}=24\).
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