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An insurance company states that it settles \(85 \%\) of all life insurance claims within 30 days. A consumer group asks the state insurance commission to investigate. In a sample of 250 life insurance claims, 203 were settled within 30 days. a. Test whether the true proportion of all life insurance claims made to this company that are settled within 30 days is less than \(85 \%,\) at the \(5 \%\) level of significance. b. Compute the observed significance of the test.

Short Answer

Expert verified
a. Reject the null hypothesis; the proportion is less than 0.85. b. The p-value is approximately 0.041.

Step by step solution

01

State the Hypotheses

We start by stating the null and alternative hypotheses. The null hypothesis is that the true proportion \( p \) is equal to 0.85 (i.e., \( H_0: p = 0.85 \)), and the alternative hypothesis is that the true proportion is less than 0.85 (i.e., \( H_a: p < 0.85 \)).
02

Calculate the Test Statistic

The test statistic for a proportion can be calculated using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] where \( \hat{p} = \frac{203}{250} = 0.812 \), \( p_0 = 0.85 \), and \( n = 250 \). Substituting these values gives: \[ z = \frac{0.812 - 0.85}{\sqrt{\frac{0.85 \times 0.15}{250}}} \approx -1.741 \].
03

Determine the Critical Value

Since we are using a 5% significance level for a one-tailed test, we find the critical value for \( z \) that corresponds to a cumulative probability of 0.05. This critical value is approximately \( z = -1.645 \).
04

Compare Test Statistic with Critical Value

If the calculated \( z \) statistic is less than the critical value, we reject the null hypothesis. Here, \( z \approx -1.741 \), which is less than \( -1.645 \). Thus, we reject the null hypothesis that the proportion is equal to 0.85.
05

Compute the Observed Significance Level (p-value)

To find the observed significance level or the p-value, we look up the probability corresponding to \( z = -1.741 \). This results in a p-value of approximately 0.041.

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Key Concepts

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

Understanding the Null Hypothesis
The concept of a null hypothesis is foundational in hypothesis testing. It is the statement being tested, usually denoting that there is no effect or no difference in a particular situation. For instance, in the context of the exercise given, the null hypothesis asserts that the true proportion of life insurance claims settled within 30 days is precisely 85% (i.e., \( H_0: p = 0.85 \)).

The null hypothesis serves as a default or status quo statement. It's what you expect if nothing significant is happening. Hypothesis testing assesses whether there is enough evidence to reject this statement in favor of an alternative, offering insight into the process or effect being studied.

Understanding when and why the null hypothesis is rejected is key in various fields, including medical trials, manufacturing quality checks, and social sciences. It forms the bedrock of statistical inference, driving decisions based on data.
Exploring the Alternative Hypothesis
The alternative hypothesis represents what you suspect might be true instead of the null hypothesis. It is usually considered when the evidence might suggest an effect or difference. In the given exercise, the alternative hypothesis is that the true proportion of life insurance claims settled within 30 days is less than 85% (i.e., \( H_a: p < 0.85 \)).

The alternative hypothesis is important because it guides the direction of the test. There are three types of alternative hypotheses :
  • Two-sided, where you are interested in any change — increase or decrease.
  • One-sided, where you specify if the change is specifically an increase or decrease.

The choice between these alternatives influences the statistical methods used and determines how results are interpreted. When the evidence strongly suggests refuting the null hypothesis, the alternative hypothesis is considered more plausible based on the data.
Deciphering the P-value
The p-value is a crucial metric in hypothesis testing. It indicates the probability of observing the data, or something more extreme, assuming the null hypothesis is true. In simpler terms, it helps measure the strength of the evidence against the null hypothesis.

Using the p-value, you determine whether to reject the null hypothesis. In our exercise, the computed p-value is approximately 0.041. To interpret this:
  • If the p-value is less than or equal to the significance level (commonly 0.05), the null hypothesis is rejected in favor of the alternative hypothesis.
  • A smaller p-value suggests stronger evidence against the null hypothesis.

  • The p-value does not measure the size of an effect or the importance of a result. It merely indicates how consistent the data is with the null hypothesis. Hence, in the context of the exercise, since the p-value (0.041) is below the 0.05 significance level, the null hypothesis might be less likely to hold true.
Significance Level and Its Implications
In hypothesis testing, the significance level, often denoted as \( \alpha \), is a threshold set for deciding when to reject the null hypothesis. It represents the risk of a Type I error, i.e., rejecting a true null hypothesis.

For the exercise provided, the significance level is set at 5% or \( 0.05 \). This means there is a 5% risk of mistakenly rejecting the null hypothesis when it is actually true. The choice of significance level is critical in research for balancing the risk of errors:
  • Lower significance levels reduce Type I error risk but increase the risk of Type II errors.
  • Higher levels lead to more lenient tests, potentially increasing false positives.

Adjusting \( \alpha \) depends on the context and consequences of errors in your specific field. In our scenario, a 5% significance level provides a reasonable balance, helping decide if the evidence against the null hypothesis is strong enough to consider the alternative hypothesis more likely.

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