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The mean income per person in the United States is \(\$ 50,000,\) and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of \(\$ 60,000\) with a standard deviation of \(\$ 10,000 .\) At the .05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

Short Answer

Expert verified
Yes, there is enough evidence to conclude that Wilmington residents earn more than the national average.

Step by step solution

01

Hypothesis Setup

Start by setting up the null and alternative hypotheses. The null hypothesis, denoted as \( H_0 \), is that the mean income of Wilmington residents is equal to the national average (\( \mu = 50,000 \)). The alternative hypothesis, \( H_a \), is that the mean income of Wilmington residents is greater than the national average (\( \mu > 50,000 \)) since we want to find evidence if it is higher.
02

Significance Level

Identify the level of significance for the test, which is given as \( \alpha = 0.05 \). This value helps determine the critical region for rejecting the null hypothesis.
03

Calculate the Test Statistic

Use the formula for the test statistic: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} = 60,000 \) is the sample mean, \( \mu = 50,000 \) is the population mean according to the null hypothesis, \( s = 10,000 \) is the sample standard deviation, and \( n = 10 \) is the sample size. Solving this gives:\[t = \frac{60,000 - 50,000}{10,000/\sqrt{10}} \approx \frac{10,000}{3162.28} \approx 3.16\]
04

Determine the Critical Value

Find the critical value from the t-distribution table at \( n-1 = 9 \) degrees of freedom for a one-tailed test at \( \alpha = 0.05 \). The critical t-value is approximately 1.833.
05

Make a Decision

Compare the computed test statistic to the critical value. Since 3.16 > 1.833, we reject the null hypothesis.
06

Conclusion

Conclude that there is enough statistical evidence at the 0.05 significance level to support the claim that the mean income of residents of Wilmington, Delaware, is greater than the national average.

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Key Concepts

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

t-distribution
The t-distribution is a type of probability distribution that is used frequently in statistics, especially for estimating population parameters when the sample size is small and the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails, which means it gives more probability to extreme values.
  • It's particularly useful when dealing with small sample sizes (usually less than 30).
  • The shape of the t-distribution depends on the degrees of freedom, which are determined by the sample size ( n-1).
In the context of hypothesis testing, we use the t-distribution to find the critical value that will help us decide whether to reject the null hypothesis. For our exercise, we calculated the t-value and compared it against a critical value sourced from the t-distribution table using 9 degrees of freedom (since the sample size was 10). This comparison is what ultimately informs our decision on the hypothesis test.
null hypothesis
The null hypothesis (symbolized as H0) is a fundamental concept in hypothesis testing that assumes no effect or no difference between groups or conditions being studied. It serves as the starting assumption for statistical testing.
  • In empirical research, the null hypothesis typically states that there's no change, no difference, or no effect.
  • The purpose of hypothesis testing is to determine if there's enough evidence to reject the null hypothesis in favor of an alternative hypothesis.
In our example, the null hypothesis is that the mean income of Wilmington, Delaware, residents is equal to the national average of $50,000. This sets the baseline we test against to see if there's a statistically significant increase in the residents' income compared to the national average.
alternative hypothesis
The alternative hypothesis (denoted as Ha) is a statement that indicates the presence of an effect or a difference. It represents what the experimenter wants to prove.
  • If the data provides strong enough evidence, we can reject the null hypothesis in favor of the alternative hypothesis.
  • In hypothesis testing, the alternative hypothesis is crafted to reflect what is being tested or what is thought to exist.
For the Wilmington income study, the alternative hypothesis suggests that the mean income of the residents is greater than the national average, specifically over $50,000. Testing against this hypothesis helps determine whether there's supporting evidence from the sample data (mean income of $60,000 in this case) that the residents earn more.
level of significance
The level of significance, depicted as α, is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It defines the probability of rejecting the null hypothesis when, in fact, it is true (type I error).
  • Common levels of significance include 0.05, 0.01, and 0.10.
  • A lower level of significance implies more stringent criteria for rejecting the null hypothesis.
In our scenario, the level of significance is set at 0.05. This means there is a 5% risk of concluding that a difference exists when there isn't one. It also specifies the critical region for the decision-making process. If the calculated test statistic falls into this region, the null hypothesis is rejected in favor of the alternative.

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Most popular questions from this chapter

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