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According to the analysis of Federal Reserve statistics and other government data, American households with credit card debts owed an average of \(\$ 15,706\) on their credit cards in August 2015 (www.nerdwallet.com). A recent random sample of 500 American households with credit card debts produced a mean credit card debt of \(\$ 16,377\) with a standard deviation of \(\$ 3800 .\) Do these data provide significant evidence at a \(1 \%\) significance level to conclude that the current mean credit card debt of American households with credit card debts is higher than \(\$ 15,706 ?\) Use both the \(p\) -value approach and the critical-value approach.

Short Answer

Expert verified
Compute the test statistic and compute the p-value. For the critical value approach, compare critical value and the test statistic. Using either approach, one decides whether to reject the null hypothesis that the mean credit card debt is less than or equal to $15,706, and favour the alternate hypothesis that it's greater. The steps must be followed correctly to reach a statistically sound decision.

Step by step solution

01

Calculate the test statistic

The test statistic for a sample mean, when the population standard deviation is unknown, follows a t-distribution. However, because the sample size is large (n = 500), this can be approximated with a z-distribution. The test statistic \(z\) is calculated using the formula: \[ z = \frac{\bar{x} - μ0}{σ / \sqrt{n}} \] where \(\bar{x} = 16,377$ is the sample mean, \(μ0 = 15,706$ is the null hypothesis mean, \(σ = 3,800$ is the sample standard deviation, and \(n = 500\) is the sample size.
02

Compute the p-value

The p-value is the probability that we would observe a test statistic as extreme as the one calculated, under the null hypothesis. For a right-tailed test, this is the probability of observing a value greater than the calculated test statistic. P-value can be found by looking up the calculated z in the z-table (found in most statistics textbooks) or by using a calculator or statistical software.
03

Compare p-value with significance level

The decision rule for the p-value approach is to reject the null hypothesis if the p-value is smaller than the significance level. Here, the significance level (α) is 1%.
04

Compute the critical value

For the critical value approach, a critical value corresponds to the given level of significance is found from the z-table. A critical region is then identified. If the test statistic falls within this critical region, the null hypothesis is rejected.
05

Compare test statistic with critical value

Compare your test statistic with the critical value. If the z-score is higher than the critical value, reject the null hypothesis.

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Key Concepts

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

z-distribution
In hypothesis testing, the choice of distribution hinges on the sample size and the information available about the population. The z-distribution is a kind of normal distribution that plays an important role when testing hypotheses with large samples. It is particularly useful when the sample size is greater than 30, or when the population standard deviation is known. However, in many practical scenarios, we approximate the population distribution using the sample data.
A z-distribution is defined by:
  • a mean of zero
  • a standard deviation of one
This standard normal distribution helps standardize your data so you can better understand where your sample mean lies relative to the population mean. In the example problem, even though the population standard deviation is not given, a large sample size allows us to approximate the distribution using a z-distribution. This is why we calculate a z-statistic to determine if the sample mean of $16,377 is significantly different from the null hypothesis mean of $15,706.
p-value
The p-value is a fundamental concept in statistical hypothesis testing. It represents the probability of obtaining a test statistic as extreme as the observed one, assuming that the null hypothesis is true. It essentially helps us determine the evidence against the null hypothesis.
Calculating the p-value involves comparing the observed data with what would be expected under the null hypothesis. If the p-value is very low, it suggests that such extreme test results would be uncommon, leading us to question the null hypothesis. For this hypothesis test, we calculate a p-value from the z-statistic determined earlier.
In the context of our original exercise, a low p-value (less than the significance level of 1%) indicates that the observed sample mean is significantly different from the hypothesized population mean, thus providing evidence to reject the null hypothesis.
critical value
When deciding whether to reject a null hypothesis, we use the critical value approach as one of the tools. The critical value is a point in the distribution of the test statistic that is compared to the calculated statistic. Depending on the significance level and the direction of the test (one-tailed or two-tailed), this value helps define the threshold for decision-making.
To find the critical value for a specific significance level, you often use a z-table. In our exercise, with a 1% level of significance for a one-tailed test, the critical value separates the region where we reject the null hypothesis from where we do not. If the test statistic exceeds this critical value, it falls into the rejection region, indicating that the null hypothesis is unlikely to be true. In our problem, since we’re looking at a right-tailed test, if our calculated z-value is greater than the critical value, we reject the null hypothesis.
null hypothesis
The null hypothesis is a statement used in hypothesis testing that asserts no effect or no difference exists. It's what you seek to test against with your data. In the context of our problem, the null hypothesis posits that the mean credit card debt of American households has not increased since 2015 and remains at $15,706.
Symbolically, the null hypothesis can be represented as:
  • \( H_0: \mu = 15,706 \)
By setting up the null hypothesis, we frame the statistical problem and prepare to use our data to test this claim. A critical part of this process is determining the evidence we would need to reject it. The hypothesis testing process involves looking at how likely it is, under this hypothesis, for the observed sample statistics to occur. If it seems improbable, we may reject the null hypothesis and consider an alternative hypothesis, suggesting the mean debt has indeed increased.

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Most popular questions from this chapter

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has an approximate normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to 15 ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the altemative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) ounces. a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and adjust it if she chooses the maximum probability of a Type I error to be .01? What if the maximum probability of a Type I error is 05 ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\). Does the machine need to be adjusted? What if \(a=.05 ?\)

Consider \(H_{0}: \mu=55\) versus \(H_{1}: \mu \neq 55\). a. What type of error would you make if the null hypothesis is actually false and you fail to reject it? b. What type of error would you make if the null hypothesis is actually true and you reject it?

The manager of a restaurant in a large city claims that waiters working in all restaurants in his city earn an average of \(\$ 150\) or more in tips per week. A random sample of 25 waiters selected from restaurants of this city yielded a mean of \(\$ 139\) in tips per week with a standard deviation of \(\$ 28\). Assume that the weekly tips for all waiters in this city have a normal distribution. a. Using a \(1 \%\) significance level, can you conclude that the manager's claim is true? Use both approaches. b. What is the Type I error in this exercise? Explain. What is the probability of making such an error?

Briefly explain the procedure used to calculate the \(p\) -value for a two- tailed and for a one-tailed test, respectively.

Consider the null hypothesis \(H_{0}: \mu=100 .\) Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of \(.01\), show the rejection and nonrejection regions and find the critical value(s) of \(t\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 100\) b. \(H_{1}: \mu>100\) c. \(H_{1}: \mu<100\)

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