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A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all long- distance calls made by residential customers is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does your conclusion change if \(\alpha=.05 ?\)

Step by step solution

01

Identify the Hypotheses

First, the null and alternative hypotheses need to be established. The null hypothesis (\(H_0\):) is that the mean phone call duration is 10 minutes (\(\mu_0 = 10\)). It's the initial claim that the company makes. The alternative hypothesis (\(H_a\):) is that the mean phone call duration is different from 10 minutes, hence (\(\mu ≠ 10\)).

02

Calculating the Test Statistics

Thereafter, the z-test statistics need to be calculated: \(Z = \frac{\bar{x} - \mu_0}{\frac{σ}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean (9.20 minutes), \(\mu_0\) is the assumed population mean (10 minutes), \(σ\) is given as the known population standard deviation (3.80 minutes), and \(n\) is the sample size (100). Substituting, the z-score is calculated, resulting in a z-score {-1.052632}.

03

Calculating the p-value

The p-value is the probability that, if the null hypothesis is true, we would observe a statistic at least as extreme as the one actually observed. This involves employing the standard normal distribution table to find the probability associated with the computed z-score. For a two-tailed test (because \(H_a\) considers the mean call duration different, not only greater or lesser), the p-value is twice the one-tailed area. The p-value is calculated to be {0.292386}.

04

Test of Hypothesis/Concluding Remarks

Initially, a significant level of α=0.02 is considered. If the p-value is smaller than α, then we reject the null hypothesis. Otherwise, there is insufficient evidence to reject \(H_0\). The calculated p-value here (0.292386) is greater than α (0.02), so there is insufficient evidence to reject the null hypothesis. That means the hypothesis that the mean duration of long-distance calls made by residential customers is 10 minutes remains valid. When α is changed to 0.05, the conclusion remains the same, because the p-value is still greater than α.

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Key Concepts

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

p-value calculation

The concept of the p-value is central to hypothesis testing. It measures how extreme the sample results are when the null hypothesis is true. To calculate the p-value for an observed statistic, we use the standard normal distribution. In our example, we perform a two-tailed test, because we are checking if the mean call duration is simply different from 10 minutes, not more or less specific.

Here's how it works:

• After determining the z-score from our test (z = -1.052, for our example), we find the p-value by looking at how probable this z-score is under the normal distribution.
• Because it's a two-tailed test, we multiply the one-tail area probability by two. This accounts for both the possibilities of the sample mean being greater or lesser than the population mean of 10 minutes.

If the p-value ends up being less than the significance level (\( \alpha \)) we choose (like 0.02 or 0.05), we reject the null hypothesis. In our case, the p-value was 0.292, which is greater than both levels, meaning we do not reject the null hypothesis.

z-test

The z-test is an essential procedure for hypothesis testing when the population standard deviation is known, and we're concerned with the mean of a large sample. In the scenario given, a z-test helps us compare the sample mean to the population mean to check if there is any significant difference.

The process involves:

• Calculating the z-score, which is the number of standard deviations our sample mean is from the population mean. This is a straightforward division: \( Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \).
• Where \( \bar{x} = 9.20 \) (our sample mean), \( \mu_0 = 10 \) (population mean), \( \sigma = 3.80 \) (standard deviation), and \( n = 100 \) (sample size). After plugging these values into the formula, we get a z-score of approximately -1.052.

The calculated z-score can then be used to determine the p-value, helping us assess the evidence against the null hypothesis.

critical-value approach

The critical-value approach offers another way to make decisions in hypothesis testing. Instead of directly using the p-value, we determine if our test statistic (z-score) falls within a certain critical region, which is defined based on significance level.

• First, we find the critical z-values for our chosen alpha levels (such as 0.02 and 0.05). These are the z-scores that mark the boundaries of our acceptance region for the null hypothesis. For a two-tailed test and alpha = 0.02, the critical values could be approximately -2.33 and 2.33.
• If the z-score from our test falls outside this critical region, it indicates that the sample result is improbable under the null hypothesis. Thus, we reject the null hypothesis.

In our particular data, the z-score of -1.052 does not reach these critical values for either significance level, indicating a failure to reject the null hypothesis: the mean call duration likely does dwell around 10 minutes.

null hypothesis

The null hypothesis is a fundamental concept in the realm of hypothesis testing. It's an assumption made at the beginning, usually proposing that there is no effect or no difference. In this exercise, the null hypothesis (\( H_0 \)) proposes that the mean duration of all long-distance calls is precisely 10 minutes.

• It's akin to a starting point for examining whether observed data are different enough to warrant the alternative hypothesis.
• Rejecting the null hypothesis means showing sufficient evidence that the sample data do not align with this presumed population mean.

Maintaining this hypothesis implies that any differences observed in sample statistics might be due to random sampling variability, rather than a true effect. Our task is to collect evidence through test statistics and p-values to judge whether to keep or reject this baseline assumption.