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A random sample of 18 observations produced a sample mean of \(9.24\). Find the critical and observed values of \(z\) for each of the following tests of hypothesis using \(\alpha=.05 .\) The population standard deviation is known to be \(5.40\) and the population distribution is normal. a. \(H_{0}: \mu=8.5\) versus \(\quad H_{1}: \mu \neq 8.5\) b. \(H_{0}: \mu=8.5\) versus \(\quad H_{1}: \mu>8.5\)

Short Answer

Expert verified
a. For \(H_{1}: \mu \neq 8.5\), critical value = \(\pm 1.96\), observed value = \(0.7337\). We do not reject the null hypothesis. b. For \(H_{1}: \mu > 8.5\), critical value = \(1.645\), observed value = \(0.7337\). We do not reject the null hypothesis.

Step by step solution

01

Identify given(s)

Sample size (\(n\)) = 18, sample mean (\(\bar{X}\)) = 9.24, population standard deviation (\(\sigma\)) = 5.40, and level of significance (\(\alpha\)) = 0.05.
02

Step 2a: Calculate Critical Value for \(H_{1}: \mu \neq 8.5\)

The critical value of \(z\) for a two-tailed test with \(\alpha = .05\) is found by looking up in a standard normal distribution table which gives \(\pm 1.96\).
03

Step 3a: Calculate Observed Value for \(H_{1}: \mu \neq 8.5\)

Using the formula for a test statistic \(z = (\bar{X} - \mu) / (\sigma / \sqrt{n})\), substituting the given values yields \(z = (9.24 - 8.5) / (5.40 / \sqrt{18}) \approx 0.7337 .\)
04

Step 4a: Compare Observed and Critical Values for \(H_{1}: \mu \neq 8.5\)

Since the observed value \(0.7337\) falls within the critical region of \(\pm 1.96\), we do not reject the null hypothesis.
05

Step 2b: Calculate Critical Value for \(H_{1}: \mu > 8.5\)

The critical value of \(z\) for a one-tailed test (right-tail) with \(\alpha = .05\) is found by looking up in a standard normal distribution table which gives \(1.645 .\)
06

Step 3b: Calculate Observed Value for \(H_{1}: \mu > 8.5\)

The observed value is the same as calculated in Step 3a, \(z \approx 0.7337 .\)
07

Step 4b: Compare Observed and Critical Values for \(H_{1}: \mu > 8.5\)

Since the observed value \(0.7337\) is less than the critical value \(1.645\), we do not 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-test
A z-test is a statistical method used to determine whether there is a significant difference between sample and population means. This is particularly useful when the population standard deviation is known. To conduct a z-test, the sample mean, population mean, standard deviation, and sample size must be identified.

The z-test is categorized into two types based on the research hypothesis: one-tailed and two-tailed tests. In a two-tailed test, the hypothesis is that the population mean is not equal to a specified value, whereas a one-tailed test considers the mean is greater than or less than the value.

Calculating the z-test involves finding the z-score, which indicates how many standard deviations the sample mean is from the population mean. The formula is:

\[ z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]

Where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
  • A z-score closer to zero indicates that the sample mean is near the population mean.
  • A significant z-score suggests a significant difference between sample and population means, based on the chosen level of significance.
critical values
Critical values are thresholds for significance in hypothesis testing. They help to decide whether to reject or not the null hypothesis. You determine these from the standard normal distribution (z-distribution) table.

For different tests and significance levels, the values vary:
  • In a two-tailed test with \( \alpha = 0.05 \), the critical values are ±1.96.
  • For a one-tailed test with the same alpha, the critical value is either 1.645 or -1.645, depending on whether it is a right or left-tailed test.


These critical values create a boundary for the acceptance region of a hypothesis test. If the test statistic lies outside this range, the null hypothesis is rejected, indicating a significant result.
observed values
Observed values, in the context of hypothesis testing, refer to the test statistic result calculated from the sample data. This observed z-value tells us how far and in what direction the sample mean deviates from the population mean, taking into account the sample size and population standard deviation.

To compute the observed value using the given formula, substitute the known values into:

\[ z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]

Observed values are crucial for decision making in hypothesis testing. You compare them with the critical values to determine whether the observed data fall within the acceptance region or rejection region of the null hypothesis.
  • If the observed value lies within the critical region, the null hypothesis is rejected.
  • If outside, we do not reject the null hypothesis.
level of significance
The level of significance, often denoted by \(\alpha\), is a threshold for determining how unlikely a result must be to reject the null hypothesis. Commonly used values are \(0.05\), \(0.01\), and \(0.10\).

A lower alpha level means a stricter criterion for rejecting the null hypothesis. Conversely, a higher alpha level is more lenient.

Choosing an appropriate level of significance depends on the context of the test and the consequences of making a Type I error (incorrectly rejecting a true null hypothesis).
  • At \(\alpha = 0.05\), there is a 5% risk of incorrectly rejecting the null hypothesis.
  • This level strikes a balance between sensitivity and specificity in many research fields.


In hypothesis testing, the level of significance determines the critical values, which form the boundaries for the acceptance region of the null hypothesis.

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Most popular questions from this chapter

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 longdistance calls made by residential customers of this company 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 \(\alpha=.05\) ?

According to Moebs Services Inc., the average cost of an individual checking account to major U.S. banks was \(\$ 380\) in 2013 (www. moebs.com). A bank consultant wants to determine whether the current mean cost of such checking accounts at major U.S. banks is more than \(\$ 380\) a year, A recent random sample of 150 such checking accounts taken from major U.S. banks produced a mean annual cost to them of \(\$ 390\). Assume that the standard deviation of annual costs to major banks of all such checking accounts is \(\$ 60 .\) a. Find the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, would you reject the null hypothesis if the maximum probability of Type I error is to be \(05 ?\) What if the maximum probability of Type I error is to be \(.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05\). Would you reject the null hypothesis? What if \(\alpha=.01 ?\) What if \(a=0\) ?

What are the five steps of a test of hypothesis using the critical value approach? Explain briefly,

According to the U.S. Bureau of Labor Statistics, all workers in America who had a bachelor's degree and were employed earned an average of \(\$ 1224\) a week in 2014 . A recent sample of 400 American workers who have a bachelor's degree showed that they earn an average of \(\$ 1260\) per week. Suppose that the population standard deviation of such earnings is \(\$ 160\). a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the current mean weekly earning of American workers who have a bachelor's degree is higher than \(\$ 1224\). Will you reject the null hypothesis at \(\alpha=.025 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.025\).

A real estate agent claims that the mean living area of all singlefamily homes in his county is at most 2400 square feet. A random sample of 50 such homes selected from this county produced the mean living area of 2540 square feet and a standard deviation of 472 square feet. a. Using \(\alpha=.05\), can you conclude that the real estate agent's claim is true? b. What will your conclusion be if \(\alpha=.01 ?\) Comment on the results of parts a and \(\mathrm{b}\).

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