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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

An earlier study claimed that U.S. adults spent an average of 114 minutes per day with their family. A recently taken sample of 25 adults from a city showed that they spend an average of 109 minutes per day with their family. The sample standard deviation is 11 minutes. Assume that the times spent by adults with their families have an approximate normal distribution. a. Using a \(1 \%\) significance level, test whether the mean time spent currently by all adults with their families in this city is different from 114 minutes a day. b. Suppose the probability of making a Type I error is zero. Can you make a decision for the test of part a without going through the five steps of hypothesis testing? If yes, what is your decision? Explain.

By rejecting the null hypothesis in a test of hypothesis example, are you stating that the alternative hypothesis is true?

A random sample of 8 observations taken from a population that is normally distributed produced a sample mean of \(44.98\) and a standard deviation of \(6.77\). Find the critical and observed values of \(t\) and the range for the \(p\) -value for each of the following tests of hypotheses, using \(\alpha=.05\). a. \(H_{0}: \mu=50\) versus \(H_{1}: \mu \neq 50\) b. \(H_{0}: \mu=50\) versus \(H_{1}: \mu<50\)

According to an article in Forbes magazine of April 3, 2014 , \(57 \%\) of students said that they did not attend the college of their first choice due to financial concerns (www.forbes.com). In a recent poll of 1600 students, 864 said that they did not attend the college of their first choice due to financial concerns. Using a \(1 \%\) significance level. perform a test of hypothesis to determine whether the current percentage of students who did not attend the college of their first choice due to financial concerns is lower than \(57 \%\). Use both the \(p\) -value and the critical-value approaches.

Shulman Steel Corporation makes bearings that are supplied to other companies. One of the machines makes bearings that are supposed to have a diameter of 4 inches. The bearings that have a diameter of either more or less than 4 inches are considered defective and are discarded. When working properly, the machine does not produce more than \(7 \%\) of bearings that are defective. The quality control inspector selects a sample of 200 bearings each week and inspects them for the size of their diameters. Using the sample proportion, the quality control inspector tests the null hypothesis \(p \leq .07\) against the alternative hypothesis \(p>.07\), where \(p\) is the proportion of bearings that are defective. He always uses a \(2 \%\) significance level. If the null hypothesis is rejected, the machine is stopped to make any necessary adjustments. One sample of 200 bearings taken recently contained 22 defective bearings.

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