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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 \(H_{1}: \mu \neq 8.5\) b. \(H_{0}=\mu=8.5\) versus \(H_{1}: \mu>8.5\)

Short Answer

Expert verified
The critical values are approximately -1.96 and 1.96 for hypothesis a, and approximately 1.645 for hypothesis b. The observed value is approximately 0.70 for both cases. For both hypotheses, the observed value does not fall into the rejection regions, so we fail to reject the null hypothesis in both cases at the 0.05 significance level. That is, there's not sufficient evidence to prove that the population mean deviates from 8.5.

Step by step solution

01

Calculate the Observed Values for Both Hypotheses

We can calculate the observed z-score using the formula: \( Z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}}\). Here, \(\bar{x}\) = sample mean = 9.24, \(\mu_{0}\) = hypothesized population mean = 8.5, \(\sigma\) = population standard deviation = 5.40, and \(n\) = sample size = 18. For both hypotheses a and b, the observed z-score would be the same, let's calculate: \( Z = \frac{9.24 - 8.5}{\frac{5.40}{\sqrt{18}}}\) which equals approximately 0.70.
02

Calculate the Critical Values for Hypotheses a and b

The critical value corresponds to the value of z dividing the rejection region(s) from the non-rejection region(s). - For hypothesis a (\(H_{1}: \mu \neq 8.5\)), it is a two-tailed test. Looking at the z-table for \(\alpha/2=0.025\) (the significance level divided by 2 because it's a two-tailed test), we get two critical values: \(z_{\alpha/2} = -1.96\) (for the left tail) and \(z_{1-\alpha/2} = 1.96\) (for the right tail). - For hypothesis b (\(H_{1}: \mu > 8.5\)), it is a one-tailed test. Looking at the z-table for \(\alpha=0.05\), we get one critical value: \(z_{1-\alpha} = 1.645\).
03

Make Conclusions

We compare the observed values with the critical values. - For hypothesis a, since the observed value 0.70 lies between the critical values -1.96 and 1.96, we fail to reject \(H_{0}\). This means there's not enough evidence at the 0.05 significance level to suggest that the population mean is different from 8.5. - For hypothesis b, since the observed value 0.70 is less than the critical value 1.645, we fail to reject \(H_{0}\). This implies that the data does not offer sufficient evidence at the 0.05 level of significance to conclude that the population mean is greater than 8.5.

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Key Concepts

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

z-score calculation
The z-score is a crucial component in hypothesis testing as it measures how many standard deviations an element is from the mean. This tells us where our sample mean falls in the distribution. Knowing where your sample mean lies can help determine how far it is from the hypothesized mean and whether it supports your hypothesis. To calculate the z-score, use the formula: \[ Z = \frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}} \] where:
  • \(\bar{x}\) is the sample mean,
  • \(\mu_{0}\) is the hypothesized population mean,
  • \(\sigma\) is the population standard deviation,
  • \(n\) is the sample size.
For a sample mean of 9.24, hypothesized mean 8.5, and a standard deviation of 5.40 with a sample size of 18, calculate: \[ Z = \frac{9.24 - 8.5}{\frac{5.40}{\sqrt{18}}} \] This results in a z-score of approximately 0.70. This value places the observed sample mean relative to the hypothesized mean in the distribution space. The z-score is then used to determine if the observed sample mean is significantly different from the population mean under the null hypothesis.
critical values
Critical values determine the boundary at which you decide to reject the null hypothesis. They separate the rejection region from the non-rejection region in a statistical test. To find the critical values, refer to a z-table or standard normal distribution table. The location of these critical values depends on the chosen significance level \( \alpha \), which in turn defines the probability of rejecting a true null hypothesis.For a two-tailed test with \( \alpha = 0.05 \), split the significance level into \( \alpha/2 = 0.025 \) for each tail. From the z-table, this gives critical values at \(-1.96\) and \(1.96\). These values define the cutoff points where if your z-score falls beyond them, you reject the null hypothesis. In a one-tailed test scenario with \( \alpha = 0.05 \), the critical value is determined directly from \( \alpha \), resulting in a single cutoff point at \(1.645\) or \(-1.645\) based on the test direction. The choice between a one-tailed and two-tailed test directly affects the selection and interpretation of critical values.
two-tailed test
A two-tailed test is used when you are testing for the possibility of an effect in either direction. In other words, you are checking if there is a significant deviation from the hypothesized mean, regardless of the direction. Use a two-tailed test when the alternative hypothesis is of the form \( H_1: \mu eq \mu_0 \). Here, the critical regions are located in both tails of the distribution. This means you have two critical values: one on the left and one on the right.Consider a significance level of \( 0.05 \), split into two critical regions with probabilities of \( 0.025 \) in each tail. A z-score that lands outside the range of \(-1.96\) to \(1.96\) (critical values for \(\alpha/2 = 0.025\)) would lead you to reject the null hypothesis.This approach ensures you remain unbiased by testing for extremes in both directions. Two-tailed tests are commonly used when initial predictions about the direction of the effect or deviation are not appropriate or unknown.
one-tailed test
One-tailed tests focus on finding an effect in a specific direction, either greater than or less than the hypothesized parameter. This focus can offer more power to detect an effect in that specified direction since the entire significance level is dedicated to that one side.If your alternative hypothesis is structured like \( H_1: \mu > \mu_0 \) or \( H_1: \mu < \mu_0 \), use a one-tailed test. This setup tests for the possibility of the value being either significantly greater or significantly less than the hypothesized mean.With a significance level of \( \alpha = 0.05 \), the critical value corresponds to \( 1.645 \) (or \(-1.645\), depending on the test's direction). If your calculated z-score surpasses this critical value (in the desired direction), reject the null hypothesis.One-tailed tests are beneficial when you have a clear prediction about the direction of the effect or simply want to test for an increase or decrease, allowing for a more focused statistical insight.

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Most popular questions from this chapter

For each of the following examples of tests of hypothesis about \(\mu\), show the rejection and nonrejection regions on the \(t\) distribution curve. a. A two-tailed test with \(\alpha=.01\) and \(n=15\) b. A left-tailed test with \(\alpha=.005\) and \(n=25\) c. A right-tailed test with \(\alpha=.025\) and \(n=22\)

Make the following tests of hypotheses. a. \(H_{0}: \mu=25, \quad H_{1}: \mu \neq 25, \quad n=81, \quad \bar{x}=28.5, \quad \sigma=3, \quad \alpha=.01\) b. \(H_{0}=\mu=12, \quad H_{1}: \mu<12, \quad n=45, \quad \bar{x}=11.25, \quad \sigma=4.5, \quad \alpha=.05\) c. \(H_{0}=\mu=40, \quad H_{1}: \mu>40, \quad n=100, \quad \bar{x}=47, \quad \sigma=7, \quad \alpha=.10\)

Brooklyn Corporation manufactures DVDs. The machine that is used to make these DVDs is known. to produce not more than \(5 \%\) defective DVDs. The quality control inspector selects a sample of \(200 \mathrm{DVDs}\) each week and inspects them for being good or defective. Using the sample proportion, the quality con trol inspector tests the null hypothesis \(p \leq .05\) against the alternative hypothesis \(p>.05\), where \(p\) is th proportion of DVDs that are defective. She always uses a \(2.5 \%\) significance level. If the null hypothesi: is rejected, the production process is stopped to make any necessary adjustments. A recent sample of 200 DVDs contained 17 defective DVD: Using a \(2.5 \%\) significance level, would you conclude that the prod" should be stoppe to make necessary adjustments b. Perform the test of part a using a \(1 \%\) significance level. Is your decision different from the one in.par Comment on the results of parts a and \(b\)

A company claims that the mean net weight of the contents of its All Taste cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that \(\sigma=.25\) ounce.

The mean balance of all checking accounts at a bank on December 31,2011, was \(\$ 850 .\) A random sample of 55 checking accounts taken recently from this bank gave a mean balance of \(\$ 780\) with a standard deviation of \(\$ 230 .\) Using a \(1 \%\) significance level, can you conclude that the mean balance of such accounts has decreased during this period? Explain your conclusion in words. What if \(\alpha=.025\) ?

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