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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 50 \\ H_{\mathrm{a}}: \mu>50 \end{array} \\] A sample of 60 is used and the population standard deviation is \(8 .\) Use the critical value approach to state your conclusion for each of the following sample results. Use \\[ \alpha=.05 \\] a. \(\bar{x}=52.5\) b. \(\bar{x}=51\) c. \( \bar{x}=51.8\)

Short Answer

Expert verified
Reject \( H_0 \) for (a) and (c); fail to reject \( H_0 \) for (b).

Step by step solution

01

Define the Hypotheses

We are testing the null and alternative hypotheses given by:\[\begin{align*}H_0: & \quad \mu \leq 50 \H_a: & \quad \mu > 50\end{align*}\]This is a one-tailed test because the alternative hypothesis is testing whether the mean is greater than 50.
02

Identify the Test Statistic

Since the population standard deviation is known and the sample size is large, we use the Z-test for the test statistic. The formula is given by:\[Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}\]where \( \bar{x} \) is the sample mean, \( \mu_0 = 50 \) is the value under the null hypothesis, \( \sigma = 8 \) is the population standard deviation, and \( n = 60 \) is the sample size.
03

Determine the Critical Value

For a significance level \( \alpha = 0.05 \) in a one-tailed test, the critical value of \( Z \) is found using a standard normal distribution table. We find that the critical value is about 1.645. Thus, if the calculated \( Z \)-value exceeds 1.645, we will reject the null hypothesis.
04

Calculate the Test Statistic for (a)

Using the sample mean \( \bar{x} = 52.5 \), calculate the test statistic:\[Z = \frac{52.5 - 50}{8/\sqrt{60}} \approx \frac{2.5}{1.03} \approx 2.43\]Since \( Z = 2.43 > 1.645 \), we reject the null hypothesis.
05

Calculate the Test Statistic for (b)

Using the sample mean \( \bar{x} = 51 \), calculate the test statistic:\[Z = \frac{51 - 50}{8/\sqrt{60}} \approx \frac{1}{1.03} \approx 0.97\]Since \( Z = 0.97 < 1.645 \), we fail to reject the null hypothesis.
06

Calculate the Test Statistic for (c)

Using the sample mean \( \bar{x} = 51.8 \), calculate the test statistic:\[Z = \frac{51.8 - 50}{8/\sqrt{60}} \approx \frac{1.8}{1.03} \approx 1.75\]Since \( Z = 1.75 > 1.645 \), we 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 if there is a significant difference between the sample mean and the population mean. This test is especially useful when the sample size is large (usually over 30), or the population standard deviation is known. It helps us decide whether to accept or reject a null hypothesis.In the given situation, the population standard deviation is 8, and our sample size is 60. Therefore, the z-test is appropriate. Here's how it works:- We calculate the z-score using the formula \( Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} \). - \( \bar{x} \) is the sample mean, \( \mu_0 \) is the hypothesized population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size.By computing this z-score, we can conclude if our sample mean statistically supports our null hypothesis.
Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that there is no effect or no difference. It is the default or original status quo that we assume to be true unless evidence suggests otherwise. For this exercise, the null hypothesis \( H_0 \) posits that the population mean is less than or equal to 50 \( (\mu \leq 50) \). It assumes that any observed difference is due to random sampling variability, rather than a true effect.We try to gather evidence against the null hypothesis by analyzing the sample data. If the data presents a convincing case, we reject the null hypothesis. If not, we fail to reject it. Not rejecting doesn't confirm the null hypothesis; it merely indicates insufficient evidence to disprove it under our test conditions.
Critical Value
The critical value in hypothesis testing is the threshold that the test statistic must exceed for us to reject the null hypothesis. It is determined by the chosen significance level \( \alpha \). With a significance level of \( \alpha = 0.05 \) for a one-tailed test, like in our case:
  • We look up a z-table to find the critical z-value that corresponds to this level.
  • For \( \alpha = 0.05 \), the critical value is approximately 1.645.
  • In practical terms, if the z-score we calculate from our sample exceeds this critical value, we reject the null hypothesis. If it doesn't, we don't have enough evidence to make that decision. The critical value helps us control the risk of making a Type I error, which is wrongly rejecting a true null hypothesis.
    Sample Mean
    The sample mean \( \bar{x} \) is an average of the values in a sample, providing an estimate of the population mean \( \mu \). It is calculated by summing all observed values, then dividing by the number of observations.In our problem, the sample means were given for different cases: \( \bar{x} = 52.5 \), \( \bar{x} = 51 \), and \( \bar{x} = 51.8 \). Each of these means represents the central value around which the sample data is concentrated.The sample mean is utilized in the z-test formula to determine how much it deviates from the hypothesized population mean (null hypothesis). By comparing sample means with the population mean, we gauge the significance of our results and decide if observed variations are consistent with the null hypothesis or if they indicate an effect present in the population.

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    Most popular questions from this chapter

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