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Make the following tests of hypotheses. a. \(H_{0}: \mu=80, \quad H_{1}: \mu \neq 80, \quad n=33, \quad \bar{x}=76.5, \quad \sigma=15, \quad \alpha=.10\) b. \(H_{0}=\mu=32, \quad H_{1}: \mu<32, \quad n=75, \quad \bar{x}=26.5, \quad \sigma=7.4, \quad \alpha=.01\) c. \(H_{0}=\mu=55, \quad H_{1}: \mu>55, \quad n=40, \bar{x}=60.5, \quad \sigma=4, \quad \alpha=.05\)

Short Answer

Expert verified
To answer the exercise, z-scores must be calculated for each set and compared with the respective critical z-values. The null hypothesis is rejected if the calculated z-score lies outside the range defined by the z-critical values for a given significance level, otherwise, it is retained.

Step by step solution

01

Step 1- Compute the Test Statistic for first hypothesis

Here, the null hypothesis is H鈧: 渭 = 80 and the alternative hypothesis is H鈧: 渭 鈮 80. Using the Z-score formula, calculate the z-score: z = (76.5 - 80) / (15 / 鈭33).
02

Step 2- Decide to Reject or Retain the First Null Hypothesis

For a two-tailed test with a significance level of 0.10, the z-critical values are -1.645 and +1.645. If the calculated z-score lies outside of this range, the null hypothesis will be rejected in favor of the alternative hypothesis.
03

Step 3- Compute the Test Statistic for Second Hypothesis

The null hypothesis is H鈧: 渭 = 32 and the alternative is H鈧: 渭 < 32. Calculate the z-score: z = (26.5 - 32) / (7.4 / 鈭75).
04

Step 4- Decide to Reject or Retain the Second Null Hypothesis

For a left-tailed test, the z-critical value at a 0.01 significance level is -2.33. If the calculated z-score is less than -2.33, the null hypothesis will be rejected.
05

Step 5- Compute the Test Statistic for Third Hypothesis

The null hypothesis is H鈧: 渭 = 55 and the alternative hypothesis is H鈧: 渭 > 55. Calculate the z-score: z = (60.5 - 55) / (4 / 鈭40).
06

Step 6- Decide to Reject or Retain the Third Null Hypothesis

For a right-tailed test, the z-critical value with a 0.05 significance level is 1.645. If the calculated z-score is greater than 1.645, then the null hypothesis should be rejected.

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Key Concepts

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

Understanding Z-score Calculation
The z-score is a statistical figure that indicates how many standard deviations an element is from the mean. In hypothesis testing, it helps determine if the observed data significantly deviates from the null hypothesis.
To calculate the z-score, we use the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where:
  • \(\bar{x}\) is the sample mean
  • \(\mu\) is the population mean under the null hypothesis
  • \(\sigma\) is the population standard deviation
  • \(n\) is the sample size
By plugging these values into the formula, we get a z-score that helps determine whether the observed mean is significantly different from the hypothesized mean. If the z-score is far from zero, it indicates a higher degree of deviation from the null hypothesis mean.
In our examples, we compute the z-score for each hypothesis to see how the sample mean compares to what is expected under the null hypothesis.
Decoding Critical Value
The critical value is an essential part of hypothesis testing. It marks the boundary or cutoff point beyond which the null hypothesis will be rejected.
Critical values differ depending on the type of test being conducted (two-tailed, left-tailed, or right-tailed) and the significance level \(\alpha\).
  • In a two-tailed test, critical values form a range, with both positive and negative cutoffs.
  • In left or right-tailed tests, there is only one critical value to be considered.
For the significance level \(\alpha\), we determine how confident we want to be about our decision to reject or accept the null hypothesis.
For example, in our exercise:
  • With a two-tailed test at \(\alpha = 0.10\), the critical values are \(-1.645\) and \(+1.645\).
  • In a left-tailed test at \(\alpha = 0.01\), the critical value is \(-2.33\).
  • In a right-tailed test at \(\alpha = 0.05\), the critical value is \(+1.645\).
Comparing the calculated z-score with these critical values guides the decision on whether to reject the null hypothesis.
Grasping the Null Hypothesis
The null hypothesis \(H_0\) is a default statement that assumes no effect or no difference in the population parameter being tested. It is the hypothesis that researchers typically aim to challenge or test against.
\(H_0\) generally posits that there is no change, no difference, or no effect. In our exercise examples, the null hypotheses suggest certain population means, such as \(\mu = 80\), \(\mu = 32\), and \(\mu = 55\).
In the context of hypothesis testing, we either retain the null hypothesis when the data supports it, or reject it when sufficient evidence suggests an alternative scenario. The null hypothesis serves as a critical baseline for statistical testing.
Exploring the Alternative Hypothesis
The alternative hypothesis \(H_1\) explains what the null hypothesis does not. It suggests that there is a significant effect, difference, or change.
\(H_1\) is what researchers desire to prove. It indicates the presence of conditions such as a mean different from a certain value, a mean less than a particular number, or a mean greater than a particular number.
  • In our two-tailed test example, \(H_1: \mu eq 80\) indicates mean different from 80.
  • For the left-tailed test, \(H_1: \mu < 32\) anticipates a mean less than 32.
  • In the right-tailed test scenario, \(H_1: \mu > 55\) hypothesizes a mean greater than 55.
The role of the alternative hypothesis is pivotal, as it guides decision-making in statistical testing. Rejecting the null hypothesis in favor of the alternative influences conclusions drawn from the analysis.

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Most popular questions from this chapter

A paint manufacturing company claims that the mean drying time for its paints is not longer than 45 minutes. A random sample of 20 gallons of paints selected from the production line of this company showed that the mean drying time for this sample is \(49.50\) minutes with a standard deviation of 3 minutes. Assume that the drying times for these paints have a normal distribution. a. Using a \(1 \%\) significance level, would you conclude that the company's claim is true? b. What is the Type I error in this exercise? Explain in words. What is the probability of making such an error?

For each of the following examples of tests of hypotheses about \(\mu\), show the rejection and nonrejection regions on the sampling distribution of the sample mean assuming that it is normal. a. A two-tailed test with \(\alpha=.05\) and \(n=40\) b. A left-tailed test with \(\alpha=.01\) and \(n=20\) c. A right-tailed test with \(\alpha=.02\) and \(n=55\)

food company is planning to market a new type of frozen yogurt. However, before marketing thit yogurt, the company wants to find what percentage of the people like it. The company's management has decided that it will market this yogurt only if at least \(35 \%\) of the people like it. The company's researcl department selected a random sample of 400 persons and asked th taste this yogurt. Of these 400 persons, 112 said they liked a. Testing at a \(2.5 \%\) significance level, can you conclude that the company should market this yogurt b. What will your decision be in part a if the probability of making a Type I error is zero? Explain Make the test of part a using the \(p\) -value approach.min

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

Consider the null hypothesis \(H_{0}: \mu=5 .\) A random sample of 140 observations is taken from a population with \(\sigma=17\). Using \(\alpha=.05\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(z\) for the following. \(\begin{array}{lll}\text { a. a right-tailed test } & \text { b. a left-tailed test } & \text { c. a two-tailed test }\end{array}\)

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