### Statistical Hypothesis Testing ExampleA random sample of 100 observations from a population with a standard deviation of 39 yielded a sample mean of 107. Complete parts a through c below.**a. Test the null hypothesis that µ = 100 against the alternative hypothesis that µ > 100, using α = 0.05. Interpret the results of the test.**#### What is the value of the test statistic?\[ z = \square \text{ (Round to two decimal places as needed.)} \]This example illustrates hypothesis testing using a sample mean and known population standard deviation. The null hypothesis ($H_0$) states that the population mean ($µ$) is equal to 100, while the alternative hypothesis ($H_1$) states that the population mean ($µ$) is greater than 100. ### Step-by-Step Solution1. **State the Hypotheses:** - Null Hypothesis ($H_0$): $µ = 100$ - Alternative Hypothesis ($H_1$): $µ > 100$2. **Find the Test Statistic:** The test statistic is calculated using the formula for the z-score: \[ z = \frac{\overline{x} - µ}{\frac{σ}{\sqrt{n}}} \] Where: - $\overline{x}$ is the sample mean (107) - $µ$ is the population mean under the null hypothesis (100) - $σ$ is the population standard deviation (39) - $n$ is the sample size (100)3. **Calculate the Test Statistic:** \[ z = \frac{107 - 100}{\frac{39}{\sqrt{100}}} \] **Note:** Use this formula and given values to compute the z-score.4. **Compare the Test Statistic to the Critical Value:** For a one-tailed test with $α = 0.05$, the critical value (z-critical) is approximately 1.645.5. **Interpret the Results:** If the calculated z-value is greater than the z-critical value, we reject the null hypothesis.**Detailed Explanation:**- **Random Sample Size (n):** 100 observations- **Population Standard Deviation (σ

Name: ### Statistical Hypothesis Testing ExampleA random sample of 100 obse
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: ### Statistical Hypothesis Testing ExampleA random sample of 100 observations from a population with a standard deviation of 39 yielded a sample mean of 107. Complete parts a through c below.**a. Test the null hypothesis that µ = 100 against the alt

Answered: Image /qna-images/answer/dc691793-84b4-4347-ad34-ec402ea518bc.jpg

Math

Statistics

A random sample of 100 observations from a population with standard deviation 39 yielded a sample mean of a. Test the null hypothesis thatu = 100 against the alternative hypothesis that u > 100, using a =0.05. Interpret the results of the test. What is the value of the test statistic? (Round to two decimal places as needed.)

A random sample of 100 observations from a population with standard deviation 39 yielded a sample mean of a. Test the null hypothesis thatu = 100 against the alternative hypothesis that u > 100, using a =0.05. Interpret the results of the test. What is the value of the test statistic? (Round to two decimal places as needed.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

What’s the Z0 and Pvalue ?
Select the correct answer below: p-value >0.10 0.05< p-value <0.10 0.01< p-value <0.05 p-value <0.01
Many people believe that the average number of Facebook friends is 150. The population standard deviation is 37.6. A random sample of 46 high school students in a particular county revealed that the average number of Facebook friends was 159. At =α0.01, is there sufficient evidence to conclude that the mean number of friends is greater than 150? A. State the hypotheses and identify the claim with the correct hypothesis. :H0 ▼claim or not claim :H1 ▼claim or not claimThis hypotheses test is a ▼one-tailed or two-tailed test B. Find the critical value(s) (c)Compute the test value. z= (d)Make the decision. reject or accept (e)Summarize the results. is there enough evidence to support the claim
pls. answer using 7 steps:-Parameter of interest-Null hypothesisAlternative hypothesis -Test statistic -Reject H0 if: -Computations: -Conclusion
The test statistic, t, is - 22.67. (Round to two decimal places as needed.) The P-value is 0.000. (Round to three decimal places as needed.) State the conclusion for the test. A. Reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. C. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. D. Reject the null hypothesis. There is not sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda.
answer 3
In a sample of 12 randomly selected high school seniors, the mean score on a standardized test was 1182 and the standard deviation was 161.6.Further research suggests that the population mean score on this test for high school seniors is 1012. Does the t-value for the original sample fall between −t0.95 and t0.95? Assume that the population of test scores for high school seniors is normally distributed. What is The t-value? And does or does not it not fall between −t0.95 and t0.95 because t0.95=? (Round to two decimal places as needed.)
T3.6
The output below is for a t-test for the hypothesis: Individuals living in bad neighborhoods commit more crime than those living in good neighborhoods. The data in the Group Statistics section provides you with the average number of crimes committed by individuals living in good and bad neighborhoods and the standard deviation of this same variable. The findings of significance are located in the Independent Samples Test section. Pay particular attention to the "t" column and the "Sig (2-tailed)"column. The "t" column is the t value that you would have calculated by hand (as we learned in class). The "Sig (2-tailed)"column provides you with the p value (the level of significance of this relationship). In this column, any value below .05 indicates that you would have rejected your null hypothesis. 1. Can someone tell me about the relationship of all these data? I'm trying to determine if my hypothesis, listed below, can be proven or disproven? 2. What is the independent variable and…
An obstetrician read that a newborn baby loses on average 7 ounces in the first two days of his or her life. He feels that in the hospital where he works, the average weight loss of a newborn baby is less than 7 ounces. A random sample of 30 newborn babies has a mean weight loss of 6.4 ounces. The population standard deviation is 1.6 ounces. Is there enough evidence at =α0.01 to support his claim? Assume that the variable is normally distributed. Use the critical value method with tables. hello the question askas to find the critical value compute the test value and select the hypothesis
A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.02 pounds. At a = 0.06, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. OA. Ho: H>3.1 Ha: HS3.1 D. Ho: H=3.3 H₂:μ*3.3 OB. Ho: ≤3.3 H:H>3.3 OE. Ho: >3.3 Ha: 53.3 (b) Identify the standardized test statistic. z = (Round to two decimal places as needed.) OC. Ho: * 3.1 Ha: 3.1 OF. Ho: $3.1 H₂:μ>3.1
An online survey company claims their surveys take on average less than 20 minutes to complete. Before running a new ad campaign, they need data to prove that the average completion time is actually less than 20 minutes. The company randomly samples 15 surveys taken in the last week and found a sample mean of 17 minutes with a sample standard deviation of 5 minutes. What is the alternative hypothesis of this test. Be sure to include symbol and number needed to form the hypothesis. HA : µ Find the test statistic for this test. (round answer to 3 decimal places) Find the p-value for this test. (round answer to 3 decimal places)