Post Two Of The Six Exercises From The List For Hypothesis T
Post Two Of The Six Exercises From The List For Hypothesis Testing P
Post two of the six exercises from the list for Hypothesis Testing. Post both exercises in a single thread for grading. Include the chapter number and exercise number in the title of the posting, e.g., 9.5.75 & 9.5.115. Hypothesis Testing-Single Sample: 9.5.74, 9.5.75, 9.5.78, 9.5.79, 9.5.80, 9.5.81 Use a level of 5% unless otherwise stated. If the sample size is less than n=30, use a "t" distribution instead of "Z." Do not compute p-value for "t"distribution problems.
Just compare the test value of "t" with the critical value corresponding to the given level. See examples in Videos-Topics In Stat 230. Click on link below for example of p-value for lower tail test. Lower Tail Hypothesis Test.JPG Exercise 9.5.74 A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000.
A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim? Exercise 9.5.75 From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years.
A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? Exercise 9.5.78 The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure.
They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten? Exercise 9.5.79 In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week.
Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?
Paper For Above instruction
Hypothesis testing is a foundational statistical method used to determine whether there is enough evidence in a sample of data to infer that a certain condition holds for the entire population. It involves formulating two competing hypotheses: the null hypothesis (H₀), which represents a skeptical perspective or status quo, and the alternative hypothesis (H₁), which represents a new claim or what the researcher aims to support. This paper explores the application of hypothesis testing through two examples: the analysis of tire lifespan and the evaluation of the average work week for women, demonstrating the practical implementation of statistical tests at the 5% significance level.
Example 1: Tire Lifespan Analysis
The claim by a tire company that its deluxe tires last at least 50,000 miles provides a basis for testing the null hypothesis that the population mean lifespan μ is greater than or equal to 50,000 miles. Given the standard deviation (σ) of 8,000 miles and a sample size of 28 tires, with a sample mean (x̄) of 46,500 miles, we perform a one-sample z-test to assess the validity of the company's claim. The null hypothesis is H₀: μ ≥ 50,000, and the alternative hypothesis is H₁: μ
The test statistic is calculated as:
z = (x̄ - μ₀) / (σ / √n) = (46,500 - 50,000) / (8,000 / √28) ≈ -3.07
The critical value for a 5% level of significance (α=0.05) in a left-tailed test is approximately -1.645. Since the calculated z-value of -3.07 is less than -1.645, we reject the null hypothesis, suggesting that the actual mean lifespan is significantly less than 50,000 miles, thereby challenging the company's claim.
Example 2: Women's Work Week Study
Investigating whether the average work week for women has increased involves testing the null hypothesis H₀: μ = 80 hours against the alternative H₁: μ > 80 hours, indicating a right-tailed test. A sample of 81 women yields a mean of 83 hours and a standard deviation of 10 hours. Since the sample size exceeds 30, we employ a z-test.
The test statistic is:
z = (x̄ - μ₀) / (s / √n) = (83 - 80) / (10 / √81) = 3 / (10 / 9) = 3 / 1.111 ≈ 2.70
The critical value for a one-sided test at α=0.05 is approximately 1.645. Because 2.70 > 1.645, we reject H₀, indicating there is statistically significant evidence that the mean work week has increased for women.
Conclusion
In hypothesis testing, the comparison of the calculated test statistic with the critical value determines the decision regarding the null hypothesis. Both examples demonstrate the application of z-tests in practical scenarios—assessing tire durability claims and evaluating changes in work habits—highlighting the importance of statistical inference in research and decision-making processes.
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