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B. Analyze and solve the following problem. The researcher wants to determine wheter the mean lifespan of his specimen of 28 in number is significantly different from the average lifespan of the population which is 95 days. The mean and standard deviation of the lifespan of his specimen is 90 days and 10 days, respectively. Assume a 90% confidence level and interpret the result.

B. Analyze and solve the following problem. The researcher wants to determine wheter the mean lifespan of his specimen of 28 in number is significantly different from the average lifespan of the population which is 95 days. The mean and standard deviation of the lifespan of his specimen is 90 days and 10 days, respectively. Assume a 90% confidence level and interpret the result.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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II. Learning Activity: The t-distribution, just like the z-distribution or standard normal curve, is bell-shaped and unimodal. It is symmetric about = 0. However, it is flatter in the middle and has more area in its tail than that of the standard normal curve. Its shape depends on the sample size . As the sample size becomes larger, the t - distribution gets closer to the standard normal distribution. Flatter V Wider area of tails standardi normal curve t-distribution curve The t-distribution is used with small samples taken from population that is approximately normal. The z - statistic is used when ≥ 30 while t - statistic is used when <30. The t - statistic below uses the sample deviation especially when the population variance is unknown. #U The t - distribution formula is: = where: - sample mean -population mean - sample size - standard deviation of the sample mean Example # 1 A sample of size = 20 is a random sample selected from a normally distributed population. Find the value of such that the shaded area to the left of - is 0.05 Solution: Find the degree of freedom 0. 0.05 0.05 y one-tail/= 0.50 0.25 0.20 0.05 two-tails / a 1.00 0.50 0.40 0.10 df 1 0.000 1.000 1.376 6.314 2 0.000 0.816 1.061 2.920 3 0.000 0.765 0.978 2.353 *** www 19 0.000 0.688 0.861 1.729 20 0.000 0.687 0.860 1.725 A portion of the t - table Since the distribution is symmetric about 0, then the area to the right of is 0.05 also. In the Table of t- Critical Values, move down the first column headed df until = 19. Move to the right until the column headed 0.05 (area in one-tail) or 0.10 (area in two-tails). Therefore, D.=0.000 with 00 = 00. Example # 2 For a t-distribution with 27 degrees of freedom, find the value of such that the area between - and is 0.90. Solution: a. The degree of freedom = 27. b. (1)100% = 90% (1)1 = 0.90 0.90 = 0.10 (two-tails) == 0.05 (one-tail) -t t In the Table of t-Critical Values, move down the first column headed df until = 27. Move to the right until the column headed 0.05 (area in one-tail) or 0.10 (area in two-tails). Hence, D.00 = 0.000 with DD = 00. 1-0 =00 00=20-1 □□ = 19 NO -t t
Transcribed Image Text:II. Learning Activity: The t-distribution, just like the z-distribution or standard normal curve, is bell-shaped and unimodal. It is symmetric about = 0. However, it is flatter in the middle and has more area in its tail than that of the standard normal curve. Its shape depends on the sample size . As the sample size becomes larger, the t - distribution gets closer to the standard normal distribution. Flatter V Wider area of tails standardi normal curve t-distribution curve The t-distribution is used with small samples taken from population that is approximately normal. The z - statistic is used when ≥ 30 while t - statistic is used when <30. The t - statistic below uses the sample deviation especially when the population variance is unknown. #U The t - distribution formula is: = where: - sample mean -population mean - sample size - standard deviation of the sample mean Example # 1 A sample of size = 20 is a random sample selected from a normally distributed population. Find the value of such that the shaded area to the left of - is 0.05 Solution: Find the degree of freedom 0. 0.05 0.05 y one-tail/= 0.50 0.25 0.20 0.05 two-tails / a 1.00 0.50 0.40 0.10 df 1 0.000 1.000 1.376 6.314 2 0.000 0.816 1.061 2.920 3 0.000 0.765 0.978 2.353 *** www 19 0.000 0.688 0.861 1.729 20 0.000 0.687 0.860 1.725 A portion of the t - table Since the distribution is symmetric about 0, then the area to the right of is 0.05 also. In the Table of t- Critical Values, move down the first column headed df until = 19. Move to the right until the column headed 0.05 (area in one-tail) or 0.10 (area in two-tails). Therefore, D.=0.000 with 00 = 00. Example # 2 For a t-distribution with 27 degrees of freedom, find the value of such that the area between - and is 0.90. Solution: a. The degree of freedom = 27. b. (1)100% = 90% (1)1 = 0.90 0.90 = 0.10 (two-tails) == 0.05 (one-tail) -t t In the Table of t-Critical Values, move down the first column headed df until = 27. Move to the right until the column headed 0.05 (area in one-tail) or 0.10 (area in two-tails). Hence, D.00 = 0.000 with DD = 00. 1-0 =00 00=20-1 □□ = 19 NO -t t
B. Analyze and solve the following problem. The researcher wants to determine wheter the mean lifespan of his specimen of 28 in number is significantly different from the average lifespan of the population which is 95 days. The mean and standard deviation of the lifespan of his specimen is 90 days and 10 days, respectively. Assume a 90% confidence level and interpret the result.
Transcribed Image Text:B. Analyze and solve the following problem. The researcher wants to determine wheter the mean lifespan of his specimen of 28 in number is significantly different from the average lifespan of the population which is 95 days. The mean and standard deviation of the lifespan of his specimen is 90 days and 10 days, respectively. Assume a 90% confidence level and interpret the result.
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