1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what are the critical values for testing equality of mean weights ? A. 2.18 B. -2.18 and 2.18 C. -1.78 D. -1.78 and 1.78 2. What is your decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D. The computed test statistic does not fall in the critical region and we reject the null hypothesis. 3.What is the best conclusion based on problem 2? A. At 5% level, the two varieties of lettuce have exactly the same average weight. B. At 5% level, the two varieties of lettuce do not have significantly different average weight. C. At 5% level, the two varieties of lettuce have significantly different average weight. D. At 5% level, one variety of lettuce has significantly higher average weight.

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

Related questions

Concept explainers

Question

1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what are the critical values for testing equality of mean weights ?

A. 2.18

B. -2.18 and 2.18

C. -1.78

D. -1.78 and 1.78

2. What is your decision using critical value approach in problem 1?

A. The computed test statistic falls in the critical region and we do not reject the null hypothesis.

B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis.

C. The computed test statistic falls in the critical region and we reject the null hypothesis.

D. The computed test statistic does not fall in the critical region and we reject the null hypothesis.

3.What is the best conclusion based on problem 2?

A. At 5% level, the two varieties of lettuce have exactly the same average weight.

B. At 5% level, the two varieties of lettuce do not have significantly different average weight.

C. At 5% level, the two varieties of lettuce have significantly different average weight.

D. At 5% level, one variety of lettuce has significantly higher average weight.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1

Given mean sample weights of two varieties of lettuce
For lettuce-1
Mean = 3.259
Standard deviation = 0.4
Sample size = 9

For lettuce-2
Mean = 1.413
Standard deviation = 0.220
Sample size = 6

1)
Here two sample pooled t test is conducted to test for equality of mean weights.
Degree of freedom = 9 + 6 - 2 = 13
With reference to t table for 13 degree of freedom and 95% confidence interval critical values are -2.16 and 1.26

Step by step

Solved in 2 steps with 1 images

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.

Similar questions