Assume that the differences are normally distributed. Complete parts (a) through (d) below. Observation 1 2 3 4 5 6 7 8 Upper X Subscript iXi 52.152.1 47.447.4 47.647.6 47.347.3 47.347.3 47.447.4 48.848.8 49.249.2 Upper Y Subscript iYi 53.953.9 48.148.1 49.749.7 51.251.2 48.548.5 49.849.8 52.152.1 48.448.4 (a) Determine d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi for each pair of data. Observation 1 2 3 4 5 6 7 8 di negative −1.8 negative −0.7 negative −2.1 negative −3.9 negative −1.2 negative −2.4 negative −3.3 0.8 (Type integers or decimals.) (b) Compute d overbard and s Subscript dsd. d overbardequals=negative 1.825 (Round to three decimal places as needed.) s Subscript dsdequals=1.485 (Round to three decimal places as needed.) (c) Test if mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. What are the correct null and alternative hypotheses? A. Upper H 0H0: mu Subscript dμdgreater than>0 Upper H 1H1: mu Subscript dμdless than<0 B. Upper H 0H0: mu Subscript dμdequals=0 Upper H 1H1: mu Subscript dμdless than<0 Your answer is correct. C. Upper H 0H0: mu Subscript dμdless than<0 Upper H 1H1: mu Subscript dμdequals=0 D. Upper H 0H0: mu Subscript dμdless than<0 Upper H 1H1: mu Subscript dμdgreater than>0 P-valueequals=0.0050.005 (Round to three decimal places as needed.) Choose the correct conclusion below. A. Do not rejectDo not reject the null hypothesis. There is insufficientinsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. B. RejectReject the null hypothesis. There is insufficientinsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. C. Do not rejectDo not reject the null hypothesis. There is sufficientsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. D. RejectReject the null hypothesis. There is sufficientsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. Your answer is correct. (d) Compute a 95% confidence interval about the population mean difference mu Subscript dμd. The lower bound is nothing. The upper bound is nothing. (Round to two decimal places as needed.)

Assume that the differences are normally distributed. Complete parts (a) through (d) below. Observation 1 2 3 4 5 6 7 8 Upper X Subscript iXi 52.152.1 47.447.4 47.647.6 47.347.3 47.347.3 47.447.4 48.848.8 49.249.2 Upper Y Subscript iYi 53.953.9 48.148.1 49.749.7 51.251.2 48.548.5 49.849.8 52.152.1 48.448.4 (a) Determine d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi for each pair of data. Observation 1 2 3 4 5 6 7 8 di negative −1.8 negative −0.7 negative −2.1 negative −3.9 negative −1.2 negative −2.4 negative −3.3 0.8 (Type integers or decimals.) (b) Compute d overbard and s Subscript dsd. d overbardequals=negative 1.825 (Round to three decimal places as needed.) s Subscript dsdequals=1.485 (Round to three decimal places as needed.) (c) Test if mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. What are the correct null and alternative hypotheses? A. Upper H 0H0: mu Subscript dμdgreater than>0 Upper H 1H1: mu Subscript dμdless than<0 B. Upper H 0H0: mu Subscript dμdequals=0 Upper H 1H1: mu Subscript dμdless than<0 Your answer is correct. C. Upper H 0H0: mu Subscript dμdless than<0 Upper H 1H1: mu Subscript dμdequals=0 D. Upper H 0H0: mu Subscript dμdless than<0 Upper H 1H1: mu Subscript dμdgreater than>0 P-valueequals=0.0050.005 (Round to three decimal places as needed.) Choose the correct conclusion below. A. Do not rejectDo not reject the null hypothesis. There is insufficientinsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. B. RejectReject the null hypothesis. There is insufficientinsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. C. Do not rejectDo not reject the null hypothesis. There is sufficientsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. D. RejectReject the null hypothesis. There is sufficientsufficient evidence that mu Subscript dμdless than<0 at the alphaαequals=0.05 level of significance. Your answer is correct. (d) Compute a 95% confidence interval about the population mean difference mu Subscript dμd. The lower bound is nothing. The upper bound is nothing. (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

Related questions

Concept explainers

Topic Video

Question

Assume that the differences are normally distributed. Complete parts (a) through (d) below.

Observation	1	2	3	4	5	6	7	8
Upper X Subscript iXi	52.152.1	47.447.4	47.647.6	47.347.3	47.347.3	47.447.4	48.848.8	49.249.2
Upper Y Subscript iYi	53.953.9	48.148.1	49.749.7	51.251.2	48.548.5	49.849.8	52.152.1	48.448.4

(a) Determine

d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi

for each pair of data.

Observation	1	2	3	4	5	6	7	8
di	negative −1.8	negative −0.7	negative −2.1	negative −3.9	negative −1.2	negative −2.4	negative −3.3	0.8

(Type integers or decimals.)

(b) Compute

d overbard

and

s Subscript dsd.

d overbardequals=negative 1.825

(Round to three decimal places as needed.)

s Subscript dsdequals=1.485

(Round to three decimal places as needed.)

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

What are the correct null and alternative hypotheses?

Upper H 0H0:

mu Subscript dμdgreater than>0

Upper H 1H1:

mu Subscript dμdless than<0

Upper H 0H0:

mu Subscript dμdequals=0

Upper H 1H1:

mu Subscript dμdless than<0

Your answer is correct.

Upper H 0H0:

mu Subscript dμdless than<0

Upper H 1H1:

mu Subscript dμdequals=0

Upper H 0H0:

mu Subscript dμdless than<0

Upper H 1H1:

mu Subscript dμdgreater than>0

P-valueequals=0.0050.005

(Round to three decimal places as needed.)

Choose the correct conclusion below.

Do not rejectDo not reject

the null hypothesis. There is

insufficientinsufficient

evidence that

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

RejectReject

the null hypothesis. There is

insufficientinsufficient

evidence that

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

Do not rejectDo not reject

the null hypothesis. There is

sufficientsufficient

evidence that

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

RejectReject

the null hypothesis. There is

sufficientsufficient

evidence that

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

Your answer is correct.

(d) Compute a 95% confidence interval about the population mean difference

mu Subscript dμd.

The lower bound is

nothing.

The upper bound is

nothing.

(Round to two decimal places as needed.)

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

Trending now

This is a popular solution!

Step by step

Solved in 7 steps with 6 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Centre, Spread, and Shape of a Distribution$

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.