Question

35 33 34 11. (10 points) Ten randomly selected families are tested for the number of gallons of water a day they use before and after viewing a conservation video. The results are shown below. Before 33 38 33 35 40 40 40 31 After 28 25 28 35 33 31 28 35 33 Assume that the paired differences are normally distributed. At the 1% significance level, do the data provide sufficient evidence to conclude that the mean amount of water use after the viewing differs from the mean amount of water use before the viewing? a) State the null and alternative hypotheses symbolically. b) Check assumptions and choose an appropriate test procedure. c) Find the paired sample differences and use 1-var-stats to find the mean and the standard deviation of the paired sample differences. d) Compute the value of the test statistic. e) Determine the P-value. Please show your work. f) Decide on the null hypothesis (reject H, or do not reject H.). 8) Interpret the results of the hypothesis test.

Answer 1

11.)a) Hypothesis to be tested:

HO: Ud=0
Vs
Hi :d +0

b) Assumptions:


Histogram of C1 Normal 3.0 Mean -4.8 StDev 5.245 N 10 2.5 2.0 - Frequency 1.5 - 1.0 0.5 0.0 -16 -12 -8 0 4 -4 c1

We draw this graph in "Minitab'16"
Steps:

1. Enter the data in "Minitab'16".
2. Click on "Graph".
3. Select "Histogram"
4. Choose "with fit".
5. Click on "OK".

Test Statistics:

d T = - tn-1 in

where,

P
Mean sample paired difference

Ps
S.D. of sample paired difference
n: Sample Size
Test Procedure:
We reject the Null hypothesis at $\alpha$ level of significance iff-

T| > tn-1, 2

where,

tn-1,9
is the
a
th percentile value of "t-distribution" with n-1 degrees of freedom.


c) Table-1(Table showing "paired difference")

Before	After	Difference(di)
33	34	1
33	28	-5
38	25	-13
33	28	-5
35	35	0
35	33	-2
40	31	-9
40	28	-12
40	35	-5
31	33	2
Total	-	-48

Mean sample paired difference:

d, = -4.8 i=1

S.D. of sample paired difference -

n Sd (di – d)2 = 5.2451 n i=1

d) Test Statistics:

Ꭲ ; -4,8 5.2451 10 = -2.8939

e) P-value:

p = 2 * Min{P(t- 1-|T), P(n-1> |T]}


2 * Min 0.0089, 0.0089

  
=20,00089


== 0.0178

f) As, P-value>$\alpha$, so we fail to reject the Null Hypothesis.

g) So, on the basis of the given data we can conclude that, there is no difference in the usage of mean ammount of water of the 10 families before and after watching the conservation video.

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