H0 : =150.00 kPaH1 :>150.00 kPab) α= 0.01;sample variance : 1300 (freq. table )sample deviation : 36.0555 (freq. table )Calculation of Data Mean of the results= = 6700/50 =134Mid Range of the results= (217.32+46.32)/2= 131.82Range = (217.32-46.32) = 171Modal Class= Class 121-145 with the highest frequency of 15Median of the sorted set of data = 132.25Mode = there are not clear modes In the samples. Sample variance S o= 1300Sample deviation: S =36.55 As a quality control manager, you have been measuring the pressure level provided by a newlyinstalleddundwhich are purchased from new manufacturer. The mean pressure maintained by theold pump was 150.00 kPa. The manufacturer of the new pump claims that the pressure will beimproved after installing the new pump. You have been tasked to ensure that the new pump isbetter.DATA96.17128.32169.54167.06108.02125.95152.66106.58156.05134.51108.77129.1993.88139.16119.73176.95175.1131.27182.63123.95116.5691.28133.55117.12217.31147.25104.32159.9699.02104.7124.9131.3198.1866.54172.0146.32140.98158.39cumula RelativeMid point (X m) =freque tiveinterval boundary ncy (f) frequenClassClassfrequencyf x Xmlower + upperсуEf246+7046-7045.5-70.5= 0. 0450= 582 x 58 = 1162471-9570.5-95.5= 0. 0850833321296-12095.5-120.5 1218= 0. 2450108129615121-145120.5-145.5 1533= 0. 305013319959146-170= 0. 18501581422145.5-170.5 9425171-195170.5-195.5 547= 0. 1050183915196-220195.5-220.5 35030.0620862450II D) CREATE HYPOTHESIS TESTINGSAMPLE MEAN FROM (FREQUENCY TABLE)SIGNIFICANCE LEVEL: 0.01SAMPLE STANDARD DEVIATION (FREQUENCY TABLE)CONFIDENCE INTERVAL 90%THIS DATA HAS NO OUTLIERS(EXPLAIN HYPOTHESIS)SHOW CRITICAL VALUE AND TEST RESULTS

Answered: As a quality control manager, you have…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

H0 : =150.00 kPa

H1 :>150.00 kPa

b) α= 0.01;

sample variance : 1300 (freq. table )

sample deviation : 36.0555 (freq. table )

Calculation of Data

Mean of the results= = 6700/50 =134

Mid Range of the results= (217.32+46.32)/2= 131.82

Range = (217.32-46.32) = 171

Modal Class= Class 121-145 with the highest frequency of 15

Median of the sorted set of data = 132.25

Mode = there are not clear modes In the samples.

Sample variance S o= 1300

Sample deviation: S =36.55

As a quality control manager, you have been measuring the pressure level provided by a newly
installed
dund
which are purchased from new manufacturer. The mean pressure maintained by the
old pump was 150.00 kPa. The manufacturer of the new pump claims that the pressure will be
improved after installing the new pump. You have been tasked to ensure that the new pump is
better.
DATA
96.17
128.32
169.54
167.06
108.02
125.95
152.66
106.58
156.05
134.51
108.77
129.19
93.88
139.16
119.73
176.95
175.1
131.27
182.63
123.95
116.56
91.28
133.55
117.12
217.31
147.25
104.32
159.96
99.02
104.7
124.9
131.31
98.18
66.54
172.01
46.32
140.98
158.39
cumula Relative
Mid point (X m) =
freque tive
interval boundary ncy (f) frequen
Class
Class
frequency
f x Xm
lower + upper
су
Ef
2
46+70
46-70
45.5-70.5
= 0. 04
50
= 58
2 x 58 = 116
2
4
71-95
70.5-95.5
= 0. 08
50
83
332
12
96-120
95.5-120.5 12
18
= 0. 24
50
108
1296
15
121-145
120.5-145.5 15
33
= 0. 30
50
133
1995
9
146-170
= 0. 18
50
158
1422
145.5-170.5 9
42
5
171-195
170.5-195.5 5
47
= 0. 10
50
183
915
196-220
195.5-220.5 3
50
3
0.06
208
624
50
II

D) CREATE HYPOTHESIS TESTING
SAMPLE MEAN FROM (FREQUENCY TABLE)
SIGNIFICANCE LEVEL: 0.01
SAMPLE STANDARD DEVIATION (FREQUENCY TABLE)
CONFIDENCE INTERVAL 90%
THIS DATA HAS NO OUTLIERS
(EXPLAIN HYPOTHESIS)
SHOW CRITICAL VALUE AND TEST RESULTS

Definition Definition Measure of central tendency that is the value that occurs most frequently in a data set. A data set may have more than one mode if multiple categories repeat an equal number of times. For example, in a data set with five item—3, 5, 5, 29, 473—the mode is 5 because it occurs twice and no other value occurs more than once. On a histogram or bar chart, the element with the highest bar represents the mode. Therefore, the mode is sometimes considered the most popular option. The mode is useful for nominal or categorical data (e.g., the most common color car that users purchase), but it is problematic for continuous data because it is more likely not to have any value that is more frequent than the other.

Expert Solution

Step 1

Claim:- The pressure will be improved after installing the new pump.

The hypothesis is,

$H_{0} : µ = 150 k P a V S H_{1} : µ > 150 k P a$

Step 2

The test statistic is,

$t = \frac{x - µ}{S / \sqrt{n}} w h e r e, x = \frac{Σ x}{n} = 130.3997 S = s a m p l e s \tan d a r d d e v i a t i o n = \sqrt{\frac{Σ {(x - x)}^{2}}{n - 1}} = 34.10135 n = 38$

So,

$t = \frac{130.3997 - 150}{34.10135 / \sqrt{38}} = - 3.5431$

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