1. A company manufactures car batteries with an average life span of 3 or more years. An engineer believes this value to be less. Using 15 samples, he measures the average life span to be 2.5 years with a standard deviation of 0.25. At 99% confidence level, is there enough evidence to discard the null hypothesis?Ho: На: a. Level of significance a= с. Туре of test. Choices: one-mean, two-tailed t-test two-mean, one-tailed t-test one-mean, one-tailed z-test two-mean, two-tailed z-test d.Tabulated value and computed value e. Decision f. Interpretation

MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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It is a t-test. The other picture is the guide. Pls help me answer this. Thank you 

1. A company manufactures car batteries with an
average life span of 3 or more years. An engineer
believes this value to be less. Using 15 samples, he
measures the average life span to be 2.5 years with
a standard deviation of 0.25. At 99% confidence
level, is there enough evidence to discard the null
hypothesis?Ho:
На:
a. Level of significance
a=
с. Туре of test.
Choices: one-mean, two-tailed t-test
two-mean, one-tailed t-test
one-mean, one-tailed z-test
two-mean, two-tailed z-test
d.Tabulated value and computed value
e. Decision
f. Interpretation
Transcribed Image Text:1. A company manufactures car batteries with an average life span of 3 or more years. An engineer believes this value to be less. Using 15 samples, he measures the average life span to be 2.5 years with a standard deviation of 0.25. At 99% confidence level, is there enough evidence to discard the null hypothesis?Ho: На: a. Level of significance a= с. Туре of test. Choices: one-mean, two-tailed t-test two-mean, one-tailed t-test one-mean, one-tailed z-test two-mean, two-tailed z-test d.Tabulated value and computed value e. Decision f. Interpretation
The mean score obtaingd by OLFU students in entrance examination is 87. A group of 25 freshman students
scored an average of 85 with a standard deviation of 5. Based on this result the admission officeasserts
that the group's average score is lower than 87. if you were one of those students, would you agree?
Make a necessary statistical analysis to support your answer. Use 0.05 level of significance.
The group's average score is
at least 87.
4. Area of Rejection t-tu
1. Ho:
HZ87
df =n -1
The group's average score is
lower than 87.
df = 25 - 1
df = 24
V Ha:
H< 87
nj et th
2. Level of significance:
0.05
-1.7110
Decision Rule:
3. Test Statistic:
Reject H, if the computed value is
< -1.7110
25
-test; one -mean/s; one.tailed test
Otherwise do not reject H
,.
CRITICAL VALUES of t
Level of significance for one-tailed test
Level of significance for one-tailed test
0.005
df
0.10
0.05
0.025
0.01
.0005
df
0.10
0.05
0.025
0.01
0.005
.0005
Level of significance for two-tailed test
Level of significance for two-tailed test
0.05
df
0.20
0.10
0.02
0.01
0.001
df
0.20
0.10
0.05
0.02
0.01
0.001
16
1.3370
1.7460
2.1200
2.5830
2.9210
4.0150
17
1.3330
1.7400
2.1100
2.5670
2.8980
3.9650
3.0780
1.8860
1
6.3140
12.7060
31.8210
63.6570
636.6190
18
1.3300
1.7340
2.1010
2.5520
2.8780
2,8610
3.9220
2.9200
2.3530
2.1320
1.3030
6.9650
1.3260
1.3250
9.9250
31.5980
2.0930
19
20
1.7290
2.5390
3.8830
3.8500
1.7250
12.9410
8.6100
3
1.6380
3.1820
4.5410
5.8410
2.0860
2.5280
2.8450
4
1.5330
2.7760
3.7470
4.6040
21
1.3230
1.7210
2.0800
2.5180
2.8310
3.6190
5
1.4760
2.0150
2.5710
3.3650
4.0320
6.8590
22
1.3210
1.7170
2.0740
2.5080
2.8190
3.7920
23
1.3190
1.7140
2.0690
2.5000
2.8070
3.7670
6.
1.4400
1.9430
2.4470
3.1430
3.7070
5.9590
24
1.3180
1.7110
2.0640
2.4920
2.7970
3.7450
7
1.4150
1.8950
2.3650
2.9980
3.4990
5.4050
25
1.3160
1.7080
2.0600
2.4850
2.7870
3.7250
1.3970
1.8600
2.3060
2.8960
3.3550
5.0410
26
1.3150
1.7060
2.0560
2.4790
2.7790
3.7070
9
1.3830
1.8330
2.2620
2.8210
3.2500
4.7810
27
1.3140
1.7030
2.0520
2.4730
2.7710
1.3720
2.2280
3.6900
3.6740
10
1.8120
2.7640
3.1690
4.5870
28
1.3130
1.7010
2.0480
2.4670
2.7630
29
1.3110
1.6990
2.0450
2.4620
2.7560
3.6590
11
1.3630
1.7960
2.2010
2.7180
3.1060
4.4370
30
1.3100
1.6970
2.0420
2.4570
2.7500
3.6460
1.3560
1.3500
1.3450
1.3410
2.6810
2.6500
2.6240
2.6020
12
1.7820
2.1790
3.0550
4.3180
4.2210
13
1.7710
2.1600
3.0120
40
1.3030
1.6840
2.0210
2.4240
2.7040
3.5510
2.6600
2.6170
2.5760
60
1.2960
1.6710
2.0000
2.3900
3.4600
1.7610
1.7530
2.1450
2.1310
14
2.9770
4.1400
120
1.2890
1.6580
1.9800
2.3580
3.3730
15
2.9470
4.0730
1.2820
1.6450
1.9600
2.3260
3.2910
Example # 4
The mean score obtained by OLFU students in entrance examination is 87. A group of 25 freshman
students scored an average of 85 with a standard deviation of §. Based on this result the
admission office asserts that the group's average score is lower than 87. if you were one of those
students, would you agree?
Make a necessary statistical analysis to support your answer. Use 0.05 level of significance.
5. Compute for the value:
アー!
t =
7. Interpretation: R+ Ha NX + Ma
85 - 87
Rejection
+ the group's average score is lower than 87
-2
of the null hypothesis (Ho) means that
%3D
V25
base on the sample of 25 students_using 0.05
of significance. Therefore, the claim of_ Admission office
level
Caleu: (85 - 87) + (5+ 25)
6. Decision:
-1 < -17)
Since, the computed t-value L
-2
correct
) is
is
less than
the tabular value (-1.711).
Therefore,
reject
the null hypothesis
(Ho) at 0.05 level of significance.
Fajact
Transcribed Image Text:The mean score obtaingd by OLFU students in entrance examination is 87. A group of 25 freshman students scored an average of 85 with a standard deviation of 5. Based on this result the admission officeasserts that the group's average score is lower than 87. if you were one of those students, would you agree? Make a necessary statistical analysis to support your answer. Use 0.05 level of significance. The group's average score is at least 87. 4. Area of Rejection t-tu 1. Ho: HZ87 df =n -1 The group's average score is lower than 87. df = 25 - 1 df = 24 V Ha: H< 87 nj et th 2. Level of significance: 0.05 -1.7110 Decision Rule: 3. Test Statistic: Reject H, if the computed value is < -1.7110 25 -test; one -mean/s; one.tailed test Otherwise do not reject H ,. CRITICAL VALUES of t Level of significance for one-tailed test Level of significance for one-tailed test 0.005 df 0.10 0.05 0.025 0.01 .0005 df 0.10 0.05 0.025 0.01 0.005 .0005 Level of significance for two-tailed test Level of significance for two-tailed test 0.05 df 0.20 0.10 0.02 0.01 0.001 df 0.20 0.10 0.05 0.02 0.01 0.001 16 1.3370 1.7460 2.1200 2.5830 2.9210 4.0150 17 1.3330 1.7400 2.1100 2.5670 2.8980 3.9650 3.0780 1.8860 1 6.3140 12.7060 31.8210 63.6570 636.6190 18 1.3300 1.7340 2.1010 2.5520 2.8780 2,8610 3.9220 2.9200 2.3530 2.1320 1.3030 6.9650 1.3260 1.3250 9.9250 31.5980 2.0930 19 20 1.7290 2.5390 3.8830 3.8500 1.7250 12.9410 8.6100 3 1.6380 3.1820 4.5410 5.8410 2.0860 2.5280 2.8450 4 1.5330 2.7760 3.7470 4.6040 21 1.3230 1.7210 2.0800 2.5180 2.8310 3.6190 5 1.4760 2.0150 2.5710 3.3650 4.0320 6.8590 22 1.3210 1.7170 2.0740 2.5080 2.8190 3.7920 23 1.3190 1.7140 2.0690 2.5000 2.8070 3.7670 6. 1.4400 1.9430 2.4470 3.1430 3.7070 5.9590 24 1.3180 1.7110 2.0640 2.4920 2.7970 3.7450 7 1.4150 1.8950 2.3650 2.9980 3.4990 5.4050 25 1.3160 1.7080 2.0600 2.4850 2.7870 3.7250 1.3970 1.8600 2.3060 2.8960 3.3550 5.0410 26 1.3150 1.7060 2.0560 2.4790 2.7790 3.7070 9 1.3830 1.8330 2.2620 2.8210 3.2500 4.7810 27 1.3140 1.7030 2.0520 2.4730 2.7710 1.3720 2.2280 3.6900 3.6740 10 1.8120 2.7640 3.1690 4.5870 28 1.3130 1.7010 2.0480 2.4670 2.7630 29 1.3110 1.6990 2.0450 2.4620 2.7560 3.6590 11 1.3630 1.7960 2.2010 2.7180 3.1060 4.4370 30 1.3100 1.6970 2.0420 2.4570 2.7500 3.6460 1.3560 1.3500 1.3450 1.3410 2.6810 2.6500 2.6240 2.6020 12 1.7820 2.1790 3.0550 4.3180 4.2210 13 1.7710 2.1600 3.0120 40 1.3030 1.6840 2.0210 2.4240 2.7040 3.5510 2.6600 2.6170 2.5760 60 1.2960 1.6710 2.0000 2.3900 3.4600 1.7610 1.7530 2.1450 2.1310 14 2.9770 4.1400 120 1.2890 1.6580 1.9800 2.3580 3.3730 15 2.9470 4.0730 1.2820 1.6450 1.9600 2.3260 3.2910 Example # 4 The mean score obtained by OLFU students in entrance examination is 87. A group of 25 freshman students scored an average of 85 with a standard deviation of §. Based on this result the admission office asserts that the group's average score is lower than 87. if you were one of those students, would you agree? Make a necessary statistical analysis to support your answer. Use 0.05 level of significance. 5. Compute for the value: アー! t = 7. Interpretation: R+ Ha NX + Ma 85 - 87 Rejection + the group's average score is lower than 87 -2 of the null hypothesis (Ho) means that %3D V25 base on the sample of 25 students_using 0.05 of significance. Therefore, the claim of_ Admission office level Caleu: (85 - 87) + (5+ 25) 6. Decision: -1 < -17) Since, the computed t-value L -2 correct ) is is less than the tabular value (-1.711). Therefore, reject the null hypothesis (Ho) at 0.05 level of significance. Fajact
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