A factory has a machine that dispenses 90ml of fluid in a bottle. An employee claims that the average amount is not 90 mL. Using 28 samples, he measures the average amount dispensed by the machine to be 88 ml with a standard deviation of 2.5. At 95% confidence level, is there enough evidence to support the idea that the machine is not working properly?
A factory has a machine that dispenses 90ml of fluid in a bottle. An employee claims that the average amount is not 90 mL. Using 28 samples, he measures the average amount dispensed by the machine to be 88 ml with a standard deviation of 2.5. At 95% confidence level, is there enough evidence to support the idea that the machine is not working properly?
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It is a t-test. The other picture is the guide. Pls help me answer this. Thank you
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
Transcribed Image Text:1. A factory has a machine that dispenses 90ml of
fluid in a bottle. An employee claims that the
average amount is not 90 mL, Using 28 samples, he
measures the average amount dispensed by the
machine to be 88 ml with a standard deviation of
2.5. At 95% confidence level, is there enough
evidence to support the idea that the machine is
not working properly?
а. Но:
На:
b. 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
www
e. Decision
f. Interpretation
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