1. A factory has a machine that dispenses 75ml of fluid in a bottle. An employee believes that the average amount is less than 75 ml. Using 40 samples, he measures the average amount dispensed by the machine to be 70 mL with a standard deviation of 5.5. At 95% confidence level, is there enough evidence to support the idea that the machine is not working properly? Answer the following: а. Но: На: b. Level of significance a= c. Type 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

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Pls help me to answer those. It is my last remaining ticket question. Thank you so much. The one picture is a guide. It is a z-test.

tus tfePecision:
Example # 1
A Barangay Captain from a certain barangay, in Valenzuela City olaims that the average monthly income of
families with five members from his vicinity is P12.000. But when the City Statisties Office (CSO) conducted a
survey to 10 families with fiye members, to his borangay randomly they found out that they only have an
average monthly income of 10,800 with a standard deviation of 1.500. with this information the Cso assert
that the claim is not true. Using 0.05 level of significance test the claim of the Brgy. Captain.
The average montly weomo of fundos with
VHo H = P12, 000 fve members from hs woinily w P12,000
4. Area of Rejection
The average monithly inoome of families with
Ha H# P12, 000 he members from his vionty is not P12,000
2. Level of significance
0.05
-1.96
1.96
Decision Rule:
3. Test Statistic:
h =
Reject H, if the computed valueg is
> +1.96 on < -1.96
100
I test; one mean/s; two tailed test
Otherwise do not reject H,
Example # 1
A Barangay Coptain from a certain barangay in Valenzuela City claims that the
families with five members from his vicinity is P12.000. But when the City Statist
survey to 10b families with five members, to his barangay randomly they found
average monthly income of 10.800 with a standard deviation of 1,50. With this
that the claim is not true. Using 0.05 level of significance test the claim of the I
5. Compute for the
-value:
7. Interpretation:
10800 – 12000
-8-
Rejection
of the ni
1500
V100
the average monthly inc
members from his vicini
Calcu: (10800 - 12000) + (1500 +y100)
tuo titePecision:
Since, the computed
base on the sample of 100 f
significance. Therefore, the
-value -8
is
the tabular value (_-1.96_).
(reject
less than
not correct
Therefore,
the null hypothesis
(Ho) at,
0.05 level of significance.
Transcribed Image Text:tus tfePecision: Example # 1 A Barangay Captain from a certain barangay, in Valenzuela City olaims that the average monthly income of families with five members from his vicinity is P12.000. But when the City Statisties Office (CSO) conducted a survey to 10 families with fiye members, to his borangay randomly they found out that they only have an average monthly income of 10,800 with a standard deviation of 1.500. with this information the Cso assert that the claim is not true. Using 0.05 level of significance test the claim of the Brgy. Captain. The average montly weomo of fundos with VHo H = P12, 000 fve members from hs woinily w P12,000 4. Area of Rejection The average monithly inoome of families with Ha H# P12, 000 he members from his vionty is not P12,000 2. Level of significance 0.05 -1.96 1.96 Decision Rule: 3. Test Statistic: h = Reject H, if the computed valueg is > +1.96 on < -1.96 100 I test; one mean/s; two tailed test Otherwise do not reject H, Example # 1 A Barangay Coptain from a certain barangay in Valenzuela City claims that the families with five members from his vicinity is P12.000. But when the City Statist survey to 10b families with five members, to his barangay randomly they found average monthly income of 10.800 with a standard deviation of 1,50. With this that the claim is not true. Using 0.05 level of significance test the claim of the I 5. Compute for the -value: 7. Interpretation: 10800 – 12000 -8- Rejection of the ni 1500 V100 the average monthly inc members from his vicini Calcu: (10800 - 12000) + (1500 +y100) tuo titePecision: Since, the computed base on the sample of 100 f significance. Therefore, the -value -8 is the tabular value (_-1.96_). (reject less than not correct Therefore, the null hypothesis (Ho) at, 0.05 level of significance.
1. A factory has a machine that dispenses 75mL of fluid in a bottle.
An employee believes that the average amount is less than 75
ml. Using 40 samples, he measures the average amount
dispensed by the machine to be 70 mL with a standard deviation
of 5.5. At 95% confidence level, is there enough evidence to
support the idea that the machine is not working properly?
Answer the following:
а. Но:
Ha:
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
e. Decision
f. Interpretation
Transcribed Image Text:1. A factory has a machine that dispenses 75mL of fluid in a bottle. An employee believes that the average amount is less than 75 ml. Using 40 samples, he measures the average amount dispensed by the machine to be 70 mL with a standard deviation of 5.5. At 95% confidence level, is there enough evidence to support the idea that the machine is not working properly? Answer the following: а. Но: Ha: 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 e. Decision f. Interpretation
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