A school principal claims that the numeracy rate of the senior high school students in his school is 60%. To test the claim, a researcher conducted a numeracy assessment on 500 senior high school students chosen through random sampling. It showed that 200 out of 500 students passed the assessment. Test if the claim is different at a= 0.05 level.
A school principal claims that the numeracy rate of the senior high school students in his school is 60%. To test the claim, a researcher conducted a numeracy assessment on 500 senior high school students chosen through random sampling. It showed that 200 out of 500 students passed the assessment. Test if the claim is different at a= 0.05 level.
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Solve the population proportion word problem by following the 5 step procedure:
1. A school principal claims that the numeracy rate of the senior high school students in his school is 60%. To test the claim, a researcher conducted a numeracy assessment on 500 senior high school students chosen through random sampling. It showed that 200 out of 500 students passed the assessment. Test if the claim is different at a= 0.05 level.
Attached is the 5-step procedure
![Solution:
STEP 1: State the null and alternative hypothesis.
Ho: p - 0.10
H.: p* 0.10
STEP 2: Choose a level of significance. a = 0.05
STEP 3: Compute the test statistic.
Given: x - 125p - 0.10
p =
n - 780
Zcom =
125
0.16-0.10
Zcom =
780
a.10(1-0.10)
780
0.06
p = 0.16
Zcom
0.03
Z com
= 5.59
STEP 4: Determine the critical value.
Since the
2 which will represent the shaded region. Using the Area Under the Normal Curve Table, critical
values at 0.05 tevel of significance for two-tailed test are t 1.96.
ernative hypothesis is non-directional, the two-tailed test shall be used. Divide a by
a/2 = 0.025
a/2 = 0.025
Non-rejection Region
-1.96
1.96
Rejection Region
Rejection Region
STEP 5: Make a decision whether to reject or fail to reject the null hypothesis. Draw a
conclusion.
DECISION: Since the computed test statistic zeom = 5.59 is greater than the critical value or
it falls in the rejection region, reject the null hypothesis.
CONCLUSION: Therefore, we conclude that at 0.05 level of significance, there is enough
evidence that the percentage of Grade 7 students who are underweight is different from 10%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0efed85e-a400-4ca8-9f24-0c8b8c5e1dbe%2Fe5caedbd-9f97-4590-a278-ab74695ed83d%2F0dvxw_processed.png&w=3840&q=75)
Transcribed Image Text:Solution:
STEP 1: State the null and alternative hypothesis.
Ho: p - 0.10
H.: p* 0.10
STEP 2: Choose a level of significance. a = 0.05
STEP 3: Compute the test statistic.
Given: x - 125p - 0.10
p =
n - 780
Zcom =
125
0.16-0.10
Zcom =
780
a.10(1-0.10)
780
0.06
p = 0.16
Zcom
0.03
Z com
= 5.59
STEP 4: Determine the critical value.
Since the
2 which will represent the shaded region. Using the Area Under the Normal Curve Table, critical
values at 0.05 tevel of significance for two-tailed test are t 1.96.
ernative hypothesis is non-directional, the two-tailed test shall be used. Divide a by
a/2 = 0.025
a/2 = 0.025
Non-rejection Region
-1.96
1.96
Rejection Region
Rejection Region
STEP 5: Make a decision whether to reject or fail to reject the null hypothesis. Draw a
conclusion.
DECISION: Since the computed test statistic zeom = 5.59 is greater than the critical value or
it falls in the rejection region, reject the null hypothesis.
CONCLUSION: Therefore, we conclude that at 0.05 level of significance, there is enough
evidence that the percentage of Grade 7 students who are underweight is different from 10%.
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