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%.
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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