Solve the word problems involving population proportion using 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.

2. For the past year, the rate of spread of the COVID-19 virus to the patients of a hospital is 85%. The director of the hospital clarifies that the rate of the spread of the virus to the patients this year is not 85% anymore. To test the claim, a researcher selected 500 patients from the hospital through simple random sampling. Out of 500 patients, 410 were infected of the virus. Test the hypothesis that the rate of the spread of the virus is not equal to 85% by using ? = 0.05 as the level of significance.

The 5-step procedure (example):

Solution: STEP 1: State the null and alternative hypothesis. Ho: p = 0.10 Ha: p = 0.10 STEP 2: Choose a level of significance. a = 0.05 STEP 3: Compute the test statistic. Given: x - 125p = 0.10 n = 780 Zoom 125 0.16-0.10 780 Zcom 010 1 0.10) p = 0.16 Zcom 0.06 0.03 Zcom = 5.59 STEP 4: Determine the critical value. Since the alternative hypothesis is non-directional, the two-tailed test shall be used. Divide a by 2 which will represent the shaded region. Using the Area Under the Normal Curve Table, critical values at 0.05 level of significance for two-tailed test are + 1.96. a/2 - 0.025 a/2 - 0.025 Non-rejection Region -1.96 0 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 Zoom - 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%