Question

a competency test is modified so that the proportion of test - takers that pass is more than 75%. a simple random sample of test - takers is collected to determine if the proportion of passers is greater than 75%. the results of the sample are shown below.
competency test sample
passed: 80
observations: 100
confidence level: 0.9
critical value: 1.28
test statistic (z): 1.15
p - value: 0.124
n = ex: 9
number passed = ex: 9
sample proportion \\(\hat{p}\\) = ex: 1.234

Explanation:

Step1: Determine the sample size \( n \)

The number of observations is given as 100, so \( n = 100 \).

Step2: Determine the number passed

From the table, the number of test - takers who passed is 80.

Step3: Calculate the sample proportion \(\hat{p}\)

The formula for the sample proportion is \(\hat{p}=\frac{\text{Number of successes}}{\text{Sample size}}\). Here, the number of successes (number of test - takers who passed) is 80 and the sample size \(n = 100\). So \(\hat{p}=\frac{80}{100}=0.8\).

Step4: Identify the test statistic \( z \)

From the table, the test statistic \( z = 1.15 \).

Step5: Identify the p - value

From the table, the p - value is 0.124.

Answer:

• \(n=\boldsymbol{100}\)
• Number passed \(=\boldsymbol{80}\)
• Sample proportion \(\hat{p}=\boldsymbol{0.8}\)
• \(z=\boldsymbol{1.15}\)
• p - value \(=\boldsymbol{0.124}\)