It is claimed that 90% of the students in STSA 1 class will pass their test. After the test a random sample of 50 students shows that 46 passed the test. Use an appropriate test on a 0,1 level of significance to investigate the claim. Determine the table value.
To investigate the claim, we can use a one-sample proportion test to determine if the proportion of students who passed the test is significantly different from the claimed proportion of 90%.
The null hypothesis (H0) for this test is that the proportion of students who pass the test is equal to 90% (p = 0.9).
The alternative hypothesis (Ha) is that the proportion is different from 90% (p ≠ 0.9).
Given that the sample size (n) is 50, and the number of students who passed the test (x) is 46, we can calculate the sample proportion (p̂) as:
p̂ = x/n = 46/50 = 0.92
Next, we can calculate the test statistic (z) using the formula:
z = (p̂ - p) / √(p*(1-p)/ n)
Substituting the values into the formula, we get:
z = (0.92 - 0.90) / √(0.90*(1-0.90)/50)
z ≈ 0.02 / 0.042
Calculating the test statistic, we find:
z ≈ 0.476
To determine the critical value from the table, we need to know the level of significance and the type of test (two-tailed, one-tailed left, or one-tailed right).
In this case, the level of significance is given as 0.1, and since the alternative hypothesis is two-tailed, we need to find the critical value for an alpha level of 0.05 (half of the 0.1 level).
Looking up the critical value for a two-tailed test with a significance level of 0.05 in the standard normal distribution table, we find a critical value of approximately ±1.96.
Therefore, the table value for this test is approximately ±1.96.
To investigate the claim, we can use a one-sample proportion test. The null hypothesis (H0) assumes that the true proportion of students passing the test is equal to 90% or 0.9. The alternative hypothesis (Ha) assumes that the true proportion is not equal to 90%.
Let's denote:
p = proportion of students passing the test (to be tested)
p0 = claimed proportion of students passing the test (0.9)
n = sample size (50)
x = number of students who passed the test in the sample (46)
To perform the test, we will calculate the test statistic z using the formula:
z = (x - np0) / sqrt(np0(1-p0))
Let's plug in the values:
x = 46
p0 = 0.9
n = 50
z = (46 - 50(0.9)) / sqrt(50(0.9)(1-0.9))
z = (46 - 45) / sqrt(40.5)
z = 1 / sqrt(40.5)
Now, we need to determine the critical value of z at a 0.1 level of significance. This corresponds to a two-tailed test.
Using a standard normal distribution table, the critical z-value is approximately 1.645.
Therefore, the table value (critical z-value) for a 0.1 level of significance is 1.645.