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 0,01 level of significance to investigate the claim. Determine the critical value.
To investigate the claim, we can use a hypothesis test. Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The proportion of students passing the test is 90% (P = 0.9)
Alternative hypothesis (Ha): The proportion of students passing the test is not equal to 90% (P ≠ 0.9)
We will use the z-test for proportions since the sample size is large (n = 50) and the data is binomial (pass or fail). The test statistic for the z-test is calculated as:
z = (p̂ - P) / √(P * (1-P) / n)
Where p̂ is the sample proportion, P is the hypothesized proportion, and n is the sample size.
From the given data, p̂ (sample proportion) = 46/50 = 0.92.
P (hypothesized proportion) = 0.9.
n (sample size) = 50.
Calculating the test statistic:
z = (0.92 - 0.9) / √(0.9 * (1-0.9) / 50)
= 0.02 / √(0.09 / 50)
= 0.02 / √0.0018
= 0.02 / 0.0424264
= 0.472
To find the critical value, we need to determine the z-score corresponding to a 0.01 level of significance. For a two-tailed test, the critical value is z = ±2.575.
Therefore, the critical value for this test is 2.575.
To determine the critical value for the hypothesis test, we need to specify the hypothesis being tested. In this case, the claim is that 90% of the students will pass the test, so we can set up the following hypotheses:
Null hypothesis (H0): The actual proportion of students who pass the test is equal to 0.9 (p = 0.9)
Alternative hypothesis (Ha): The actual proportion of students who pass the test is less than 0.9 (p < 0.9)
Since the alternative hypothesis is one-sided (less than), we will be conducting a one-tailed test. The significance level is given as 0.01.
To determine the critical value for this test, we can use a standard normal distribution table or calculator.
Calculate the z-score for a 0.01 level of significance:
- z = invNorm(0.01) = -2.33 (approximated)
The critical value is -2.33.
Please note that if you are using a statistical software or calculator, it will provide the exact critical value based on the significance level and type of test.