Problem 2: Suppose that a 3 Ã 5 coefficient matrix A has a pivot in each of the three rows. If we augment A to represent a linea
1 answer:
You might be interested in
Using the z-distribution, since the p-value of the test is of 0.057 > 0.05, there is not enough evidence that the population proportion of successes does not equal 0.50.
<h3>What are the hypotheses tested?</h3>At the null hypotheses, we test if the proportion of successes equals 0.5, hence:
At the alternative hypotheses, we test if it does not equal, hence:
The test statistic is given by:
In which:
is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
For this problem, the parameters are given by:
Hence the test statistic is given by:
z = (0.35 - 0.5)/(0.5/sqrt(40))
z = -1.9.
<h3>What is the decision?</h3>Using a z-distribution calculator, considering a two-tailed test, as we are testing if the proportion is different of a value, with z = -1.9, we get that the p-value of the test is of 0.057.
Since the p-value of the test is of 0.057 > 0.05, there is not enough evidence that the population proportion of successes does not equal 0.50.
More can be learned about the z-distribution at brainly.com/question/16313918
#SPJ4