Question

an educational organization is analyzing whether there is a difference in student quiz scores between traditional teaching methods and flipped classrooms. independent random samples of students from each teaching style were assessed, with results summarized below:

do the data provide evidence that the traditional teaching method results in a different average quiz score compared to the flipped classroom method? use \\( \alpha = 0.05 \\).

a. state the null and alternative hypotheses for this test. use \\( \mu_t \\) for the traditional style and \\( \mu_f \\) for the flipped classroom style.

\\( \bigcirc \\) a. \\( h_0: \mu_t - \mu_f = 0, h_a: \mu_t - \mu_f \
eq 0 \\)
\\( \bigcirc \\) b. \\( h_0: \mu_t - \mu_f = 0, h_a: \mu_t - \mu_f > 0 \\)
\\( \bigcirc \\) c. \\( h_0: \mu_t - \mu_f = 0, h_a: \mu_t - \mu_f < 0 \\)

b. compute the test statistic and the p - value. (round your test statistic to two decimal places and your p - value to four decimal places.)
test statistic: \\( -3.90 \\)
p - value:

Explanation:

Step1: Determine the type of test

Since we are testing for a difference in means (\(H_a:\mu_t-\mu_f
eq0\)), this is a two - tailed test.

Step2: Use the formula for the p - value in a two - tailed z or t test (assuming large enough sample or normal distribution, and since the test statistic is given)

For a two - tailed test with test statistic \(z=- 3.90\) (if we assume a z - test, for large samples or known variances; the formula for the p - value of a two - tailed test is \(p = 2P(Z\lt|z|)\) when \(z\) is the test statistic.
We know that for \(z = 3.90\), using the standard normal table (or a calculator with a normal distribution function, e.g., in R: 2pnorm(-3.90) or in Excel: 2NORMSDIST(-3.90)).
The probability \(P(Z\lt - 3.90)\approx0.00005\)

Step3: Calculate the p - value

\(p=2\times0.00005 = 0.0001\)

Answer:

\(0.0001\)