Using traditional methods it takes 9898 hours to receive an advanced flying license. a new training technique using computer aided instruction (cai) has been proposed. a researcher used the technique on 100 students and observed that they had a mean of 97 hours. assume the population variance is known to be 49.
1. is there evidence at the 0.05 level that the technique performs differently than the traditional method?
2. find the value of the test statistic. round your answer to two decimal places.

1. Yes, the technique performs differently than the traditional method.

2. 0.20

Step-by-step explanation:

The null hypothesis  

H0 = mean is 97 hours

Ha = mean > 97

Let´s make the z test

z = (x – mean) / standard deviation / √sample size

z = (97 – 98) / 49 / √100

z = -0.204

Confidence level of 95%, that means the the siginficance level α is 1 – p

α = 1 – 0.95 = 0.05

Z(α /2) = Z(0.05/2) = Z (0.025)

Using a z table Z = 1.96

since our calculated z values is smaller than the critical value, the null hypothesis can be rejected. Sample mean is different from sample population so the technique performs differently than the traditional method.

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.