What would the inter-rater reliability be for a 50-item measure in which the number of agreements between Rater 1 and Rater 2 was 45?
C) 0.90
Answer:
Test statistic = 1.3471
P-value = 0.1993
Accept the null hypothesis.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4
Sample mean,
= 4.8
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation, s = 2.3
First, we design the null and the alternate hypothesis
We use two-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have

Now, we calculate the p-value.
P-value = 0.1993
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.
Answer:
B. ab+ac=d
Step-by-step explanation:
It would be 20 points bcuz it says each week it costs 20 points for unlimited bus rides
Step 2 9y - 24 = 10 + 3y because 3 times 3y its not 6y nor is 3 times 8 equal to 5