16% of what equals 44?
1 answer:
You might be interested in
Answer:
A) t = 1.73
B) p-value = 0.0558
C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05
D) The decision means that the design specifications are not met.
E) Type II error
Step-by-step explanation:
The hypotheses are:
H₀: μ = 20
H₁: μ > 20
A) Formula for the test statistic is;
t = (x' - μ)/(s/√n)
Now, we are given;
x' = 21.5
μ = 20
s = 3
n = 12
Thus;
t = (21.5 - 20)/(3/√12)
t = 1.73
B) we have our t-value as 1.73
Now, Degree of freedom(DF) = n - 1
So,DF = 12 - 1 = 11
Using significance level of α = 0.05, t-value = 1.73 and DF = 11, one tailed hypothesis, from online P-value calculator attached, we have;
p-value = 0.0558
C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05
D) We will not reject the null hypothesis. The decision means that the design specifications are not met.
E) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, then the null hypothesis is false.
Since we did not reject the null hypothesis even though it is false, the error that was committed was therefore a type II error.
