Answer:
x + 3y = 21
Step-by-step explanation:
Equation of the line: y =mx+ b
Multiply the equation by 3
3y = -x + 21
x + 3y = 21
Answer:
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
Step-by-step explanation:
We formulate null and alternate hypotheses are
H0 : u1 < u2 against Ha: u1 ≥ u 2
Where u1 is the group tested after they were awake for 24 hours.
The Significance level alpha is chosen to be ∝ = 0.05
The critical region t ≥ t (0.05, 13) = 1.77
Degrees of freedom is calculated df = υ= n1+n2- 2= 5+10-2= 13
Here the difference between the sample means is x`1- x`2= 35-24= 11
The pooled estimate for the common variance σ² is
Sp² = 1/n1+n2 -2 [ ∑ (x1i - x1`)² + ∑ (x2j - x`2)²]
= 1/13 [ 120²+360²]
Sp = 105.25
The test statistic is
t = (x`1- x` ) /. Sp √1/n1 + 1/n2
t= 11/ 105.25 √1/5+ 1/10
t= 11/57.65
t= 0.1908
The calculated value of t= 0.1908 does not lie in the critical region t= 1.77 Therefore we accept our null hypothesis that fatigue does not significantly increase errors on an attention task at 0.05 significance level
$63.90 (total)
$5.40 (cards)
$11.70/per (bags)
$63.90 - $5.40 = $58.50
$58.50 / $11.70 = 5
She invited 5 guests (b).
Answer:
The correct answer is D. It is not true that cluster sampling uses randomly selected clusters and samples everyone within each cluster.
Step-by-step explanation:
Cluster sampling is a method of collecting samples and statistical data, by means of which a certain group formed by people, things, events, etc., is taken as a sample, which are not considered individually but as part of a whole, which is in turn a proportional representation of the universality of samples available in the field.
Now, since this type of sampling allows to embrace large groups of sample units, data are not always obtained from all the components of the cluster, but from those necessary to be able to quantify the desired statistics.