Answer:
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
Step-by-step explanation:
Let 'M' be the event of selecting males n(M) = 12
Number of ways of choosing 3 students From all males and females
Number of ways of choosing 3 students From all males
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
<u><em>Final answer</em></u>:-
The probability that all are male of choosing '3' students
P(E) = 0.067 = 6.71%
Answer:
1) B. The differences are normally distributed or the sample size is large
C. The sample size mus be large
E. The sampling method results in an independent sample
2) The null hypothesis H₀: =
The alternative hypothesis Hₐ: <
Test statistic, t = -0.00693
p- value = 0.498
Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers at 0.10 level of significance
Step-by-step explanation:
1) B. The differences are normally distributed or the sample size is large
C. The sample size mus be large
E. The sampling method results in an independent sample
2) The null hypothesis H₀: =
The alternative hypothesis Hₐ: <
The test statistic for t test is;
The mean
Height of Father, h₁, Height of Son h₂
72.4, 77.5
70.6, 74.1
73.1, 75.6
69.9, 71.7
69.4, 70.5
69.4, 69.9
68.1, 68.2
68.9, 68.2
70.5, 69.3
69.4, 67.7
69.5, 67
67.2, 63.7
70.4, 65.5
= 69.6
s₁ = 1.58
= 69.9
s₂ = 3.97
n₁ = 13
n₂ = 13
(We reversed the values in the square root of the denominator therefore, the sign reversal)
t = -0.00693
p- value = 0.498 by graphing calculator function
P-value > α Therefore, we do not reject the null hypothesis
Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers at 0.10 lvel of significance