Answer & Step-by-step explanation:
The confidence interval formula is:
I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]
alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%
σ= standard deviation. In this case 5
mean= 37
n= number of observations. In this case, 15
(a)
Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645
Then, the confidence interval (90%):
I 90%(μ)= 37+- [1.645*(5/sqrt(15))]
I 90%(μ)= 37+- [2.1236]
I 90%(μ)= [37-2.1236;37+2.1236]
I 90%(μ)= [34.8764;39.1236]
(b)
Z(alpha/2)= Z(2.5%)= 1.96
Then, the confidence interval (90%):
I 95%(μ)= 37+- [1.96*(5/sqrt(15))
]
I 95%(μ)= 37+- [2.5303]
I 95%(μ)= [37-2.5303;37+2.5303]
I 95%(μ)= [34.4697;39.5203]
(c)
Z(alpha/2)= Z(0.5%)= 2.5758
Then, the confidence interval (90%):
I 99%(μ)= 37+- [2.5758*(5/sqrt(15))
I 99%(μ)= 37+- [3.3253]
I 99%(μ)= [37-3.3253;37+3.3253]
I 99%(μ)= [33.6747;39.3253]
(d)
C. The interval gets wider as the confidence level increases.
