Answer:
96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].
Step-by-step explanation:
We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.
Firstly, the Pivotal quantity for 96% confidence interval for the population mean is given by;
P.Q. = ~
where, = sample average desired retirement age = 55 years
= sample standard deviation = 3.4 years
n = sample of seniors = 101
= true mean retirement age of all college students
Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 96% confidence interval for the population mean, is ;
P(-2.114 < < 2.114) = 0.96 {As the critical value of t at 100 degree
of freedom are -2.114 & 2.114 with P = 2%}
P(-2.114 < < 2.114) = 0.96
P( < < ) = 0.96
P( < < ) = 0.96
96% confidence interval for = [ , ]
= [ , ]
= [54.30 , 55.70]
Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].
Answer:
A is correct (1/5)
Not 1/4
Not 5
Not 4
Step-by-step explanation:
Took edgenuity test
Answer:
Step-by-step explanation:
-3
Answer:
m = 0.421
Step-by-step explanation:
5.4(m-2)= -2(3m+3) (expand parentheses by distributive property)
m(5.4) -2(5.4) = 3m(-2) + 3(-2)
5.4m - 10.8 = -6m - 6 (add 6m to both sides)
5.4m - 10.8 + 6m = - 6
11.4m - 10.8 = - 6 (add 10.8 to both sides)
11.4m = -6 + 10.8
11.4m = 4.8 (divide both sides by 11.4)
m = 4.8 / 11.4
m = 0.421
Answer:
m = 8/19
Step-by-step explanation :
5.4 (m-2) = -2 (3m +3)
54/10 (m-2) = (3m +3) as n = 1
2.3^3 / 2.5 (m-2) = (3m + 3)
4.868 (m-2) = (3m +3)
4.868 + 3 = (3m^2 * m-2)
8 / 9.5 * 2
m = 8 / 19