SOLUTION: A sociologist found that in a sample of 50 retired men, the average number of jobs they had during their lifetimes was 7.2. The population standard deviation is 2.1.
a) Find the b
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Question 1193182: A sociologist found that in a sample of 50 retired men, the average number of jobs they had during their lifetimes was 7.2. The population standard deviation is 2.1.
a) Find the best point estimate of the mean.
b) Find the 95% confidence interval of the mean number of jobs.
c) Find the 99% confidence interval of the mean number of jobs.
Found 2 solutions by Boreal, math_tutor2020:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Best point estimate is from the sample, 7.2 jobs
95% half-interval is z(0.975)*sigma/sqrt(n)
=1.96*2.1/sqrt(50)
=0.58 hours
interval is (6.62, 7.78) hours
-
half-interval is 2.576*2.1/sqrt(50)
=0.765 hours
(6.43, 7.97) hours
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
Part (a)
The best point estimate of the population mean is the sample mean. It will be handy later to form the confidence intervals.
We can use the notation xbar to indicate the sample mean.
Answer: 7.2
==============================================================
Part (b)
n = 50 = sample size
sigma = 2.1 = population standard deviation
At 95% confidence, the z critical value is roughly z = 1.960
You'll need to either memorize this value or have it on a reference table. A specialized calculator is also another option.
E = margin of error
E = z*sigma/sqrt(n)
E = 1.960*2.1/sqrt(50)
E = 0.58209030227277
E = 0.582090
L = lower bound of confidence interval
L = xbar - E
L = 7.2 - 0.582090
L = 6.61791
L = 6.62
U = upper bound of confidence interval
U = xbar + E
U = 7.2 + 0.582090
U = 7.78209
U = 7.78
Notice how the sample mean xbar = 7.2 is the center of the confidence interval.
Answer: (6.62, 7.78)
==============================================================
Part (c)
We use the same idea as part (b)
This time we have z = 2.576 as the critical value for a 99% confidence interval.
Every other variable stays the same.
E = margin of error
E = z*sigma/sqrt(n)
E = 2.576*2.1/sqrt(50)
E = 0.76503296870134
E = 0.765033
L = lower bound of confidence interval
L = xbar - E
L = 7.2 - 0.765033
L = 6.434967
L = 6.43
U = upper bound of confidence interval
U = xbar + E
U = 7.2 + 0.765033
U = 7.965033
U = 7.97
The point estimate is the same as last time. The only thing that changed is the margin of error.
The higher the confidence level, the larger the margin of error will be. This in turn creates a wider confidence interval.
It's like trying to catch an elusive fish. The wider the net (i.e. the wider the interval), the more confident we are in catching the fish.
Answer: (6.43, 7.97)
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