Answer:
Step-by-step explanation:
Hello!
Be the variable of interest:
X: Number of weeks it takes a worker aged 55 plus to find a job
Sample average X[bar]= 22 weeks
Sample standard deviation S= 11.89 weeks
Sample size n= 40
a)
The point estimate of the population mean is the sample mean
X[bar]= 22 weeks
It takes on average 22 weeks for a worker aged 55 plus to find a job.
b)
To estimate the population mean using a confidence interval, assuming the variable has a normal distribution is
X[bar] ± [tex]t_{n_1; 1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]
[tex]t_{n-1; 1-\alpha /2}= t_{39; 0.975}= 2.023[/tex]
The structure of the interval is "point estimate" ± "margin of error"
d= [tex]t_{n_1; 1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]= 2.023*[tex](\frac{11.89}{\sqrt{40} })[/tex]= 3.803
c)
The interval can be calculated as:
[22 ± 3.803]
[18.197; 25.803]
Using s 95% confidence level, you'd expect the population mean of the time it takes a worker 55 plus to find a job will be within the interval [18.197; 25.803] weeks.
d)
Job Search Time (Weeks)
21 , 14, 51, 16, 17, 14, 16, 12, 48, 0, 27, 17, 32, 24, 12, 10, 52, 21, 26, 14, 13, 24, 19 , 28 , 26 , 26, 10, 21, 44, 36, 22, 39, 17, 17, 10, 19, 16, 22, 5, 22
To study the form of the distribution I've used the raw data to create a histogram of the distribution. See attachment.
As you can see in the histogram the distribution grows gradually and then it falls abruptly. The distribution is right skewed.
