Statistics Problem

Question

Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 4 weeks. Suppose you would like to select a sample of 50 unemployed individuals for a follow-up study.

(a)

Show the sampling distribution of x,

 the sample mean average for a sample of 50 unemployed individuals.

A bell-shaped curve is above a horizontal axis labeled weeks.

1. The horizontal axis ranges from about -1.8 to about 1.8.
2. The curve enters the viewing window near -1.8 just above the horizontal axis, curves up to the right, and reaches a maximum near 0.
3. The curve then curves down and to the right until it leaves the viewing window at the same height it entered near 1.8.

 

A bell-shaped curve is above a horizontal axis labeled weeks.

1. The horizontal axis ranges from about 6.5 to about 30.5.
2. The curve enters the viewing window near 6.5 just above the horizontal axis, curves up to the right, and reaches a maximum near 18.5.
3. The curve then curves down and to the right until it leaves the viewing window at the same height it entered near 30.5.

 

A bell-shaped curve is above a horizontal axis labeled weeks.

1. The horizontal axis ranges from about 48.2 to about 51.8.
2. The curve enters the viewing window near 48.2 just above the horizontal axis, curves up to the right, and reaches a maximum near 50.
3. The curve then curves down and to the right until it leaves the viewing window at the same height it entered near 51.8.

 

A bell-shaped curve is above a horizontal axis labeled weeks.

1. The horizontal axis ranges from about 16.7 to about 20.3.
2. The curve enters the viewing window near 16.7 just above the horizontal axis, curves up to the right, and reaches a maximum near 18.5.
3. The curve then curves down and to the right until it leaves the viewing window at the same height it entered near 20.3.

(b)

What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1 week of the population mean? (Round your answer to four decimal places.)

(c)

What is the probability that a simple random sample of 50 unemployed individuals will provide a sample mean within 1

2

 week

 of the population mean? (Round your answer to four decimal places.)

Answer 1

Hi Steele,

(a) Note two things. First, we are looking at weeks an individual is unemployed. Nothing else. So any graph that does not include weeks should be eliminated. Second, recall the 68-95-99.7 rule for normal distributions. 68% of data falls within one standard deviation, 95% within two, and 99.7% within three. Now, for sample means, we have to compute a metric called standard error and substitute it for the standard deviation. This is:

SE=sigma/sqrt (n)

sigma=standard deviation

n=sample size

Here,

sigma=4

n=50

SE=4/sqrt(50)

SE=0.6

Therefore, looking at the final graph, we see that almost all of the data falls within 0.6 weeks of 18.5 in either direction, which means this is the most likely graph to satisfy our 68-95-99.7 rule. Answer is final graph.

(b) Here, we need to compute two z-scores, but we need a sample-specific formula to do it. That formula is:

z=(x-bar-mu)/SE

Breaking this down,

x-bar=sample mean

mu=population mean

SE=standard error

For this problem, we were asked for sample means within one week of the population mean, so our ultimate goal is to find:

P(17.5 < x-bar < 19.5)

since our population mean was 18.5 weeks.

So, we need z for both of those x-bar values--17.5 and 19.5 weeks. I will compute for 17.5 as z₁ and 19.5 as z₂.

z₁₌(x-bar-mu)/SE

x-bar=17.5

mu=18.5

SE=0.6, computed in part (a)

z₁=(17.5-18.5)/0.6

z₁= -1.67

Keep this value in mind. We still need to find z₂.

z₂=(x-bar - mu)/SE

x-bar=19.5

mu=18.5

SE=0.6, from part (a)

z₂=(19.5-18.5)/0.6

z₂=1.67

Now, we can standardize. A property of the normal distribution and z-scores is that:

P(-1.67 < z < 1.67)= P(17.5 < x < 19.5)

z-table gives "less than" probabilities, so we can get both P(z< -1.67) and P(Z<1.67) From table:

P(Z< -1.67)=0.0475

P(Z< 1.67)=0.9525

Now, we need to subtract those values, since we want the value between the two zs, equivalent to the value between the two xs:

P=0.9525 - 0.0475

P=0.9050

(c) This is a little unclear, but is the question asking for within 1/2 week of mu? If so, we want to use the same formula as above but with different sample means. I will still use z₁ and z₂. Our desired probability:

P(18 < x-bar < 19)

Now, we can compute our z-scores as above:

z=(x-bar-mu)/SE

x-bar=18

mu=18.5

SE=0.6

z₁=(18-18.5)/0.6

z₁= -0.83

Now, keep that value in mind, but we still need z₂.

z₂=(x-bar-mu)/SE

x-bar=19

mu=18.5

SE=0.6

z₂=(19-18.5)/0.6

z₂= 0.83

Now, we standardize again: P(-0.83 < z < 0.83) = P(18 < x-bar < 19)

From table:

P(z< -0.83)=0.2033

P(z<0.83)=0.7967

Again, we do the subtraction:

P=0.7967-0.2033

P=0.5934

I hope this helps.