Question

19. Baldf aced hornets are actually a type of yellowjacket, and not a hornet. They are distinguished from other elowackets by their white and black coloring. They have white or a "baldfaced" head, which is the source ot its colloquial namesake. These wasps also h bodies. They are considerably ave three white stripes at the end of their ger than other species of Dolichovespula. Adults average 19 millimeters with a standard deviation of 2.9 millimeters i n length. Suppose a biologist studying Baldfaced hornets in the northern part of Idaho collected 43 of these wasps from the same area within 10,000 acres of Spruce Lake for study.

Identify the sampling distribution of the sample mean of length.

Between what two values would you expect to find the average lengths of "all adult Baldfaced hornets based on this same sample size?

How does this differ from 'all" the lengths?

d) Identify the 95h percentile for the average adult body lengths based on this same sample size.

(e) Suppose the biologist found the average adult average body length was 17.98 mm for this sest. Should the biologist consider this an unusal find? Why?

(f) what average body lengths would the top 10% of all average body lengths would you expect the Ballade hornets have (based on the same sample size)?

g) Suppose the sample size was 53, rather than 43. Between what two values would you expect to find the average lengths of "all" adult Baldfaced hornets based on this sample size? Why are these expected upper and lower bounds so different than those you calculated in part (b) above? What is happening?

Answer 1

from the given data of information

Baldf aced hornets are actually a type of yellow jacket, and not a hornet. They are distinguished from other elowackets by their white and black coloring.

They are considerably ave three white stripes at the end of their greater than other species of Dolichovespula

from the information we consider the following results

(a) The sampling distribution of sample mean of lengths is a Normal distribution with mean as sample mean and variance as sample variance divided by the sample size.

(b) We are 95% confident

the true average length will lie within 1.96 standard deviations.

Standard deviation of mean = 2.9 / sqrt(43)

= 0.4422459
Margin of error = 1.96 * 0.4422459

= 0.8668
So, the true average length will lie between(19 - 0.8668, 19 + 0.8668)
= (18.1332, 19.8668)

(c)  95% of times, average of all the lengths will lie

(18.1155, 19.8845)

(d)  Z value for 95% percentile is 1.645
95% percentile for the average adult body length = 19 + 1.645 * 0.4422459

= 19.7275

(e) In part (b), we see that 95% confidence interval does not contain the value 17.98 mm.

So, the biologist conside this as unusual find.

(f)    For top 10%, the percentile = 100 - 10 = 90%
Z value for 90% percentile is 1.28
95% percentile for the average adult body length = 19 + 1.28 * 0.4422459

= 19.5661 mm

(g)   sample size n = 53
Standard deviation of mean = 2.9 / sqrt(53)

= 0.3983456
Margin of error = 1.96 * 0.3983456

= 0.7808
So, the true average length will lie between(19 - 0.7808, 19 + 0.7808)
= (18.2192, 19.7808)

The lower and upper limits are different because the standard error of the mean decreases with the sample size and thus, the margin of error and length of confidence interval decreases