Question

3.114. Cicadas Forsyth (1976) estimates the abundance of 17-year cicadas in eastern Ohio in 1968 (42). A sample of 30 locations in the Mineral Ridge oak forest had mean number of emergence holes per square foot of 4.5 with standard deviation 5.9. Continue as if this is a random sample. (a) The minimum posssible number of emergence holes is obviously 0. With that in mind, what does the comparison of the mean to the standard deviation suggest about whether the shape of the 30 data values is symmetric, skewed right, or skewed left? Explain. (b) Write out the hypotheses (step 2), find the test statistic (step 3c), find the p-value (step 4b), and make a decision using a = 0.01 (step 4e) to test if the mean number of emergence holes in the Mineral Ridge oak forest is less than 6.

Answer 1

Answer:

Given that:

Cicadas frosyth (1979 )estimate the ambundance of 17 -year cicadas in eastern Ohio in 1968 42.

Consider X-is the

X : number of emergence holes in the mineral ridge

Ok forest

From the information

n=sample size =30

x=sample= mean=4.5

s= standard deviation =5.9

a) given minimum possible value =0

Since the 99.73% values lies in the

=(
X - 3*5, X +3*5
)

=(4.5-3 *5.9,4.5+3*5.9)

= (-13.2 , 22.2)

maximum possible value = 4.5+3*5.9

=22.2

The distance of minimum from mean is less than

the distance of maximum from mean

The right tail is long relative to the left tail

the distraction is positivity skewed on right skewed

b)

let

$\mu$; mean number of emergence holes in the mineral

ridge ok forest .

Test the hypothesis

Whether or not mean number of emergence holes in the

mineral ridge ok forest is less then 6

Null hypothesis :

HO :μ = 6

Alternative hypothesis :

H, :μ<6

the population standard deviation is unknown

and sample size is small

(n <= 30

we use one sample test for testing population mean.

the value of test statistic is

+- X-H Sta

X - u t = S/Vn

t= 4.5 - 6 5.9/ 30

—1.5 t=59/5.47722

—1.5 1.077189

t = -1.3925132

the Value of test statistic is -1.3925132

level of significance

a = 0.01

degrees of freedom

(df) = n - 1

=30-1=29

P-value=P

(t<-1,392)

= 0.0872

since p-value is greater than level of significance

x, the result is not significant we fail to reject the null hypothesis at 1%

level of significance

there is no sufficient evidence to say that mean

number of emergence holes in mineral ridge ook forest is less than c.

.