Question

can someone help me answer both A and B please?

Name: Laboratory Worksheet #8 (continued) In a tree survey, densities of tree stems were determined. Using the data and formula given below, answer the following questions. Show your all calculation work. Use the backside if necessary Plot #6 Plot #4 Plot #5 31 stems 28 stems Plot #3 29 stems Plot #2 37 stems Plot #1 38 stems 36 stems Area A 35 stems 34 stems not 25 stems 32 stems 38 stems Area B sampled X- ΣΝ S- Xi-X(n-1) square root of E (xi-X)/1(n-1 )) SE S/(VN-1), where (VN-1)-square root of (N-1) t-(XA-XB)N(SEA2+ SEB), where v((SEA2 +SEn) = square root of (SEA+ SEB df- Ni+N2-2 (A) Calculate mean, standard deviation, and standard error of mean for each area. (B) Conduct t-test at a 0.05 to test statistical significance in the difference of stem densities between the two areas.

Answer 1

A -

For Area A -

Mean - 35 + .. +38 / 6

= 34.3

Standard deviation -

sqrt [(35-34.3)² +....+ (38-34.3)²]

= 3.55

Standard error -

= std dev / sqrt (n-1)

= 1.6

For Area B -

Mean - 38 + .. +28 / 5

= 31.4

Standard deviation -

= 5.07

Standard error -

= 2.53

B.

Area A

N₁: 6
df₁ = N - 1 = 6 - 1 = 5
M₁: 34.3
SS₁: 63.3 (SS = sum of squared errors = sum over all observations [(Mean - observation)²]
s²₁ = SS₁/(N - 1) = 63.3/(6-1) = 12.67


Area B

N₂: 5
df₂ = N - 1 = 5 - 1 = 4
M₂: 31.4
SS₂: 103.2
s²₂ = SS₂/(N - 1) = 103.2/(5-1) = 25.8


T-value Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂)

= ((5/9) * 12.67) + ((4/9) * 25.8)

= 18.5

s²_M1 = s²_p/N₁

= 18.5/6

= 3.08
s²_M2 = s²_p/N₂

= 18.5/5

= 3.7

t = (M₁ - M₂)/√(s²_M1 + s²_M2)

= 2.93/√6.78

= 1.13

The p-value for t = 1.13 and degrees of freedom = 11 is .289233

The result is not significant at p < .05

Hence, we do not reject the null hypothesis that -

"The mean stem densities of the trees of two areas do not differ significantly"