Invertase is an enzyme that may aid in spore germination of the fungus Colletotrichum graminicola. A botanist incubated specimens of the fungal tissue in petri dishes and then assayed the tissue for invertase activity. The specific activity values for nine petri dishes incubated at 90% relative humidity for 24 hours are summarized as follows:
Mean=5,111 units And SD = 818 units

This question is incomplete, the complete question is;

Invertase is an enzyme that may aid in spore germination of the fungus Colletotrichum graminicola.

A botanist incubated specimens of the fungal tissue in petri dishes and then assayed the tissue for invertase activity.

The specific activity values for nine petri dishes incubated at 90% relative humidity for 24 hours are summarized as follows:

Mean = 5,111 units And SD = 818 units

Assume that the data are a random sample from a normal population. Construct a 95% confidence interval for the mean invertase activity under these experimental conditions.

95% confidence interval for the mean is ( 4482.2309, 5739.7691 )

Step-by-step explanation:

Given that;

mean x" = 5,111

standard deviation σ = 818

sample size n = 9  

df = 9-1 = 8

95% confidence interval; ∝ = ( 1-0.95) = 0.05

so

tₙ₋₁,∝/2 = t₉₋₁, 0.05/2 = t₈, 0.025

t₈, 0.025 =  2.306     { from table }

so at 95% confidence interval, μ will be;

x" ±  t₀.₀₂₅ × s/√n

we substitute

5111 ±  (2.306 × 818/√9)

5111 ±  (2.306 × 272.6666)

5111 ± 628.7691

⇒ 5111 - 628.7691, 5111 + 628.7691

⇒ 4482.2309, 5739.7691

Therefore, 95% confidence interval for the mean is ( 4482.2309, 5739.7691 )

answer: idk

step-by-step explanation: sry idk

your answer would be 12