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.
Answer: 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 )