A researcher who wants to test the hypothesis that the population mean of a variable is 85, takes a sample of 50 people and obtains a sample mean of 81. A hypothesis test finds that the sample mean is not significantly different from the hypothesized population mean at 5% level of significance. What could be a 95% confidence interval for the population mean?

CI = (  28,28  - 1,96*σ ;  28,28 + 1,96*σ )

Step-by-step explanation:

Population mean      μ₀  = 85

Sample size   n  =  50

Sample mean   μ  = 81

Significance Level     α = 5 %    α  =  0,05

From Hypothesis Test

Null Hypothesis                         H₀         μ    =    μ₀

Alternative Hypothesis             Hₐ         μ    ≠   μ₀  

After a two taio-test was found no significative difference about the mean

Then  α/2  =  0,05/2   α/2 = 0,025

CI = 95 %     CI = 0,95

From z-table

z(c) = (±) 1,96

CI =  (  μ -  μ₀ ) ± z(c) * σ/√50

CI =  (  85 - 81 )*√50  ± 1,96 *σ

CI =  4*7,07   ± 1,96*σ

CI  =  (  4*7,07  - 1,96*σ ;   4*7,07  + 1,96*σ )

CI = (  28,28  - 1,96*σ ;  28,28 + 1,96*σ )

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>