Respuesta :
Answer:
c) use the student t- distribution.
t = 22.23
t= 22.23 > 1.645 at 149 degrees of freedom
Null hypothesis is rejected
Step-by-step explanation:
Step (i):-
Given sample size 'n' =150
The mean RDA of sodium is 2400mg
The mean of the Population 'μ' = 2400mg
Given mean of the sample x⁻ = 3400
The standard deviation of the sample 'S' = 550
Step(ii):-
Null hypothesis :H₀: μ = 2400
Alternative hypothesis : H₁: μ ≠2400
The test of statistic
[tex]t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{3400-2400}{\frac{550}{\sqrt{150} } } = 22.23[/tex]
degrees of freedom γ= n-1 = 150 -1 =149
t₀.₉₅ = 1.645
Conclusion:-
t= 22.23 > 1.645 at 149 degrees of freedom
Null hypothesis is rejected
The mean RDA of sodium is not equal to 2400mg
The appropriate approach is the student t distribution
How to determine the appropriate approach
The sample data are:
Sample size, n = 150
x = 3400
Sample standard deviation, [tex]\sigma[/tex] = 550
The population data is given as:
Population mean, [tex]\mu[/tex] = 2400
Notice that the population standard deviation is not given.
When the population standard deviation is not given, we make use of the student t distribution
Hence, the appropriate approach is the student t distribution
Read more about student t distribution at:
https://brainly.com/question/17305237