Respuesta :
Answer:
The value for p <0.1 i.e. 0.02088 that means ,we had success in test on potassium content
Step-by-step explanation:
Given:
Means: 138 mg
sample size =40
New mean=136.9 mg
To find : Hypothesis on test.
Solution:
We know that ,
Z=(X-mean)/standard deviation)/{(sqrt(sample size)}
Consider the standard deviation of 3.00 mg
Z=(136.9-138)/(3/sqrt(40))
=-1.1/0.4743
Z=-2.3192
two tailed test
=2{(1-p(<z))}
P value for Z=2.3192 is 0.98956
=2{1-0.98956}
=0.02088
The value for p <0.1
we had success in test on potassium content.
The value for p <0.1 is 0.02088 which testing means, we had success in the test on potassium content.
Given that,
The mean potassium content of a popular sports drink is listed as 138 mg in a 32-oz bottle.
Analysis of 40 bottles indicates a sample mean of 136.9 mg.
We have to determine,
The hypotheses for a two-tailed test of the claimed potassium content.
According to the question,
When we test a hypothesis for significance, we can test them either for the right-hand side of the mean, left-hand side of the mean, or both sides. When it is for only one side, it is called a one-tail test. If it is for both sides, we check the significance for both sides. That is a two-tail test.
The mean potassium content of a popular sports drink is listed as 138 mg in a 32-oz bottle.
The hypotheses for a two-tailed test of the claimed potassium content is determined by the formula,
[tex]\rm z = \dfrac{Mean - Standard \ deviation }{\sqrt{Sample \ size} }[/tex]
Where the value of Mean = 138 mg , Sample size = 40 and New mean = 136.9 mg.
Substitute all the values in the formula,
[tex]\rm z = \dfrac{New \ mean - Mean \times \sqrt{Sample \ size} } {Standard \ deviation }\\\\ z = \dfrac{(136.9 - 138) \times \sqrt{40} } {3 }\\\\Z = \dfrac{-6.952} {3 }\\\\ z = -2.31[/tex]
The value of p by using the two-tail test is,
P-value for Z = 2.3192 is 0.98956
P = 2{(1-p<z)
P = 2{1-0.98956}
P = 0.0208
The value for p < 0.1.
Hence, The value for p <0.1 is 0.02088 which testing means, we had success in the test on potassium content.
For more details refer to the link given below.
https://brainly.com/question/18833193
