The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an
1 answer:
Answer:
We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .
Step-by-step explanation:
Given -
The sample size is large then we can use central limit theorem
n = 50 ,
Standard deviation = 7.1
Mean = 110
1 - confidence interval = 1 - .98 = .02
= 2.33
98% confidence interval for the mean caffeine content for cups dispensed by the machine =
=
=
First we take + sign
= 112.34
now we take - sign
= 107.66
We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .
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It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}
Answer:
a) For this case the value of the significanceis and
, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:
If the calculated statistic we can reject the null hypothesis at 5% of significance
b) Where
c)
d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.
Step-by-step explanation:
Data given and notation
represent the number of people with the characteristic 1
represent the number of people with the characteristic 2
sample 1 selected
sample 2 selected
represent the proportion estimated for the sample 1
represent the proportion estimated for the sample 2
represent the pooled estimate of p
z would represent the statistic (variable of interest)
represent the value for the test (variable of interest)
significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We need to apply a z test to compare proportions, and the statistic is given by:
(1)
a.State the decision rule.
For this case the value of the significanceis and
, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:
If the calculated statistic we can reject the null hypothesis at 5% of significance
b. Compute the pooled proportion.
Where
c. Compute the value of the test statistic.
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Replacing in formula (1) the values obtained we got this:
d. What is your decision regarding the null hypothesis?
Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.