Answer:
Step-by-step explanation:
Given:
The two points on the line are and
.
The slope of the line joining two points and
is given as:
Here,
∴
Equation of line with a point and slope
is given as:
Plug in -2 for , 3 for
and -7 for
. This gives,
Therefore, the equation of the line in vertex form is .
Answer:
The p value is higher than any significance level given for example , so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.
Let put some notation
x=Auscultatory method , y = Oscillatory method
x: 74 86 84 79 70 78 79 70
y: 70 85 90 110 71 80 69 74
The system of hypothesis for this case are:
Null hypothesis:
Alternative hypothesis:
The first step is define the difference , that is given so we have:
d: 6.6, 4.2, -5.5, -3.1, 9.3, -3.9
The second step is calculate the mean difference
The third step would be calculate the standard deviation for the differences, and we got:
The fourth step is calculate the statistic given by :
The next step is calculate the degrees of freedom given by:
Now we can calculate the p value, since we have a right tailed test the p value is given by:
The p value is higher than any significance level given for example , so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.
Answer:
a) Private Colleges
Sample mean = 42.5 thousand dollars
Standard deviation = S1 = 6.62 thousand dollars.
Public colleges
Sample mean = 22.3 thousand dollars
Standard deviation = 4.34 thousand dollars
b) The difference in sample mean for both cases = 42.5 - 22.3 = 20.2 thousand dollars
The average amount of going to a Private college is 20.2 thousand dollars more than the average cost of going to public colleges
c) 95% confidence interval for a sampling distribution of the difference of the cost of private and public colleges is given as
(15.0, 25.4) thousand dollars.
Step-by-step explanation:
Private colleges.
52.8 43.2 45.0 33.3 44.0 30.6 45.8 37.8 50.5 42.0
Public colleges.
20.3 22.0 28.2 15.6 24.1 28.5 22.8 25.8 18.5 25.6 14.4 21.8
a) Calculate sample mean and standard deviation for both data set.
Mean = (Σx)/N
where N = Sample size
Σx = sum of all variables
Private colleges
Σx = (52.8+43.2+45.0+33.3+44.0+30.6+45.8+37.8+50.5+42.0) = 425
N = 10
Mean = 425/10 = 42.5 thousand dollars
Standard deviation = S1 = √[Σ(x - xbar)²/N]
Σ(x - xbar)² = (52.8-42.5)² + (43.2-42.5)²
+ (45.0-42.5)² + (33.3-42.5)²
+ (44.0-42.5)² + (30.6-42.5)² + (45.8-42.5)² + (37.8-42.5)² + (50.5-42.5)² + (42.0-42.5)² = 438.56
N = 10
Standard deviation = √[Σ(x - xbar)²/N]
Standard deviation = √(438.56/10) = 6.62 thousand dollars
Public colleges
Mean = (Σx)/N
Σx =
(20.3+22.0+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) = 267.6
N = 12
Mean = (267.6/12) = 22.3 thousand dollars
Standard deviation = √[Σ(x - xbar)²/N]
[Σ(x - xbar)²
(20.3-22.3)² + (22.0-22.3)² + (28.2-22.3)² + (15.6-22.3)² + (24.1-22.3)² + (28.5-22.3)² + (22.8-22.3)² + (25.8-22.3)² + (18.5- 22.3)² + (25.6-22.3)² +(14.4-22.3)+(21.8-22.3) = 225.96
N = 12
standard deviation = s2 = √(225.96/12) = 4.34 thousand dollars
b) The difference in sample mean for both cases = 42.5 - 22.3 = 20.2 thousand dollars
The average amount of going to a Private college is 20.2 thousand dollars more than the average cost of going to public colleges.
c. Develop a 95% confidence interval of the difference between the annual cost of attending private and pubic colleges.
95% confidence interval, private colleges have a population mean annual cost $ to $ more expensive than public colleges.
To combine the distribution in this manner,
Sample mean of difference = 20.2 thousand dollars
Combined standard deviation of the sampling distribution = √[(S1²/n1) + (S2²/n2)]
= √[(6.62²/10) + (4.34²/12)] = 2.44 thousand dollars
Confidence interval = (Sample mean) ± (Margin of error)
Sample mean = 20.2
Margin of error = (critical value) × (standard deviation of the sampling distribution)
standard deviation of the sampling distribution = 2.44
To obtain the critical value, we need the t-score at a significance level of 5%; α/2 = 0.025
we obtain the degree of freedom too
The degree of freedom, df, is calculated in the attached image.
df = 15
t (0.025, 15) = 2.13145 from the tables
Margin of error = 2.13145 × 2.44 = 5.20
Confidence interval = (Sample mean) ± (Margin of error)
= (20.2 ± 5.2) = (15.0, 25.4)
Hope this Helps!!!
