10.4.1 .WP The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at ran
1 answer:
Hello! You need to calculate a 99% confidence interval for the difference in mean lifespan between two tire brands. Each tested car was assigned one tire from each brand randomly on the rear wheels, allowing for paired sample analysis.
Brand 1 Brand 2 X₁-X₂
car 1: 36,925; 34,318; 2.607
car 2: 45,300; 42,280; 3.020
car 3: 36,240; 35,500; 0.740
car 4: 32,100; 31,950; 0.150
car 5: 37,210; 38,015; -0.0805
car 6: 48,360; 47,800; 1.160
car 7: 38,200; 37,810; 0.390
car 8: 33,500; 33,215; 0.285
n= 8
The study variable is defined as Xd= X₁-X₂, where X₁ represents the tire lifespan (in km) from Brand 1 and X₂ represents Brand 2. Thus, Xd is the difference in tire lifespan.
Xd~N(μd;δd²) (normality test p-value is 0.4640).
For calculating the confidence interval, the best statistic is the Student's t using the following formula:
t= (xd[bar] - μd)/(Sd/√n) ~t₍ₙ₋₁₎
sample mean: xd[bar]= 0.94
standard deviation: Sd= 1.29
= 3.355
xd[bar] ± 
*(Sd/√n) ⇒ 0.94 ± 3.355*(1.29/√8)
[-0.65;2.54]km.
The CI can be compared to bilateral hypothesis testing:
H₀:μd=0
H₁:μd≠0
using significance level of 0.01.
Since the confidence interval includes zero, we do not reject the null hypothesis, indicating no significant difference between the tire brands.
Hope you have a fantastic day!
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Part A
To identify the values of x that make 2x−1 positive
⇒ 2x - 1 > 0
⇒ 2x > 1
⇒ x >
Part B
To find values of y making 21−37 negative
⇒ 21-3y < 0
⇒ 21 < 3y
⇒ 7 < y
Thus, for all y values exceeding 7, the expression 21-3y is negative
Part C
To identify values of c that digit 5−3c greater than 80
⇒ 5-3c > 80
⇒ -3c > 75
⇒ -c > 25
⇒ c < -25
Therefore, for values of c less than -25, the expression 5-3c surpasses 80
Answer:
The correct selection is 1.
Step-by-step explanation:
The histogram indicates that
In the 5-10 age group, there are 7 students.
In the 11-13 age group, 5 students are represented.
In the 14-18 age category, 4 students are included.
This implies that the ages of 7 students fall within the 5-10 range, 5 students within the 11-13 range, and 4 students within the 14-18 range.
The total count in the dataset is
In options 3 and 4, the counts in the datasets are fewer than 16. Hence, options 3 and 4 are invalid.
Option 2 exclusively contains data within the 5-10 range, making it incorrect as well.
In option 1,
Class intervals elements frequency
5-10 5,6,7,8,9,10,10 7
11-13 11,11,12,12,13 5
14-18 14,14,15,16 4
Based on the above table, we can determine that data set 1 is represented in the provided histogram.
Thus, the answer is option 1.
Response:
a. As student debt rises, current investment diminishes.
b. Y= 68778.2406 - 1.9112X
For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.
c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.
d. Y= $59222.2406
e. R²= 0.9818
Step-by-step breakdown:
Hello!
Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:
Y: Current investment by an individual who graduated from college five years prior.
X: Total debt of an individual upon graduating five years ago.
a)
To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.
The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.
Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.
b)
The population regression equation is Y= α + βX +Ei
To develop this equation, estimates for alpha and beta are required:
a= Y[bar] -bX[bar]
a= 44248.55 - (-1.91)*12829.70
a= 68778.2406
b=
b=
b= -1.9112
∑X= 256594
∑X²= 4515520748
∑Y= 884971
∑Y²= 43710429303
∑XY= 9014653088
n= 20
Averages:
Y[bar]= ∑Y/n= 884971/20= 44248.55
X[bar]= ∑X/n= 256594/20= 12829.70
The estimated regression equation becomes:
Y= 68778.2406 - 1.9112X
For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.
c)
To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:
H₀: β = 0
H₁: β ≠ 0
α: 0.01
Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)
Using t= b - β = -1.91 - 0 = -31.83
Sb 0.06
The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001
Since this P-value is underneath the significance level, we reject the null hypothesis.
In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.
The P-value remains: 0.0001
Using this method, we similarly reject the null hypothesis.
This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X
I trust this is beneficial!
