We gave a national verbal proficiency test that has a mean score of 500 with a standard deviation of 75. What raw score would co
1 answer:
Answer:
519
Step-by-step explanation:
Let's assume that the scores for the national verbal proficiency test follow a normal distribution.
Mean score (μ) = 500
Standard deviation (σ) = 75
According to a Z-score table, the score for the 60th percentile is roughly z = 0.253
For any score "x", the z-score is given by:
For z = 2.53:
The raw score to the 60th percentile is 519.
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Answer:
y = x + 2
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Calculate m using the slope formula
m =
with (x₁, y₁ ) = (0, 2) and (x₂, y₂ ) = (3, 4)
m = =
The line crosses the y- axis at (0, 2) ⇒ c = 2
y = x + 2 ← equation of line
Answer:
The 95% confidence interval for the true slope is (0.03985, 0.14206).
Step-by-step explanation:
For the regression equation:
The (1 - <em>α</em>)% confidence interval for the regression coefficient or slope is:
The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:
The summary of the regression analysis is:
Predictor Coefficient SE t-stat p-value
Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹
Entering GPA 0.090953 0.022162 4.103932 0.003419
The regression coefficient and standard error are:
The critical value of <em>t</em> for 95% confidence level and 8 degrees of freedom is:
Compute the 95% confidence interval for as follows:
Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).
