Calculation and Interpretation of Covariance, Correlation Coefficient, Probability and Test Statistics

This text explains the calculation and interpretation of covariance, correlation coefficient, probability and test statistics. It includes examples and solutions for each concept. The negative covariance between salary and years of experience may be due to compressed salaries over time. The strong negative correlation between years of experience and salary indicates that salary tends to decrease as experience increases. The probability of customers holding for more than 1.5 minutes is 20.12%. The power of the test to detect a mean lifetime different from 950 hours is 34.65%. The mean filling weight is in proper adjustment based on the test statistics.

Question 1
a. The covariance is calculated as shown below:
Xi Yi
5 20 -0.5 1.5 -0.75
3 23 -2.5 4.5 -11.25
7 15 1.5 -3.5 -5.25
9 11 3.5 -7.5 -26.25
2 27 -3.5 8.5 -29.75
4 21 -1.5 2.5 -3.75
6 17 0.5 -1.5 -0.75
8 14 2.5 -4.5 -11.25
= 44 148 Σ = -89
= 5.5 = 18.5
 Cov (x, y) = = -89/ 7 = -12.7143
The covariance between x and y variables indicates a negative relationship. In other
words, the salary decreases as the years of experience increases. The magnitude of the
covariance cannot be explained.
b. The negative covariance could be because the salaries are not adjusted upwards to reflect
the changing market values, hence the salaries become compressed over time.
c. The coefficient of correlation is calculated as:
= -0.9904
The variables x and y have a very strong negative relationship which means that as the
years of experience increase, the salary tends to decrease.

d. The negative relationship may arise from salaries being compressed over time by not
adjusting them upwards for changing market values.
Question 2
a. μ = 3 =>
b.
X > 1.5 minutes => P (X > 1.5) = =
0.2012 =20.12%
c. P (X > k) = 10% = e-0.33k =>
10% of the customers will continue to hold for approximately 7 minutes
d. P (3 < X < 6) = P (X < 6) – P (X < 3)
=
= 0.8619 – 0.6284 = 0.2335 =23.35%
Question 3
H0: μ = 950 hours
H1: μ ≠ 950 hours
Sample size, n = 25; standard deviation, s = 200 hours
a. When μ = 1000 and α = 0.10, β = P (884.2 < x < 1015.8) = P ( -2.9 < z < 0.40) = 0.6535
b. Power = 1 – β = 1 – 0.6535 = 0.3465

c. The power of the test means that the probability of detecting that the mean lifetime is not
950 hours, when indeed the lifetime of the battery is 1,00 hours, is 34.65%, when the
level of significance, α = 0.10.
d. The value of β decreases, when the sample size, n increases.
Question 4
Ho: μ = 47 ounces
Ha: μ ≠ 47 ounces
Sample mean = 48.6 ounces; α = 5%; n = 36; standard deviation = 6 ounces
 Test statistics = = 1.6
 P-value = P (Z > |z|) = P(Z < -1.6) + P(Z > 1.6) = 2* P(Z < 1.6) = 1.8904
Since the p-value, 1.89, is greater that the level of significance, 0.05, we do not reject Ho.
Hence, there is sufficient evidence to conclude that the mean filling weigh is in proper
adjustment.