BUS219 / MAT 167 Online sections Please return your solutions (which means not just answers but the work) by the deadline indicated on our schedule of due dates. Acid rain has caused many lakes to...

BUS 219
BUS219 / MAT 167
Online sections
Please return your solutions (which means not just answers but the work) by the deadline indicated on our schedule of due dates.
1. Acid rain has caused many lakes to become more acidic. A lake is considered nonacidic if the PH level is greater than 6. The PH level for 15 lakes was:
7.2, 7.3, 6.1, 6.9, 6.6, 7.3, 6.3, 5.5, 6.3, 6.5, 5.7, 6.9, 6.7, 7.9, and 5.8
At “alpha” = 0.05, is there enough evidence to conclude that, on average, these lakes are nonacidic? (6 points: 1,1,2,1,1)
Solution:
Null Hypothesis (Ho): µ ≤ 6
Alternative Hypothesis (Ha): µ> 6
Sample mean, X-bar = ∑X/n = 6.6
Sample standard deviation, s = √∑ (X – mean) ²/ (n – 1) = 0.6719
Test Statistics
t = (X-bar - µ)/ (s/√n)
= (6.6 – 6)/ (0.6719/√15)
= 3.46
Degrees of freedom, df = n – 1 = 15 – 1 = 14
Using t-tables, the critical value is
t (0.05, 14) = 1.761
Since test statistics is greater than critical value, we reject the null hypothesis.
There is enough evidence to conclude that, on average, these lakes are nonacidic.
2. A used car salesman claims that the average price of a 2009 Honda Accord LX is $20,500. You, of course, distrust him (you feel that it is much lower) and find that a random sample of 14 similar cars has a mean of price of $19,850 with a standard deviation of $1084. Is there enough evidence to reject the salesman’s claim at “alpha” = 0.05? (6 points: 1,1,2,1,1)
Solution:
Null Hypothesis (Ho): µ ≥ 20,500
Alternative Hypothesis (Ha): µ < 20,500
Test Statistics
t = (X-bar - µ)/ (s/√n)
= (19,850 – 20,500)/ (1084/√14)
= -2.24
Degrees of freedom, df = n – 1 = 14 – 1 = 13
Using t-tables, the critical value is
t (0.05, 13) = -1.77
Since test statistics is less than critical value, we reject the null hypothesis.
There is enough evidence to reject the salesman’s claim at “alpha” = 0.05.
3. Some big shot Institute claims that 25% of college graduates believe that a college degree is not worth the cost. To test this hypothesis you ask a random sample of 200 graduates and 21% agree with that a degree is not worth the cost. At “alpha” = 0.10, is there enough evidence to reject the claim? (6 points: 1,1,2,1,1)
Solution:
Null Hypothesis (Ho): p = 0.25
Alternative Hypothesis (Ha): p ≠ 0.25
Test Statistics
Z = (p’ – p)/ √ p (1 – p)/n
= (0.21 – 0.25)/ √0.25 x 0.75/200
= -1.31
Using Z-tables, the critical value is
Z (α/2) = Z (0.10/2) = -1.645
Since test statistics lie outside the critical value, we reject the null hypothesis.
There is enough evidence to reject the claim.
4. A 2009 national poll of 1053 U.S. adults found that 548 favored the Governments plan for a stimulus package of a minimum of 800 billion dollars Using “alpha” = 0.05, do the data provide evidence to conclude that a majority (more than 50%) of adults favored passage? (6 points: 1,1,2,1,1)
Solution:
Null Hypothesis (Ho): p ≤ 0.50
Alternative Hypothesis (Ha): p > 0.50
We have X = 548 and n = 1053
Sample proportion, p’ = X/n = 548/1053 = 0.52
Test Statistics
Z = (p’ – p)/ √ p (1 – p)/n
= (0.52 – 0.5)/ √0.5 x 0.5/1053
= 1.33
Using Z-tables, the critical value is
Z (α) = Z (0.05) = 1.645
Since test statistics is less than critical value, we fail to reject the null hypothesis.
The data provide insufficient evidence to conclude that a majority (more than 50%) of adults favored passage.
5. A sporting goods producer claims that the variance of the strengths of a certain fishing line is 15.9. A random sample of 15 spools has a variance of 21.8. At “alpha” = 0.01, can the claim be proven wrong? Assume the population is normally distributed. (6 points: 1,1,2,1,1)
Solution:
Null Hypothesis (Ho): σ² = 15.9
Alternative Hypothesis (Ha): σ² ≠ 15.9
Test Statistics
X² = (n – 1) s²/σ²
= (15 – 1) (21.8)/ 15.9
= 19.194
Degrees of freedom, df = n – 1 = 15 – 1 = 14
Using chi-square tables, the critical value is
X² (0.01/2, 14) = 4.075 and X² (0.99/2, 14) = 31.319
Since test statistics lie within the critical values, we fail to reject the null hypothesis.
Hence, the claim is not proven to be wrong.
6. The standard deviation of the amount of fat in whole milk is claimed to be no greater than 0.50 by the Dairy Association. You do not believe this. You find that a random sample of 31 milk containers has a standard deviation of 0.53. Using “alpha” = 0.05, is there enough evidence to reject the...