1. The manager of a paint supply store wants to determine whether the mean amount of paint contained in 1-gallon
cans purchased from a nationally known manufacturer actually is 1 gallon. It is known from the manufacturer's
specifications that the standard deviation of the amount of paint is equal to 0.02 gallon. A random sample of 50 cans is
selected, and the mean of the amount of paint per 1-gallon can is found to be .995 gallon.
a. Is there evidence that the mean amount is different from 1.0 gallon (use a=0.01)?
b. Compare the p-value and interpret its meaning.
c. Construct a 99% confidence interval estimate of the population mean amount of paint.
d. Compare the results of (a) and (c). What conclusions do you reach?
(a)
n = 50
μ= 1
s = 0.02
x-bar = 0.995
(1) Formulate the hypotheses:
Ho: μ = 1
Ha: μ ≠ 1
(2) Decide the test statistic and the level of significance:
z (Two-tailed), α = 0.01
Lower Critical z- score = -2.5758
Upper Critical z- score = 2.5758
(3) State the decision Rule:
Reject Ho if |z| > 2.5758
(4) Calculate the value of test statistic:
SE = s/√n = 0.0028
z = (x-bar - μ)/SE = -1.7678
(5) Compare with the critical value and make a decision:
Since 1.7678 < 2.5758 we fail to reject Ho
Decision: There is no sufficient evidence that the average amount of paint in the can is different from 1.0 gallon
(b) p- value = 0.0771. This is the probability of rejecting a true null hypothesis.
(c)
n = 50
x-bar = 0.995
s = 0.02
% = 99
Standard Error, SE = σ/√n = 0.0028
z- score = 2.5758
Width of the confidence interval = z * SE = 0.0073
Lower Limit of the confidence interval = x-bar - width = 0.9877
Upper Limit of the confidence interval = x-bar + width = 1.0023
The confidence interval is [0.988 gallon, 1.002 gallons]
(d) 1.0 gallon lies within the confidence interval in (c). This means we can’t reject Ho. The same conclusion was reached
in (a) also. Both (a) and (c) mean the same thing.
2. If you use a 0.05 level of significance in a two tail hypothesis test, what will decide if zstat = -0.76?
Two-tail p- value for z = -0.76 is 0.4473
Since 0.4473 > 0.05, the null hypothesis will not be rejected.
3. The quality- control manager at a light bulb factory needs to determine whether the mean life of a large shipment of
light bulbs is equal to 375 hours. The population standard deviation is 100 hours. A random sample of 64 light bulbs
indicates a sample mean life of 350 hours.
a. At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours?
b. Compute the p- value and interpret its meaning.
c. Construct a 95% confidence interval estimate of the population mean life of the light bulbs.
d. Compare the results of ( a) and ( c). What conclusions do you reach?
(a)
n = 64
μ = 375
s = 100
x-bar = 350
(1) Formulate the hypotheses:
Ho: μ = 375
Ha: μ ≠ 375
(2) Decide the test statistic and the level of significance:
z (Two-tailed), α = 0.05
Lower Critical z- score = -1.9600
Upper Critical z- score = 1.9600
(3) State the decision Rule:
Reject Ho if |z| > 1.9600
(4) Calculate the value of test statistic:
SE = s/√n = 12.5000
z = (x-bar - μ)/SE = -2.0000
(5) Compare with the critical value and make a decision:
Since 2.0000 > 1.9600 we reject Ho and accept Ha
Decision: It appears that the average life of the bulbs is different from 375 hours
(b) p- value = 0.0455. This is the probability of rejecting a true null hypothesis.
(c)
n = 64
x-bar = 350
s = 100
% = 95
Standard Error, SE = σ/√n = 12.5000
z- score = 1.9600
Width of the confidence interval = z * SE = 24.4995
Lower Limit of the confidence interval = x-bar - width = 325.5005
Upper Limit of the confidence interval = x-bar + width = 374.4995
The confidence interval is [325.50 hours, 374.50 hours]
(d) 375 hours lies outside the confidence interval in (c). This means we reject Ho. The same conclusion was reached in (a)
also. Both (a) and (c) mean the same thing.