JustAnswer
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. |
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.