Download 1.) The Target Corporation personnel director wants to estimate the

1.) The Target Corporation personnel director wants to estimate the number of employees within one
year of retirement. A random sample of 120 employee records is selected, and 36 people are found to
be within one year of retirement. Formulate an 84% interval estimate for the true proportion of
employees within one year of retirement in the entire corporation.
Observed proportion = 36/120 = 0.3
Standard error = sqrt(0.3*(1-0.3)/120) = 0.041833
For 84% confidence level, z-score = 1.4051
lower limit = 0.3 - 1.4051*0.041833 = 0.24122
upper limit = 0.3+1.4051*0.041833 = 0.35878
So, confidence interval for true proportion is (0.24122, 0.35878)
2) As a quality control expert, you want to estimate the mean thickness of optical lenses produced by
your firm. A sample of 144 lenses reveals a mean of 0.52 millimeters (mm). The population standard
deviation is known to be 0.17 mm. Construct a 95% confidence interval.
Standard error = 0.17/sqrt(144) = 0.014167
For 95% confidence level, z-score = 1.96
lower limit = 0.52 - 1.96*0.01467 = 0.49125
upper limit = 0.52 + 1.96*0.01467 = 0.54875
So, confidence interval is (0.49125, 0.54875)
3) As the new manager for a CPA firm, you want to establish the confidence interval for the time to
complete a medium sized organization's tax return. A sample of 25 medium corporate accounts is
selected at random. The sample mean and standard deviation are 20 days and 4 days, respectively.
Using a confidence interval of 90%, construct the interval.
Standard error = 4/sqrt(25) = 0.8
degree of freedom = 25-1 = 24
For 90% confidence level, t-score = 1.7109
lower limit = 20 - 1.7109*0.8 = 18.6313
upper limit = 20+1.7109*0.8 = 21.3687
So, confidence interval is (18.6313, 21.3687)
4) A manufacturer of detergent claims that the mean weight of a particular box of detergent is at least
3.25 pounds. A random sample of 81 boxes reveals a sample average of 3.18 pounds and a sample
standard deviation of 0.15 pounds. Using a 0.10 level of significance, is there evidence that the
average weight of the boxes is different from at least 3.25 pounds as stated above?
Ho: Average weight is at least 3.25 pounds. (µ≥3.25)
Ha: Average weight is less than 3.25 pounds. (µ<3.25)
t-statistic = (3.18 - 3.25)/(0.15/sqrt(81)) = -4.2
critical region: t<-1.2922
Decision: Null hypothesis is rejected as test statistic lies in critical region.
There is sufficient evidence that the average weight of the boxes is different from at least 3.25 pounds
as claimed.
5) A package-filling device is set to fill cereal boxes with a mean weight of 20 ounces of cereal per box.
The population standard deviation for the filling machine is 0.5 ounces. A random sample of 25 filled
boxes is weighed, yielding a mean weight of 20.27 ounces. Test at the 0.05 significance level to
determine whether the device is working properly.
Ho: Mean weight of cereal per box = 20 ounces
Ha: Mean weight of cereal per box ≠ 20 ounces
z-statistic = (20.27 - 20)/(0.5/sqrt(25)) = 2.7
critical region: z<-1.96 or z>1.96
Decision: Null hypothesis is rejected since test statistic lies in critical region.
The device isn’t working properly as mean weight of cereal boxes is different from 20 ounces.
6) The manager of the credit department for an oil company would like to determine whether the
average monthly balance of credit card holders is greater than $75. An auditor selects a random
sample of 100 accounts and finds that the average owed is $83.40 with a population standard
deviation of $23.65. Using the 0.05 level of significance, should the auditor conclude that there is
evidence that the average balance is greater than $75?
Ho: Average balance in less than or equal to $75. (µ≤75)
Ha: Average balance is greater than $75. (µ>75)
z-statistic = (83.40-75)/(23.65/sqrt(100)) = 3.5518
critical region = z>1.6449
Decision: Null hypothesis is rejected because test statistic lies in critical region.
There is sufficient evidence that the average balance is greater than $75.
7. Linear Regression
Below is the printout for a regression of crop yield on rainfall, fertilizer, and soil acidity.
Predictor B CoefStdevBeta
Constant 3.3
Rain 0.230.1588 0.2508
Fertilizer 1.150.2772 0.7714
Acid-0.1130.1093-0.0935
S= 6.4987
r-sq = 0.9283
r-sq (adj) = 0.9087
Using the computer results for the regression analysis answer the following:
a.Write the equation to describe the model.
Crop yield = 3.3 + 0.23*Rain + 1.15*Fertilizer - 0.113*Acid
b.Explain the relationships between fertilizer and crop yield.
There is positive relationship between fertilizer and crop yield. If fertilizer is increased by 1 units, crop
yield increases by 1.15 units.
c.Is the relationship between soil acidity a direct or inverse relationship?
The relationship between soil acidity and crop yield is inverse.
d.Is this a model a good fit and justify your response?
This model is a good fit because r-squared value is 0.9283. This means about 92.83% of variation of crop
yield is explained by the predictors.
e.What is the adjusted coefficient of determination and explain what it means
Adjusted coefficient of determination is 0.9087. This is the R-squared value after adjusting for the
upward bias resulted from addition of multiple predictors in the models.