# Business Statistics

BUSINESS STATISTICS 6

BusinessStatistics

BusinessStatistics

Question1

Z-scoreis the measure of standard deviations of an observation from the meanscore.

Itmeasures the relative distance of observations (x) in a particulardata set, based on the average value (µ) generated from the samedata(Berenson, Levine, Szabat, & Krehbiel, 2012).Z-score in business assists in identifying outliers, to make properfinancial decisions. The formula for computing the Z-score is Z=(×-µ)/∂.

Acar manufacturing company plans to move the production of car partsfrom the central city to its suburb. The cost of the manufacturingplant in the city is $ 1,000,000. Given that the average price of theplant in the town is $ 1,325,000, with a standard deviation of $300,000, and the mean cost of a plant in the suburbs is $ 740,000,with a standard deviation of $ 200,000.

TheZ-score of the city is Z= (1,000,000-1,325,000)/300,000= -1.08

Therefore,the cost of the plant sold in the city is 1.08 standard deviationbelow the average price, implying the company sells the plant at alower price than the average city price.

Ifthe company purchases the plant in the suburbs at $ 1,000,000, thesuburb Z-score is (1,000,000-740,000)/200,000= 1.3. Thus, the priceof the manufacturing plant is 1.3 standard deviations above theaverage price, indicating that the company can buy a plant in thesuburbs at a relatively lower cost to match the mean cost, making itsoperations cheaper than in the city.

Question2

Populationproportion is derived from a sample size, which assists in estimatingthe exact quantity of observations from a data set of interest. Theconfidence interval accommodates the margin of error to enhanceaccurate estimation of the items in a population. A confidenceinterval of 95% implies that the chances of the actual value of apopulation in the range are 95%. Thus, a higher confidence level iscommendable since it increases the certainty that the interval hasthe correct proportion.

Π=population proportion

*p=X/n, X*being the number of items in the sample, and n the sample sizewith a particular characteristic sought for (Berenson, Levine,Szabat, & Krehbiel, 2012).

Π=p± Z √ *p*(1-*p*)/n= p – Z √ *p*(1-*p*)/n≤ π ≤ p + Z √ *p*(1-*p*)/nis the confidence interval for the population proportion. Zrepresents the critical value in a normal distribution (Berenson,Levine, Szabat, & Krehbiel, 2012)

Ifin a sample of 250 manufactured switches, 50 are faulty, one canestimate the proportion of the switches made that are faulty.

*P=*50/250=0.2

Using95% confidence, Z= 1.96. Thus, π= 0.20± (1.96) √(0.2*0.8)/250=0.20± (1.96*0.0252)

0.249≤π≤0.151.Thus, a 95% confidence that between 24.9%, and 15.1% of allmanufactured switches are faulty.

Question3

Itis rare for the sample mean to equate the population mean because itis a point estimation. The characteristics of samples from the samepopulation vary, and a question that arises is the accuracy of thesample mean in estimating the population mean(Berenson, Levine, Szabat, & Krehbiel, 2012).Since the population mean (µ) is unknown, and so is its standarddeviation (∂), the values of the sample mean are used to estimatethe interval in which the true population mean falls.

µ=_{}±Z(∂/√n) = _{}–Z (∂/√n) ≤ µ≤ _{}+ Z (∂/√n) becomes the estimated population mean interval, while_{}is the sample mean, and ∂ the sample’s standard deviation (Berenson, Levine, Szabat, & Krehbiel, 2012).

Forexample, a manufacturing power plant produces electric cables with anexpected average width of 8 mm. If in one occasion, the line managertests the conductivity of a sample of cables and notices that thereis a variance, it means that there is a disparity in their thickness,given a standard deviation of 0.2 mm. As such, the manager takes asample of 200 electric cables, with a mean of 8.5 mm to determine anyproblem in the manufacturing process by assessing if the width of thewires is 8 mm. To do this, an estimate of population mean iscalculated and using a confidence interval of 95%, µ= 8.5± (1.96)(0.2/√200) = 8.5±0.0277.

Thus8.5277≤ µ≤ 8.4723 and since 8 is outside the range of theinterval of the estimated population mean, the cables’ aredefective.

Question4

Thesample variance is an inferential point in estimating the unknownpopulation variance. Although the sample variance represents thepopulation, there is a minimal probability that the populationvariance will have the same value as the sample. Hence, the use of aninterval in determining the range within which the true populationvariance lies.

s^{2}=∑(x_{i}–_{})^{2}/n-1is the sample variance, but for purposes of estimating the populationvariance, a sampling distribution of (n-1)s^{2}/∂^{2}isused(Berenson, Levine, Szabat, & Krehbiel, 2012).n-1 is the degrees of freedom and the sample follows a chi-squaredistribution. Therefore, the interval estimate population variance is(n-1)s^{2}/*X*^{2}_{α/2}≤∂^{2}≤(n-1)s^{2}/*X*^{2}_{1-α/2}(Berenson,Levine, Szabat, & Krehbiel, 2012)_{.}

Thestock market uses an estimate of the population variance to measurethe volatility of the market of how the stock prices deviate from theaverage price.

Ifa sample of 20 commodities is taken from the stock exchange with avariance of 0.005, a population variance can be estimated using 95%confidence interval. The chi-square distribution is two-tail, and α_{2}is0.05.

19*0.005^{}/32.852≤ ∂^{2}≤19*0.005^{}/8.907

0.00289≤ ∂^{2}≤0.0107.

Thesample variance (0.005) is within the range of the estimatedpopulation variance with a confidence level of 95%. It presents atrue magnitude of the stock volatility to the traders.

Reference

Berenson,M., Levine, D., Szabat, K. A., & Krehbiel, T. C. (2012). *Basic: Concepts and Applications.*Pearson Higher Education AU.

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