business statistics report midterm project

Utilize appropriate descriptive statistics to summarize the data on transmissionfailures.Data  Data Analysis  Descriptive Statistics  Input range: A2 to A51 Choose Summary Statistics

NATIONAL ECONOMICS UNIVERSITYBUSINESS SCHOOL - E-BBA & E-BDB PROGRAM

Problem 1: Metropolitan Research, Inc.

Metropolitan Research, Inc is a consumer research organization that conductssurveys to assess various consumer products and services In a specific study,Metropolitan focused on gauging consumer satisfaction regarding the performanceof automobiles manufactured by a major Detroit-based company They distributeda questionnaire to owners of the manufacturer's full-sized cars and discoverednumerous complaints related to early transmission problems In order to delvedeeper into these transmission failures, Metropolitan obtained a sample of actualtransmission repair records from a Detroit-based transmission repair firm Thefollowing data reveals the actual number of miles driven by 50 vehicles at the timeof experiencing transmission failure.

1 Utilize appropriate descriptive statistics to summarize the data on transmissionfailures.

Data  Data Analysis  Descriptive Statistics  Input range: A2 to A51 Choose Summary Statistics

And then we have this table:

2 Construct a 95% confidence interval for the mean number of miles driven untiltransmission failure for the entire population of vehicles with transmission issues.Provide a managerial explanation of what this interval estimate implies.

Data  Data Analysis  Descriptive Statistics  Input range: A2 to A51 Choose Summary Statistics and Confidence for Mean (95%)

So, we have this table:

Confidence Level = 95%So α = 0.05

Sum of both sides = 0.05 so each side of the normal distribution curve equals 0.025Normal distribution inverse = 1.96

Confidence Value = CONFIDENCE.NORM(0,05;24898,71511;50) = 6901,445Lower Confidence Limit = Mean – Confidence Value = 66438.8551

Upper Confidence Limit = Mean + Confidence Value = 80241.7449

The expected transmission failure of the sample is between the range 66438.8551to 80241.7449 Below 66438.8551 miles, it’s not acceptable Since the interval isquite small, we can say that it is not severe issue and failures are normal in nature.We suggest that company should give warranty-compensation to those having earlyfailures.

3 Analyze the implications of your statistical findings in relation to the belief thatsome car owners encountered premature transmission failures.

Lowest: 25066Mean: 73340.3Highest: 138114

Considering the lowest interval range at 95% confidence level, it shows that carsexperiencing transmission failure before 66438.8551 miles shows prematurefailure, the first quartile is 60421 miles which shows that 25% of the repairs (12 of50 cars) were done before 60000 miles.

4 Determine the required number of repair records to be sampled if the researchorganization aims to estimate the population mean number of miles driven untiltransmission failure with a margin of error of 5000 miles, using a 95% confidencelevel.

α = 0.05 so Z(α/2) = Z(0.025) = 1.96ME = Z(0.025) x = 5000 = 1.96 x = 5000 = 2551.0204

= 24898.71511

So N = 95.2634 ≈ 96 repair records should be sampled.

5 Identify any additional information you would like to collect to comprehensivelyassess the problem of transmission failures.

 Collect information about the type of mechanical failures that the vehiclepresents simultaneously with the transmisson.

 Levels of transmission failures throughout the automotive industry isrelative to miles/time of use.

 Collect the information based on larger samples which is a morerepresentative group of the automobile population.

 When providing mileage data before the transmission failure occurs, add thetime the vehicle has been in operation.

Problem 2 Par, Inc.

Par, Inc is a prominent golf equipment manufacturer Par's management believesthat increasing their market share is possible by introducing a golf ball that is bothcut-resistant and longer-lasting Consequently, the research team at Par has beeninvestigating a novel golf ball coating that can withstand cuts and enhancedurability Initial tests on this coating have yielded positive results However, oneof the researchers expressed concerns about how the new coating might affect thedistance the golf ball can travel when hit Par's objective is to ensure that the newcut-resistant ball provides driving distances on par with their current golf ballmodel To make a comparative assessment of the driving distances between thetwo types of golf balls, they conducted distance tests on 40 balls of both the newand existing models These tests were carried out using a mechanical hittingmachine, ensuring that any difference in mean distances between the two modelscan be attributed to the models themselves The results of these tests, withdistances recorded to the nearest yard, are accessible on the textbook'saccompanying website.

1 Develop and present the rationale for a hypothesis test that Par could employ tocompare the driving distances of the current and new golf balls.

The result of the test on the durability of the improved product has been recordedand this is the effect of the new coating on driving distances 40 balls of both thenew and current models were used for the distance test They are independentsamples, and the test follows a large sample case By formulation of thesehypothesis, there is assumed that the new and current golf balls show no significantdifference to each other.

The null hypothesis and alternative hypothesis are formulated as follows:µ1: Mean distance of current-model balls

µ2: Mean distance of new cut-resistant balls

H0: µ = µ (Mean distance of current balls equals mean distance of new balls)12

H1: µ ≠ µ (Mean distance of current balls is not equal mean distance of new balls)12

This is a two-tails test.

2 Analyze the data to arrive at the conclusion of the hypothesis test What is the value for your test? What recommendation would you make to Par, Inc.?

p-Specify the level of significance, α = 0.05 so Z = 1.96

Our rejection criteria is Reject H and accept H if P < α We use the P-value01

approach.

Looking at the descriptive statistics for each model, we can initially conclude thatthe Current model has a longer range of distance based on the 40 samples with amean of 270.275 compared to 267.5 for the New one Besides, the standarddeviation of Current is 8.75 and of New is 9.89 Although the two values of meanare different, Sd is quite large so we can say that they are quite the same.

P-value = 0.1879 > 0.05 = α.Our decision rule for this problem is:

 Do not reject H0

 Mean distance of current balls equals mean distance of new balls

 The new cut-resistant balls have no different in distance compared to thecurrent-model one

Therefore, we recommend to Par, Inc that they should not launch this new productbecause the new model is not an improvement in distance compared to the currentone.

3 Provide descriptive statistical summaries for the data pertaining to each golfball model.

Data  Data Analysis  Descriptive Statistics  Input range: A2 to B41 Choose Summary Statistics and Confidence for Mean (95%)

Then we have this table:

4 What are the 95% confidence intervals for the population mean drivingdistances of each model, and what is the 95% confidence interval for the differencebetween the means of the two populations?

α = 0.05

Mean difference = 270.275 – 267.5 = 2.775Pooled Variance = 87.282

Pooled Standard Deviation = 9.342

t-value (two tails) = 1.99 ME = 4.159

 95% Confidence Interval Lower = -1.384 95% Confidence Interval Upper = 6.934

The 95% Confidence Interval for the difference between the means of the twopopulations is (-1.384;6.934).

5 Discuss whether you believe there is a need for larger sample sizes and moretesting with the golf balls.

The larger the sample size, the smaller the standard deviations which means pointestimator of mean will become more precise Hence, there is no need to take largersample size.