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 UNIVERSITY

BUSINESS SCHOOL - E-BBA & E-BDB PROGRAM

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BUSINESS STATISTICS

Report Midterm Project

Lecturer: Vuong Van Yen Subject: Business Statistics Class: EBBA 14.1 GROUP:

Trần Hoàng Ngọc Minh – 11224312

Ngô Thu Thảo – 11225901 Trần Thanh Lam – 11223210 Phan Linh Trang – 11226499

Tạ Công Vũ – 11226965 Đặng Quốc Huy – 11222759

Problem 1: Metropolitan Research, Inc.

Metropolitan Research, Inc is a consumer research organization that conducts surveys to assess various consumer products and services In a specific study, Metropolitan focused on gauging consumer satisfaction regarding the performance

of automobiles manufactured by a major Detroit-based company They distributed

a questionnaire to owners of the manufacturer's full-sized cars and discovered numerous complaints related to early transmission problems In order to delve deeper into these transmission failures, Metropolitan obtained a sample of actual transmission repair records from a Detroit-based transmission repair firm The following data reveals the actual number of miles driven by 50 vehicles at the time

of experiencing transmission failure

1 Utilize appropriate descriptive statistics to summarize the data on transmission failures.

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 until transmission 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 0.025 Normal distribution inverse = 1.96

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

to 80241.7449 Below 66438.8551 miles, it’s not acceptable Since the interval is quite 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 early failures

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

Lowest: 25066

Mean: 73340.3

Highest: 138114

Considering the lowest interval range at 95% confidence level, it shows that cars experiencing transmission failure before 66438.8551 miles shows premature failure, the first quartile is 60421 miles which shows that 25% of the repairs (12 of

50 cars) were done before 60000 miles

4 Determine the required number of repair records to be sampled if the research organization aims to estimate the population mean number of miles driven until transmission failure with a margin of error of 5000 miles, using a 95% confidence level.

α = 0.05 so Z(α/2) = Z(0.025) = 1.96

ME = 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 comprehensively assess the problem of transmission failures.

 Collect information about the type of mechanical failures that the vehicle presents simultaneously with the transmisson

 Levels of transmission failures throughout the automotive industry is relative to miles/time of use

 Collect the information based on larger samples which is a more representative group of the automobile population

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

Problem 2 Par, Inc.

Par, Inc is a prominent golf equipment manufacturer Par's management believes that increasing their market share is possible by introducing a golf ball that is both cut-resistant and longer-lasting Consequently, the research team at Par has been investigating a novel golf ball coating that can withstand cuts and enhance durability Initial tests on this coating have yielded positive results However, one

of the researchers expressed concerns about how the new coating might affect the distance the golf ball can travel when hit Par's objective is to ensure that the new cut-resistant ball provides driving distances on par with their current golf ball model To make a comparative assessment of the driving distances between the two types of golf balls, they conducted distance tests on 40 balls of both the new and existing models These tests were carried out using a mechanical hitting machine, ensuring that any difference in mean distances between the two models can be attributed to the models themselves The results of these tests, with distances recorded to the nearest yard, are accessible on the textbook's accompanying website

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

The result of the test on the durability of the improved product has been recorded and this is the effect of the new coating on driving distances 40 balls of both the new and current models were used for the distance test They are independent samples, and the test follows a large sample case By formulation of these hypothesis, there is assumed that the new and current golf balls show no significant difference 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)1 2

H1: µ ≠ µ (Mean distance of current balls is not equal mean distance of new balls)1 2

This is a two-tails test

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

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-value0 1

approach

Looking at the descriptive statistics for each model, we can initially conclude that the Current model has a longer range of distance based on the 40 samples with a mean of 270.275 compared to 267.5 for the New one Besides, the standard deviation of Current is 8.75 and of New is 9.89 Although the two values of mean are 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 the current-model one

Therefore, we recommend to Par, Inc that they should not launch this new product because the new model is not an improvement in distance compared to the current one

3 Provide descriptive statistical summaries for the data pertaining to each golf ball 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 driving distances of each model, and what is the 95% confidence interval for the difference between the means of the two populations?

α = 0.05

Mean difference = 270.275 – 267.5 = 2.775

Pooled 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 two populations is (-1.384;6.934)

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

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