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Chapter 9 – Modeling Breaking Strength with Dichotomous DataPowerPoint Presentation

Chapter 9 – Modeling Breaking Strength with Dichotomous Data

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Chapter 9 – Modeling Breaking Strength with Dichotomous Data

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Chapter 9 – Modeling Breaking Strength with Dichotomous Data

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You are a statistician working for the Cry Your Eyes Out Tissue Company.

The company wants to investigate the breaking strength distribution of

tissues produced on an experimental production line. Due to limited

funds, the equipment to make precise measurements of breaking strength

is unavailable. Hence, you will perform a dichotomous data experiment to

Investigate the breaking strength distribution, analyze the data, and make

A formal report. Based on past studies, it is believed that the breaking

Strengths may be well modeled by a logistic distribution. The logistic

distribution function is , for - < x < +, for some

constants and > 0.

The p.d.f. of the logistic distribution is .

The mean and variance of the distribution are and .

A graph of the p.d.f. is shown below for = -9 and = 1.5.

The measurement will involve determining whether a 1-ounce, egg-shaped

fishing weight dropped from a specific height falls through a two-ply tissue

that is clamped in a 7-inch embroidery hoop. The hoop is elevated on

three 12-ounce soft drink cans. Steps are given below:

1. Separate the inner and outer hoops and gently stretch the tissue over

the inner hoop. Clamp the outer hoop around the inner hoop, making

sure that the tissue remains taut over the inner hoop, and place the hoop

on top of the three cans. If the tissue tears during the process, discard it.

2. Place a ruler perpendicular to the surface of the tissue, and place the

weight next to the ruler so that the bottom of the weight is at a

predetermined distance from the tissue. Release the weight. If the

weight falls through, the measured value is “Yes”; otherwise, “No”.

3. Replace the tissue after each trial, whether or not the tissue tears.

4. Since we must have at least 2 distances to do our parameter

estimation, we will use the distances 1 cm, 2 cm, 3 cm, and 4 cm, with

16 trials (tissues) at each distance. The data will be the number of broken

tissues at each distance.

5. If we don’t get at least one tissue tear at a certain distance, then we

must add a column for drops from 5 cm. We must have at least two

data columns with both “Yes”’s and “No”’s in them.

Parameter estimation will be done using the maximum likelihood method.

The log-likelihood function for this experiment is

,

where Yi = number of breaks at distance xi.

Differentiating with respect to each parameter and setting each derivative

equal to 0, we get a nonlinear system of two equations in two unknowns:

, and

.

These two equations may be solved numerically, using Newton’s method,

provided we have a set of initial approximate values for and . These

initial values may be found by substituting for in the equation

. Once we have these initial values, Newton’s

method involves iteratively solving the equation

to find the M.L.E.’s. We stop the iterative procedure when

and are less than some criterion value, such as 0.0000001.

Generally, the M.L.E.’s will be close to the initial values calculated above.

We may use SAS PROC LOGISTIC to do the estimation of the parameters.

Sample SAS code is given next.

data one;

input x n y;

cards;

1 16 2

2 16 4

3 16 6

4 16 13

proc logistic;

model y / n = x;

run;

After obtaining the M.L.E.’s for the parameters, we may assess the fit of

the data to a logistic distribution by two general methods – graphical and

inferential.

1)Graphical: We substitute the estimates into the formula for the c.d.f.

and plot the result against distance, overlaying the points for the observed

failure proportions. These points should lie close to the graph of the c.d.f.

2)Inferential: We want to test the null hypothesis that the c.d.f. has a

logistic form against the alternative that it does not. The test statistic will

be a likelihood ratio test statistic,

,

where . Under the null hypothesis, this test statistic has an

approximate chi-square distribution with d.f. = k-2.