Chapter 9 – Modeling Breaking Strength with Dichotomous Data. 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
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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 .
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:
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
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; when
input x n y;
1 16 2
2 16 4
3 16 6
4 16 13
model y / n = x;
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
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.