ABSTRACT
In this article, we discuss a real-life problem and its statistical solution via an interesting application of sequential analysis. First we describe the scenario. Early in the winter, one of the authors experienced problems with his home-heating system and thought that it might need replacement. At a local plumbing, heating and air-conditioning superstore, he was advised to look into the possibility that his homeinsulation might not be functioning properly, particularly since the house was more than three decades old. Asked what might be a good indicator of the effectiveness of insulation, the store personnel suggested that one should check whether the indoor temperature was being affected significantly by the outdoor temperature. In other words, if the indoor temperature on a snowy and blustery day was significantly lower than that on a warmer day after the heating system stayed on for the same duration, starting from the same base temperature, it might mean
that the house needed better insulation. The author was intrigued and thus began the data-collection phase of the present study. Over a period of nearly three months, both indoor and outdoor temperatures (0F) were recorded once or twice a day with a handheld thermometer. Before recording the indoor temperature, the heating system would be turned off until the room-temperature dropped to 68°F (20°C) and then the heating system would be turned back on and kept running for nearly 3 hours. Meanwhile, the outdoor temperature at the midpoint of a three-hour interval would be measured using the same thermometer. Here is the resulting dataset:
Table 8.1.1: The Temperature Dataset
In Out In Out In Out In Out In Out
76.0 37.9 76.0 40.5 75.0 34.0 72.0 39.0 74.2 29.2 79.5 40.5 73.0 42.0 78.0 39.5 76.0 36.4 75.3 31.8 77.0 35.0 71.5 29.0 69.0 26.0 73.5 37.0 76.5 33.1 73.5 34.0 77.8 37.0 74.0 35.0 75.6 32.3 74.5 30.5 71.0 26.5 75.6 38.5 77.7 37.0 74.9 39.0 72.8 28.8 75.5 36.5 71.0 26.4 76.5 35.8 69.5 24.5 72.6 28.5 71.8 22.0 76.0 32.0 73.5 28.0 74.2 31.5 75.0 30.1 69.5 22.0 73.0 28.0 71.0 28.0 74.0 30.0 75.6 30.8 74.5 37.0 79.6 44.5 75.0 39.1 78.5 38.5 76.2 32.0 75.0 34.0 72.3 36.2 72.2 24.0 72.2 30.5 75.2 30.5 71.5 28.0 74.0 33.5 72.2 26.5 70.5 23.5 74.3 29.1 78.0 40.5 71.0 25.0 72.8 29.5 71.5 26.5 72.5 27.8 77.2 34.5 70.3 25.7 71.8 29.0 73.0 28.6. 72.8 28.5 69.7 25.0 69.5 23.2 73.1 26.5 75.0 28.0 73.6 29.0 75.8 30.4 75.5 30.0 76.6 32.0 73.4 29.2 73.4 28.7 72.8 28.0 73.0 28.0 74.8 29.5 75.3 30.8 76.1 32.2 75.9 32.1 75.9 32.0 76.0 33.5 77.0 34.4 76.5 32.8 76.8 33.5 77.0 34.0 74.5 31.7 77.6 35.1 78.0 37.1 74.8 34.1 74.1 33.6 73.8 33.0 73.8 32.8 73.4 32.2
Each day, temperatures were recorded either in the morning (outdoor around 7:30 A.M., indoor around 9:00 A.M.) or in the afternoon (outdoor around 3:00 P.M., indoor around 4:30 P.M.). Obviously, what we had in mind was a regression with the indoor temperature as the response variable (Y) and the outdoor temperature as the regressor (X). Under a regression setup, we might want to test a hypothesis that the slope is not zero. However, a confidence interval for the slope would
be more useful. Such a confidence interval would serve not only as a tool for carrying out the hypothesis test, it would actually give us an idea as to how steeply the indoor temperature rises or falls with a unit change in the outdoor temperature. For example, if a 95% confidence interval for the slope turned out to be (3.0°F, 4.5°F), it would be a more serious concern to a homeowner than a 95% interval such as (0.3°F, 1.3°F). So, we decided to construct a confidence interval rather than performing a hypothesis test. In any case, both variables X and Y were measured with errors, since the thermometer used was not digital and was calibrated only to show whole-number temperature readings. That is, a temperature reading of, say, 79.5°F is only an eyeball approximation. So, a simple linear regression model with errors in variables (EIV) would be more appropriate rather than an ordinary regression model.
