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A confidence interval in linear regression provides a range of values within which we can be certain to a specified probability that the true value of the dependent variable lies given a certain value of the independent variable. In the context of the provided exercise, we computed a 99% confidence interval for the estimated strawberry texture at a coded storage temperature of -1. The resulting interval from 0.35 to 5.65 implies that we can be 99% confident that the expected texture score would fall in this range, assuming that our model is correct and that the underlying assumptions of linear regression are met.

Constructing a confidence interval involves determining the standard error of the estimate and then using a critical value from a distribution (like the t-distribution for small sample sizes) to set the bounds of the interval. The wider the interval, the less precise the estimate, but it increases our confidence in capturing the true parameter.

A prediction interval, on the other hand, quantifies the uncertainty around the prediction of a new observation. It is usually wider than a confidence interval because it has to account for both the error in estimating the underlying regression line (just like the confidence interval) and the variability of the individual observations around that line. The exercise demonstrates this by predicting the particular value of strawberry texture, at a storage temperature of 1, along with a 99% prediction interval, which ranges from -1.57 to 3.57.

To calculate the prediction interval, one would find the standard error of the prediction and then incorporate an additional variability term to reflect the spread of individual observations. This prediction interval provides a range that you would expect to contain the actual observed value of a new data point with a chosen level of confidence (99% in the exercise).

The least squares method is a foundational mathematical approach used in linear regression to estimate the line of best fit for a set of data points. The main goal is to minimize the sum of the squared differences between the observed values and the values predicted by the line. In the exercise, the least squares method was applied to calculate the regression coefficients — namely, the slope and the intercept of the regression line.

By minimizing the sum of the squares of the vertical deviations of the points from the line, we obtain the 'least squares' estimates of the regression line. This method is advantageous because it is computationally straightforward and, under common assumptions, gives the best unbiased estimates of the coefficients.

Regression coefficients are numerical values that represent the relationship between the independent variable(s) and the dependent variable in the regression equation. In the provided exercise, we calculated the slope (b) and intercept (a) as the regression coefficients, using the least squares method. The slope indicates the change in the dependent variable (texture score) for a one-unit change in the independent variable (storage temperature). The intercept represents the value of the dependent variable when the independent variable is zero. For this exercise, the regression line equation came out to be \( y = 2 - x \), where 2 is the intercept and -1 is the slope.

In the context of the problem, the intercept coefficient indicates the estimated texture score when the storage temperature is at a baseline or coded zero level, and the slope tells us how the texture score changes for each degree change in the storage temperature.