Suppose Joy, an owner of a frozen yogurt store, wants to understand the relation

Question

Suppose Joy, an owner of a frozen yogurt store, wants to understand the relationship between temperature and frozen yogurt sales. She records the highest daily temperature, in degrees Celsius (°C), and the daily amount of frozen yogurt sold, in kg, for 30 randomly selected days. The least-squares regression equation for her data is y =5.67135x+175.18204 where x represents the highest daily temperature (°C), and y represents the predicted amount of frozen yogurt sold (kg) on a day when the highest daily temperature is x.

Suppose Joy, an owner of a frozen yogurt store, wants to understand the relationship between temperature and frozen yogurt sales. She records the highest daily temperature, in degrees Celsius (C), and the daily amount of frozen yogurt sold, in kg, for 30 randomly selected days. The least-squares regression equation for her data is y = 5.671 35x + 1 75·18204 where x represents the highest daily temperature (°C), and y represents the predicted amount of frozen yogurt sold (kg) on a day when the highest daily temperature is x. Summary statistics for the model are provided in the table. Standard error of prediction SEy 6.07970 Standard error of mean response Sample size 30 1.14639 Assuming that the requirements for a least-squares regression model have been met, compute the predicted amount of frozen yogurt sold in a day when the temperature is 27 °C. Provide your answer precise to three decimal places. Avoid rounding until the final step kg

Explanation / Answer

The predicted amount of yoghurt sold in a day when highest temperature is 27 degrees celsius can be computed by using the regression equation given above(very first equation) that connects prediction of amount of yoghurt sold to the highest daily temperature of the day. So, for the highest daily temperature of 27 degrees celsius, the predicted amount is 0 = 5.67135*(27) + 175.18204 = 328.30849 = 328.309 (three decimal places) kilograms of yoghurt.

To calculate the critical value, refer to the t-distribution where to choose a value, two parameters are needed, interval percentage (which in our case is 95 %) and the degrees of freedom. For a 1-sample t-test, one degree of freedom is spent estimating the mean, and the remaining n - 1 degrees of freedom estimate variability. Since, we have 30 data points used for regression, 29 is the number of degrees of freedom and so, the critical t value is 2.045.

Computing the upper and lower bounds for the highest daily tempearture of 27 degrees are very straightforward. th formula for bounds are = 0 +(or)- tcrit * s.e, where s.e. is the standard error of prediction(which is provided in the question as 6.0797). So, upper bound for 95% prediction interval is 328.309 + (2.045 * 6.0797) = 340.742 and lower bound for 95% prediction interval is 328.309 - (2.045 * 6.0797) = 315.876

The correct interpretation of the 95 % prediction interval (not to be confused with confidence interval) is a range of values such that the probability is 95 % that the predicted frozen yoghurt sales amount for a randomly selected day with maximum temperature of 27 degress celsius is in that range.  

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