Question

Q 7. A horticulturalist devised a scale to measure the freshness of roses that were packaged and stored for varying periods of time before transplanting. The freshness measure- ment y and the length of time in days that the rose is packaged and stored before transplanting r are given below. x5 10 15 20 25 V 15.3 13.6 98 55 1.8 (a) What is the best estimate of σ2, the variance of the random error e? (b) Construct the ANOVA table. (c) Estimate the expected freshness measurement for a storage time of 14 days with a 95% confidence interval.

Answer 1

All the parts are answered below:

The Regression Equation is:

SUMMARY OUTPUT Regression Statistics MultipleR R Square Adjusted R Square Standard Error Observations 0.99127233 0.982620833 0.976827777 0.852251919 5 ANOVA Significance F df MS Regression Residual Total 123.201 123.201 169.6205 0.000977489 2.179 0.726333 125.38 4 Coefficients Standard Error P-value Lower 95% Upper 95% Lower 95.0% t Stat Upper 95.0% 16.88537243 22.57462757 Intercept 19.73 0.893849353 22.07307 0.000204 22.5746275716.88537243 0.702 0.053901144-13.0238 0.000977 0.873537497 0.530462503 -0.873537497 0.530462503

Y = 19.73 - 0.702*X

(a) The Anova table is also shown above...

(b) The best estimate of the standard error is: 0.8522

(c) X = 14, so...

Y = 19.73 - 0.702*14 = 9.902

The provided sample mean is X = 9.902 and the sample standard deviation is s 0.8522. The size of the sample is n-5 and the required confidence level is 95%. The number of degrees of freedom are df-5-1-4, and the significance level is α-005. Based on the provided information, the critical t-value for a0.05 and df 4 degrees of freedom is te 2.776. The 95% confidence for the population mean μ is computed using the following expression Therefore, based on the information provided, the 95 % confidence for the population mean μ is 2.776 x 0.8622,9.902+ 2.776 x 0.8522 CI = ( 9.902 (9.902 1.058, 9.902 +1.058) = (8.844, 10.96)

Answered...