Question

ear lata and 9. Examples in this section in cluded actual low temperatures and low temperatures that were forecast tive days earlier. Listed below are actual high temperatures and the high temperatures that were forecast one day earlier (based on data recorded near the author's home). Assume normality Actual High High Forecast one day earlier 85 73 80 81 85 7to test the claim of a a. Use a 0.05 significance level 78 75 81 zero mean difference between the actual high temperatures and the high temperatures that were orecast one day earlier. What do the results suggest about the accuracy of the forecast temperatures? b, construct a 95% confidence interval estimate of the mean difference between actual high temperatures and high temperatures forecast one day earlier. Interpret the resulting confidence interval

Answer 1

(a)

Actual high (X)              High forecast (Y)                    d = X - Y

80                                       78                                    2

77                                      75                                     2

81                                      81                                      0

85                                    85                                        0

73                                     76                                     - 3

From the d values, the following statistics are calculated:

n = 5

$\bar{d}$ = 1/5 = 0.2

s_d = 2.0494

SE = s_d/$\sqrt{n}$

= 2.0494/$\sqrt{5}$ = 0.9165

Test statistic is:

t = $\bar{d}$ /SE

= 0.2/0.9165 = 0.2182

ndf = 5 - 1 = 4

$\alpha$ = 0.05

From Table, critical values of t = $\pm$ 2.7764

Since the calculated value of t = 0.2182 is less than critical value of t = 2.7764, Fail to reject H0.

Conclusion:

The data support the claim a zero mean difference between the actual high temperatures and the high temperatures that were forecast one day earlier. The results suggest that the forecast is accurate.

(b) Confidence interval:

$\bar{d}$$\pm$ t SE

= 0.2 $\pm$ ( 2.7764 X 0.2182)

= 0.2 $\pm$ 0.6059

= ( -0.4059, 0.8059)

Interpretation:

If repeated samples are taken and 95% confidence interval is calculated for each sample, 95% of the intervals will contain the population mean difference of zero.