Question

(((( no handwriting, please )))

The following Table consists of five actual low temperatures and the corresponding low temperatures that were predicted five days earlier. Use a 0.05 significance level to test the claim that there is a difference between the actual low temperatures and the low temperatures that were forecasted five days earlier. Table : Actual and Forecast Temperature Actual low Low forecast 5 days earlier Difference d-actual - predicted 32 36 4 35 35 0 37 36 38 35 33 34 2 4 Given that the mean and standard deviation of d using above table are d - 1.0, sq - 3.16, Critical value of t for a 0.05 (two tail) with 4 df. is +2.776 Hint: Ho: ld=0, H1: μ 0

Answer 1

We have to test the claim that there is difference between the actual low temperature and the low temperature that were forecasted five days earlier.

Find test statistic:

Decision rule: Reject H0, if absolute t test statistic value > absolute t critical value, otherwise we fail to reject H0.

Since absolute t test statistic value = t test statistic value = 0.708 is not greater than t critical value = 2.776

thus we fail to reject H0.

Conclusion: Since we failed to reject H0, we conclude that there is no sufficient evidence to support the claim that there is difference between the actual low temperature and the low temperature that were forecasted five days earlier.