Do Predicted Temperatures match Actual Temperatures?
The table shows the predicted and actual temperatures for a seven day period. Assume that the two samples are randomly selected. At the 0.1 significance level, test the claim that there is no mean difference in the predicted and actual temperatures.
(Be sure to subtract in the same direction).
Predicted Temperatures ⁰F
|
Actual Temperatures ⁰F
|
Difference ⁰F |
81 |
76 |
|
75 |
79 |
|
86 |
83 |
|
85 |
90 |
|
82 |
82 |
|
77 |
78 |
|
78 |
80 |
|
What are the correct hypotheses? (Select the correct symbols and values.)
H0: Select an answer p μ₁ s₁² μ(d) σ₁² x̄₂ μ x̄₁ ? < = > ≥ ≠ ≤ Select an answer s₁² μ 0 σ₁² μ₂ μ₁ x̄₁ p x̄₂
H1: Select an answer μ₂ x̄₁ x̄₂ p μ s₂² μ₁ μ(d) σ₂² ? ≤ ≠ > ≥ < = select an="">
Original Claim = Select an answer H₁ H₀
df =
Based on the hypotheses, find the following:
Test Statistic =(Round to three decimal places.)
Critical value(s) = ±±(Round to three decimal places.)
p-value =(Round to four decimal places.)
Shade the sampling distribution curve with the correct critical value(s) and shade the critical regions. The arrows can only be dragged tot-scores that are accurate to 1 place after the decimal point (these values correspond to the tick marks on the horizontal axis). Select from the drop down menu to shade to the left, to the right, between or left and right of the t-score(s).
Shade: Left of a valueRight of a valueBetween two values2 regions. Click and drag the arrows to adjust the values.
Decision: Select an answer Accept the null hypothesis Accept the alternative hypothesis Fail to reject the null hypothesis Reject the null hypothesis .
Conclusion: Select an answer There is not enough evidence to support There is sufficient evidence to warrant rejection of The sample data supports There is not sufficient evidence to warrant rejection of the claim that there is no mean difference in the predicted and actual temperatures.