The temperature outside an airliner was -50*F. This was 110* less than the temperature on the ground. What was the temperature o
2 answers:
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Answer:
a) The calculated value t = 5.903 > 2.572 at 0.01 level of significance
Null hypothesis is rejected at 0.01 level of significance
There is a true mean value is less than 10
b) p - value 0.00001
The p - value 0.00001 < 0.01
The result is significant at p<0.01
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given mean of the population = 10
Sample size 'n'= 35
Mean of the sample x⁻ = 14.44
Standard deviation of the sample 's' = 4.45
Level of significance ∝ = 0.01
<u><em>Step(ii)</em></u>:-
<em>Null Hypothesis H₀</em> : μ >10
<em>Alternative Hypothesis</em> : H₁ : μ < 10
Test statistic
t = 5.903
Degrees of freedom = n-1 = 35-1 = 34
tabulated value t₀.₀₁,₃₄ = 2.572
The calculated value t = 5.903 > 2.572 at 0.01 level of significance
Null hypothesis is rejected at 0.01 level of significance
There is a true mean value is less than 10
<u><em>p- value</em></u> :
p - value 0.00001
The p - value 0.00001 < 0.01
The result is significant at p<0.01