Please help! I will mark brianliest!!
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Answer:
The calculated chi -square value =5.7356 Is less than12.01 at 0.1level of significance .
The null hypothesis is accepted.
The past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2 = 23 mpg^2.
Step-by-step explanation:
<u>Step:-(1)</u>
Given data 28 25 29 25 32 36 27 24
Sample size 'n' = 8
mean of the sample (x⁻) = ∑x /n =
mean of the sample (x⁻) = 28.25
x x- x⁻ (x-x⁻ )²
28 28 - 28.25 = -0.25 0.0625
25 25 -28.25 = -3.25 10.56
29 29-28.25 = 0.75 0.5625
25 25-28.25 = -3.25 10.56
32 32-28.25 = 3.75 14.06
36 36-28.25 = 7.75 60.06
27 27-28.25 = -1.25 1.5625
24 24-28.25 = -4.25 18.06
∑(x-x⁻)² = 115.4875
<u>Step:-(2)</u>
The sample standard deviation
S² = ∑(x-x⁻)²/n-1 = = 16.49
By using χ² distribution
χ² =
By above test can be applied only if the population from which sample is drawn normal.
Given data For the past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2 = 23 mpg^2.
Population variance σ² = 23
<u>Step:-(3)</u>
Null hypothesis :H₀:σ² = 23
Alternative hypothesis :H₁:≠23
<u>The calculated chi -square value =5.7356</u>
The degrees of freedom γ=n-1 =8-1=7
The tabulated value of chi -square = 12.01 at 0.1level of significance (check table)
The calculated chi -square value =5.7356 Is less than12.01 at 0.1level of significance .
The null hypothesis is accepted.
<u>Conclusion</u>:-
The null hypothesis is accepted.
The past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2 = 23 mpg^2.
