Alicia took a friend for a birthday dinner. The total bill for dinner was $40.86 (including tax and a tip). If Alicia paid a 20.
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Answer:
|t| =|-0.96|=0.96< 2.718
since calculated' t' value is less than tabulated value 't'.
we accept null hypothesis at 1 % level of significance.
The data support the claim that the mean cost of a daily news paper is $1.00
Step-by-step explanation:
<u>Step:-(1)</u>
Given the sample size n = 12
Given Twelve costs yield a mean cost of $0.95 with a standard deviation of $0.18
mean of the sample(x⁻) = $0.95
Standard deviation of sample (S) = $0.18
Given claim that the mean cost of a daily newspaper is $1.00.
The mean of the population 'μ' = $1.00.
<u>Null hypothesis:</u>-H₀:'μ' = $1.00.
<u>Alternative hypothesis</u>: H₁:'μ' ≠ $1.00.
<u>level of significance ∝ = 0.01</u>
<u>Step2:-</u>
<u>The test statistic </u>
on calculation, we get
t = -0.96
|t| =|-0.96|=0.96
Degrees of freedom γ = n-1 = 12-1 =11
tₐ = 2.718
Calculated value t = 0.96
Tabulated value t at 0.01 level for 11 degrees for two tailed test = 2.718
since calculated' t' value is less than tabulated value 't'.
<u>conclusion</u>:-
<u>we accept null hypothesis at 1 % level of significance.</u>
<u>The data support the claim that the mean cost of a daily news paper is $1.00</u>
