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:
Step:-(1)
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.
Null hypothesis:-H₀:'μ' = $1.00.
Alternative hypothesis: H₁:'μ' ≠ $1.00.
level of significance ∝ = 0.01
Step2:-
The test statistic
[tex]t = \frac{x^{-}-u_{0} }{\frac{S}{\sqrt{n} } }[/tex]
[tex]t = \frac{0.95-1 }{\frac{0.18}{\sqrt{12} } }[/tex]
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'.
conclusion:-
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
