Question

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of $0.20. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of $0.95 with a standard deviation of $0.18. Do the data support the claim at the 1% level?

Answer 1

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

$t = \frac{x^{-}-u_{0} }{\frac{S}{\sqrt{n} } }$

$t = \frac{0.95-1 }{\frac{0.18}{\sqrt{12} } }$

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