Question

A sample of 39 cookies is taken to test the claim that each cookie contains, on average, 8 chocolate chips. The average number of chocolate chips per cookie in the sample was 8.2 with a standard deviation of 0.6. Assume the distribution of the population is normal. Use a 0.01 level of significance. Choose the t-statistic and the appropriate conclusion for testing this claim

Answer 1

Answer:

The calculated value t = 2.083 < 2.429 at 0.01 level of significance with 38 degrees of freedom.

The null hypothesis is accepted

That is  x⁻ and  'μ' do not differ significantly.

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 39

The mean of the Population 'μ' = 8

The mean of the sample x⁻ = 8.2

The standard deviation of the sample 'S' = 0.6

Level of significance ∝=0.01

Step(ii):-

Null hypothesis : H₀ : x⁻ = 'μ'

Alternative hypothesis: H₁: x⁻ ≠ 'μ'

Degrees of freedom ν = n-1 = 39-1 =38

t₀.₉₉ =    2.429 ( see t- table)

The test statistic

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

         $t = \frac{8.2-8 }{\frac{0.6}{\sqrt{39} } }$

        t = 2.083 < 2.429 at 0.01 level of significance

Step (iii):-

The calculated value t = 2.083 < 2.429 at 0.01 level of significance with 38 degrees of freedom.

The null hypothesis is accepted

That is  x⁻ and  'μ' do not differ significantly.