Question

A manufacturer wants to increase the shelf life of a line of cake mixes. Past records indicate that the average shelf life of the mix is 216 days. After a revised mix has been developed, a sample of nine boxes of cake mix had a mean of 217.222 days and a standard deviation of 1.2019 days. At the 0.025 significance level, decide if the sample data support the claim that shelf life has increased. State your decision in terms of the null hypothesis.

Answer 1

Answer:

The calculated value t = 3.050 <  3.8325 at 0.0025 level of significance

The null hypothesis is accepted

The sample data support the claim that shelf life has increased.

Step-by-step explanation:

Step(i):-

Given that the Mean of the Population = 216days

Given that sample size n=9

Given that mean of sample x⁻ = 217.222 days

Given that the Standard deviation of the sample (S) = 1.2019days

Step(ii):-

Null hypothesis:-H₀:The sample data support the claim that shelf life has increased.

Alternative Hypothesis:H₁: The sample data support the claim that shelf life has decreased

Test statistic

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

       $t = \frac{217.222-216}{\frac{1.2019}{\sqrt{9 } } }$

       t = 3.050

Degrees of freedom γ = n-1 = 9-1 =8

t₀.₀₂₅ = 3.8325

Final answer:-

The calculated value t = 3.050 <  3.8325 at 0.0025 level of significance

The null hypothesis is accepted

The sample data support the claim that shelf life has increased.