Answer to Question #152905 in Statistics and Probability for hadi

Question

4. A food production company HaHa claims that the average fat content of their food products is less than that of its main competitors. A nutritionist wants to know the truth of HaHa's allegations. He bought 11 packs of HaHa food and measured the fat content in the food. It was found that the average fat content was 31.3 gm with a standard deviation of 2.1gm. While the average fat content of the 11 main competitor food packs is 33.2gm with a standard siding of 1.8gm. Assuming a normal population with the same standard deviation, test the HaHa company claim at a significance level of 5%.

Answer 1

H₀: mean fat content of their food products = mean fat content of food products of its main competitors

H_A: mean fat content of their food products < mean fat content of food products of its main competitors

Hypothesis testing for two means (σ is unknown, and the variable is normally distributed in the population or n > 30, the same standard deviatioin)

So we use t-test with df = n1 + n2 - 2 = 20 degree of freedom:

$s_p =\sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}=1.9558\\ t = \frac{x_1-x_2}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}= \frac{31.3 - 33.2}{1.9558 \sqrt{2/11}}=-2.2783\\ t_{0.05, 20}=-1.7247\\ \\ t<t_{0.05,20}$

Hence, we reject the null hypothesis and accept the alternative hypothesis that the average fat content of their food products is less than that of its main competitors at significant level of 0.05.