Question

The average price of a sample of 12 bottles of diet salad dressing taken from different stores is $1.43. The standard deviation is $0.09. The average price of a sample of 16 low-calorie frozen desserts is $1.03. The standard deviation is $0.10. Use α = 0.05 to test that the price of salad dressings significantly higher than the price of frozen desserts. Formulate the null and alternative hypothesis.

Answer 1

Let m₁,m₂ be average prices of salad dressing and frosen deserts.

Null hypothesi is:

$H_0:m_1=m_2;$

Alternative hypothesis

$H_1:m_1>m_2;$

$\alpha=0.05$ - significance level

1) $S_p=\sqrt{\frac{(n_1-1)\cdot s_1^2+(n_2-1)\cdot s_2^2}{n_1+n_2-2}}= \sqrt{\frac{(12-1)\cdot 0.09^2+(16-1)\cdot 0.1^2}{12+16-2}}=0.096$

common standard deviation

2) $t=\frac{\overline{x_1}-\overline{x_2}}{S_p\cdot \sqrt {\frac{1}{n_1}+\frac{1}{n_2}}}=\frac{1.43-1.03}{0.096\cdot \sqrt{\frac{1}{12}+\frac{1}{16}}}=10.911$

value of t criterion.

If is not true that $m_1>m_2$ than t valuy may not be too big,threshold value

is $t_{tab}=t_{1-\alpha,n_1+n_2-2}=t_{0.95,26}=1.706$ -critical value for t statistics from statistical tables(or qt(0.95,26)=1.706 with mathcad soft).

3) Conclusion: because t=10.911>t_tab=1.706 with reliability 95% H0 hypothesis should be rejected and average price of salad dressing is significantly highter than if for frosen deserts.