Question

An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola

Beverage SampleSize SampleMean SampleSD
Strawberry Drink 10 537 22
Cola 10 559 17.
Assume the two populations are normal. Does the data suggest that the extra carbonation of cola results in a higher average compression strength?

Answer 1

The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted

The extra carbonation of cola results in a higher average compression strength

Explanation:

Step(i):-

Given data

First sample size (n₁) = 10

mean of the first sample(x₁⁻) = 537

standard deviation of the first sample (S₁) = 22

second sample size (n₂) = 10

mean of the second sample (x₂⁻) = 559

standard deviation of the second sample (S₂) = 17

Step(ii):-

Null hypothesis : H₀:- The extra carbonation of cola results in a lower average compression strength

Alternative Hypothesis :H₁

The extra carbonation of cola results in a higher average compression strength

Step(iii):-

By using student's t -test for difference of means

Test statistic

`t = \frac{x^(-) _(1)-x^(-) _(2) }{\sqrt{S^(2) ((1)/(n_(1) )+(1)/(n_(2) ) } )}`

where

`S^(2) = (n_(1)S^(2) _(1) + n_(2) S_(2) ^(2) )/(n_(1)+n_(2) -2 )`

`S^(2) = (10(22)^(2) + 10 (17) ^(2) )/(10+10 -2 ) = ( 7730)/(18) = 429.4`

`t = \frac{537-559 }{\sqrt{429.4 ((1)/(10 )+(1)/(10 ) } )}`

t = -2.375

|t| = |-2.375| = 2.375

Degrees of freedom

γ = n₁+n₂ -2 = 10+10-2 =18

`t_{(\alpha )/(2) ,n-1}=t_{((0.05)/(2) ,18)} = t_((0.025,18))}=2.1009`

The calculated value |t| = 2.375 > 2.1009 at 0.05 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted

Final answer:-

The extra carbonation of cola results in a higher average compression strength