Answer to Question #240944 in Statistics and Probability for Matilda

Question

two loading brands of potato chips were tested to compare their sodium content. ten packs of brand a yielded a mean sodium content of 80 mg/pack with a standard deviation of 5.4 mg while eight packs of brand b yielded an average content of 75 mg/pack with standard deviation of 6.1 mg. is there difference in the sodium content of the two brands at the 0.05 level of significance? (assume normality with equal variances)

Answer 1

Let the following notations denote the 2 populations,

Population 1

n1=10

$\bar{x}$₁=80

s1=5.4

Population 2

n2=8

$\bar{x}$₂ =75

s2=6.1

The hypothesis tested is,

H₀:$\mu$₁=$\mu$₂ against H₁:$\mu$₁ $\not=$ $\mu$₂ at $\alpha$ =0.05

Assuming normality and equal variances, this hypothesis is tested using the student's t distribution.

Since population variances are unknown and assumed equal, pooled sample variance is used. Its formula is given by,

$Sp^2=((n1-1)s1^2+(n2-1)s2^2)/(n1+n2-2)$

=((10-1)*5.4²+(8-1)*6.1²)/(10+8-2)

=9*(29.16)+7*(37.21)/16

=32.681875

The test statistic t^*(calculated) is given by,

t^*=( $\bar{x}$₁- $\bar{x}$₂)$/$$\sqrt{Sp^2*(1/n1+1/n2)}$

=$(80-75)/$ $\sqrt{32.681875*(1/10+1/8)}$

=5/2.71171936

=1.8438(4 decimal places)

t* is compared with the table value with n1+n2-2=16 degrees of freedom and $\alpha/2=0.025$ since it is a two sided t-test.

t_{0.025, 16}=2.120.

Considering that t*(calculated)=1.8438 is less than the table value t_{0.025, 16}=2.120, we fail to reject the null hypothesis and conclude that there is a significant evidence at the 95% level of confidence that sodium content for the two brands are not different.