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)
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.
t0.025, 16=2.120.
Considering that t*(calculated)=1.8438 is less than the table value t0.025, 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.
