A ichthyologist measured the adult body weight of two groups

A ichthyologist measured the adult body weight of two groups of fish: one group containing external parasites and another healthy group. The n 1 = 34 parasitized fish had a mean body weight of x = 194 g and a standard deviation of s 1 1 = 33 g, whereas the n 2 = 38 healthy fish yielded x = 211 g and s 2 = 25 g. Test the hypothesis of no difference in body weight between parasitized and healthy fish versus the alternative that the parasitized fish are smaller. Use a 1% level of significance.

Solution

A ichthyologist measured the adult body weight of two groups of fish: one group containing external parasites and another healthy group. The n 1 = 34 parasitized fish had a mean body weight of x = 194 g and a standard deviation of s 1 1 = 33 g, whereas the n 2 = 38 healthy fish yielded x = 211 g and s 2 = 25 g. Test the hypothesis of no difference in body weight between parasitized and healthy fish versus the alternative that the parasitized fish are smaller. Use a 1% level of significance.

H₀: ?₁ =?₂

H_A: ?₁< ?₂

Both n1 and n2 are > 30, z test is used

This is lower tail z test

Z Test for Differences in Two Means

Data

Hypothesized Difference

0

Level of Significance

0.01

Population 1 Sample

Sample Size

31

Sample Mean

194

Population Standard Deviation

33

Population 2 Sample

Sample Size

38

Sample Mean

211

Population Standard Deviation

25

Intermediate Calculations

Difference in Sample Means

-17

Standard Error of the Difference in Means= sqrt(33²/31+25²/38) =

7.1817

Z Test Statistic = -17/7.1817 =

-2.3671

Lower-Tail Test

Lower Critical Value at 0.01 level=

-2.326

Calculate z= -2.367 < -2.326 the table value of z at 0.01 level

The null hypothesis is rejected.

We conclude that the parasitized fish are smaller in body weight than the healthy fish