TEST OF HYPOTHESES FOR EQUALITY OF MEANS OF TWO POPULATIONS (POPULATION VARIANCES ARE UNKNOWN)
Step-1 : Let µX and σX2 be respectively the mean and the variance of Population -1. Also, let µY and σY2 be respectively the mean and the variance of Population -2 under study. Here σX2 and σY2 are assumed to be unknown.
Frame the null hypothesis as H0: µX = µY and choose the suitable alternative hypothesis from
(i) H1: µX ≠ µY (ii) H1: µX> µY (iii) H1: µX< µY
Step 2 : Let (X1, X2, …, Xm) be a random sample of m observations drawn from Population-1 and (Y1, Y2, …, Yn) be a random sample of n observations drawn from Population-2, where m and n are large (m ≥30 and n ≥30). Here, these two samples are assumed to be independent.
Step 3 : Specify the level of significance, α.
Step 4 : Consider the test statistic
i.e., the above test statistic is obtained from Z considered in the test described in Section 1.11 by substituting SX2 and SY2 respectively for σX2 and σY2
The approximate sampling distribution of the test statistic under H0 is the N(0,1) distribution.
Step 5 : Calculate the value of Z for the given samples (x1, x2, ...,xm) and (y1, y2, …, yn) as
Here and are respectively the values of and for the given samples.
Also, sx2 and sy2 are respectively the values of SX2 and SY2 for the given samples.
Step 6 : Find the critical value, ze, corresponding to α and H1 from the following table
Step 7 : Make decision on H0 choosing the suitable rejection rule from the following table corresponding to H1.
A Model Examination was conducted to XII Standard students in the subject of Statistics. A District Educational Officer wanted to analyze the Gender-wise performance of the students using the marks secured by randomly selected boys and girls. Sample measures were calculated and the details are presented below:
Test, at 5% level of significance, whether performance of the students differ significantly with respect to their gender.
Step 1 : Let μX and μY denote respectively the average marks secured by boys and girls in the Model Examination conducted to the XII Standard students in the subject of Statistics. Then, the null and the alternative hypotheses are
Null hypothesis: H0: µ X = µY
i.e., there is no significant difference in the performance of the students with respect to their gender.
Alternative hypothesis: H1 : µ X ≠ µY
i.e., performance of the students differ significantly with the respect to the gender. It is a two-sided alternative hypothesis.
Step 2 : Data
The given sample information are
Since m ≥ 30 and n ≥ 30, both the samples are large.
Step 3 : Level of significance
Step 4 : Test statistic
The test statistic under H0 is
The sampling distribution of Z under H0 is the N(0,1) distribution.
Step 5 : Calculation of the Test Statistic
The value of Z is calculated for the given sample informations from
Step 6 : Critical value
Since H1 is a two-sided alternative, the critical value at 5% level of significance is ze = z0.025 = 1.96.
Step 7 : Decision
Since H1 is a two-sided alternative, elements of the critical region are determined by the rejection rule |z0| ≥ z0 . Thus it is a two-tailed test. But, |z0|= 1.75 is less than the critical value ze = 1.96. Hence, it may inferred as the given sample information does not provide sufficient evidence to reject H0. Therefore, it may be decided that there is no sufficient evidence in the given sample to conclude that performance of boys and girls in the Model Examination conducted in the subject of Statistics differ significantly.