USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

Tests for Large Samples (n > 30)

(a) Testing of A Single Mean. (Inferences about a single mean)

1. Set up H0 : µ = µ0

2. Set up H1 : µ > µ0 or µ < µ or µ ≠ µ0

3. Set up the test statistic

which follows standard normal distribution.

4. Set up the level of significance a. and the critical value as Ztab from the normal table.

5. Compute the statistic, say Zcal

6. Decisions :

Note. If the population S.D. (σ) is not known for large sample we can take the sample S.D. (S) in the test statistic.

 

Example 2. The mean lifetime of I 00 picture tubes produced by a manufacturing company is estimated to be 5795 hours with a standard deviation of 150 hours. If µ be the mean lifetime of all the picture tubes produced by the company, test the hypothesis J µ = 6000 hours against µ:F. 6000 hours at 5 % level of significance.

 

Solution.

  1. H0 : µ = 6000 hours
  2. H1 : µ:F. 6000 hours

Test statistic : Here n = 100 i.e., large sample. Population S.D. is not given but the sample S.D. is given as 150 hours.

4.  α = 0.05, α /2 = 0.025

H1 shows, this is two tailed test.

Z α /2 = 1.96 so the critical values are 1.96 and -1.96.

5. Computation :

6. Decision :

Since Zcal < - 1.96 i.e., it lies in the rejection region

=> H0 is rejected

=> The claim produced by the company is not true.

 

Example 3. A tire company claims that the lives of the tires have mean of 42000 kilometers with standard deviation of 4000 kilometers. A change in the production process is believed to result in a better product. A test sample of 81 new types has a mean life of 42500 kilometers. Test at 5% level of significance that the new product is significantly better than the current one ?

 

Solution.

  1. H0 : µ= 42000 km.
  2. H1 : µ > 42000 km.

Here n = 81, Test statistic :

  1. Alt. hypothesis shows, this is right tailed test.

α= 0.05, Zα = 1.64 which is the critical value.

5. Computation :

6. Decision :

Since Zca1 < Zα => Zeal lies in the acceptance region

=> H0 is accepted

=> H1 is rejected

::::> New product is not significantly better than the current one.

 

(b) Testing of Difference of Two Means. Consider two populations having the means µ1 and µ2 and the variance σ12 and σ22 respectively.

  1. Set up Ho : µ1 - µ2 = k
  2. Set up H1 : µ1 and µ2 > k or µ1 and µ2 < k or µ1 and µ2k where k is a specified constant.
  3. Set up the test statisticwhere,             x1 = Means of sample size n1 from the first population.

    X2 = Means of sample size n2 from the second population.

    4. Set up the level of significance a. and the critical value as Z13b from the normal table.

    5. Compute the statistic, says Zcal.

    6. Decisions :

    Note. When both n1 and n2 are greater than or equal to 30, the population S.D. can be estimated by sample S.D. in the statistic.

     

     

     

    Example 4. A random sample of 100 villages was taken from a district A and the average height of the population per village was found to be 170 cm with a standard deviation of 10 cm. Another random sample of 120 villages was taken from another district B and the average height of the population per village was found to be 176 cm with a standard deviation of 12 cm. Is the difference between averages of the two populations statistically significant ?

    Solution.

     

    1. Ho : µ1 - µ2 = 0, i.e., there is no significant difference between the means of two populations.
    2. H1 : µ1 - µ2 ≠0
    3.  Test statistic4. Let the level of significance α = 0.05, α/2= 0.025,

      critical values are - 1.96 and 1.96.

      5. Computation :

      Given

      xi = 170, s1 = 10, n1 = 100

      x2 = 176, s2 = 12, n2 = 120

    6. Conclusion :

    zeal < - 1.96 i.e., zeal lies inside the critical region

    => Reject H0

    => There is significant difference between the means of two populations.