Question

1.

The mean produce of rice of sample of 150 fields yields is 200 quintals with a standard deviation of 12 quintals. Another sample of 100 fields gives the mean at 200 quintals with a SD of 10 quintals. Assuming the standard deviation of the mean field at 11 quintals for the universe, test whether the results are consistent at 1% level of significance.

2) Testing of Hypothesis [2 Marks] In a city A, out of 600 men, 325 men were found to be smokers. Does this information support the statement” Majority of men in city are smokers”?

Answer 1

1) Solution:

(1) Null and Alternative Hypotheses :
The following null and alternative hypotheses need to be tested: Ho:p1=p2 Ha:pl != p2
This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations, will be used.
(2) Rejection Region :
Based on the information provided, the significance level is (a=0.05, and the critical value for a two-tailed test is Zc =1.96.
The rejection region for this two-tailed test is R= (z:IzI>1.96)

(3) Test Statistics :The z-statistic is computed as follows: z = 0.00

(4) The decision about the null hypothesis: Since it is observed that Iz1=0 <=Zc=1.96, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p=1, and since p=1>=0.01, it is concluded that the null hypothesis is not rejected.
Confidence Interval
The 95% confidence interval for pl-p2 is -2.783<pl-p2 <2.783.
  The result are consistent at a 1% level of significance.

2) Answer:

Here proportion of smokers=325/600=0.5416.

54.16% are smokers so here not majority are smokers if that percentage(%)  more than 60% then only we can say that majority are smokers.

Here non smokers % =1-0.5416=0.4584

non smokers % =0.4584

although smokers percentage (%) are higher but we are not consider as majority are smokers because percentage(%) of smokers are not significancy higher.