Question

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.

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ1​=μ2​

Ha:μ1​ ≠  μ2​

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 α=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:∣z∣>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 ∣z∣=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 μ1​−μ2​ is −2.783<μ1​−μ2​<2.783.

The result are consistent at a 1% level of significance.

Dear Student,

I am waiting for your feedback. I have given my 100% to solve your queries.If you are satisfied with my given answer. Can you please please like it.

Thank You!!!