Question

Sample survey studies estimate wheat prices in July and September.

      Month               n                                  s

      September       45           $3.61          $0.19

      July                    90           $2.95          $0.22

      

      a.  Do a hypothesis test to conclude if the national average prices are the same?

                                 {.05 level}

        

2. Find a 95% confidence interval for the difference of the two means.

Answer 1

Answer a.

T-test for two Means – Unknown Population Standard Deviations - Equal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=3.61 and Xˉ2​=2.95 b. Sample Standard deviation: s1=0.19 and s2=0.22 c. Sample size: n1=45 and n2=90 (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 t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) The degrees of freedom Assuming that the population variances are equal, the degrees of freedom are given by n1+n2-2=45+90-2=133. (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 133. Therefore the critical value for this Two-tailed test is tc​=1.978. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Two-tailed test is |t|>1.978 i.e. t>1.978 or t<-1.978 (4)Test Statistics The t-statistic is computed as follows: (5) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0 (6) The Decision about the null hypothesis (a) Using the traditional method Since it is observed that |t|=17.1693 > tc​=1.978, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0, and since p=0≤0.05, it is concluded that the null hypothesis is rejected. (7) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is different than μ2, at the 0.05 significance level.

Answer b.

T-test for two Means Confidence Interval -Equal Variance

The following information about the sample has been provided: a. Sample Means : Xˉ1​=3.61 and Xˉ2​=2.95 b. Sample Standard deviation: s1=0.19 and s2=0.22 c. Sample size: n1=45 and n2=90 d. Level of confidence is 95% The degrees of freedom Assuming that the population variances are equal, the degrees of freedom are given by n1+n2-2=45+90-2=133. Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 133. Therefore the critical value is tc​=1.978. This can be found by either using excel or the t distribution table. Standard Error The Standard error is computed as: Confidence Interval The 95% Confidence interval is computed as: Therefore, we are 95% confident that the true difference between population means is contained by the interval (0.584, 0.736)

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