Question

Are these rides equal in terms of monetary revenue? Test at α = 0.5 significance level.

Happy Cat:                 Average daily revenue = $705.00;   standard deviation = 80;    n = 12

Vicious Mouse:          Average daily revenue = $663.00;   standard deviation = 90;    n = 17

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Answer 1

I think you mean 0.05 significance level and not 0.5

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

The following information about the samples has been provided: a. Sample Means : Xˉ1​=705 and Xˉ2​=663 b. Sample Standard deviation: s1=80 and s2=90 c. Sample size: n1=12 and n2=17 (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=12+17-2=27. (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 27. Therefore the critical value for this Two-tailed test is tc​=2.0518. 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|>2.0518 i.e. t>2.0518 or t<-2.0518 (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 0.2065 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |t|=1.2943 < tc​=2.0518, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.2065, and since p=0.2065>0.05, it is concluded that the null hypothesis is Not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population mean μ1​ is different than μ2, at the 0.05 significance level.

Just in case you did mean 0.5 significance level

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

The following information about the samples has been provided: a. Sample Means : Xˉ1​=705 and Xˉ2​=663 b. Sample Standard deviation: s1=80 and s2=90 c. Sample size: n1=12 and n2=17 (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=12+17-2=27. (3a) Critical Value Based on the information provided, the significance level is α=0.5, and the degree of freedom is 27. Therefore the critical value for this Two-tailed test is tc​=0.6837. 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|>0.6837 i.e. t>0.6837 or t<-0.6837 (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 0.2065 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |t|=1.2943 > tc​=0.6837, 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=0.2065, and since p=0.2065≤0.5, 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 0.5 significance level.