Question

Directions: Show all your work. Indicate clearly and thoroughly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations.

Will using name-brand microwave popcorn result in a greater percentage of popped kernels than using store-brand microwave popcorn? To find out, Briana and Maggie randomly selected 10 bags of namebrand microwave popcorn and 10 bags of store-brand microwave popcorn. The chosen bags were arranged in a random order. Then each bag was popped for 3.5 minutes, and the percentage of popped kernels was calculated. The results are displayed in the following table.

Namebrand 95 88 84 94 81 90 97 93 91 86

Store-brand 91 89 82 82 77 78 84 86 86 90

Do the data provide convincing evidence that using name-brand microwave popcorn will result in a greater mean percentage of popped kernels?

Answer 1

Assuming the significance level to be 5%

	Namebrand	Store-brand
Sample Sizes	10	10
Sample Means	89.9	84.5
Sample St. Deviations	5.130518709	4.813176359

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

The following information about the samples has been provided: a. Sample Means : Xˉ1​=89.9 and Xˉ2​=84.5 b. Sample Standard deviation: s1=5.1305 and s2=4.8132 c. Sample size: n1=10 and n2=10 (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 Right-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=10+10-2=18. (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 18. Therefore the critical value for this Right-tailed test is tc​=1.7341. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Right-tailed test is t>1.7341 (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 0.013 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=2.4274 > tc​=1.7341, 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=0.013, and since p=0.013≤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 greater than μ2, at the 0.05 significance level.