Question

Nate ran an experiment to test optimum power and time settings for microwave popcorn. His goal was to deliver popcorn with fewer than 11% of the kernels left unpopped, on average. He determined that power 9 at 4 minutes was the best combination. To be sure that the method was successful, he popped 8 more bags of popcorn (selected at random) at this setting. All were of high quality, with the percentages of unpopped kernels shown below. 5.1, 4.2, 4.8, 9.4, 6.5, 11.3, 8.1, 12.7 Does this provide evidence that he met his goal of an average of fewer than 11% unpopped kernels? Use 0.05 as the P-value cutoff level.

t=?

P-value=?

Answer 1

From the given data

Descriptive Statistics: Data Total Variable Count Mean StDev Variance Sum Data 8 7.76 3.16 9.98 62.10 Sum of Squares 551.89

(1) Null and Alternative Hypothesis:

H0: He met his goal of an average of not fewer than 11% unpopped kernels

H1: He met his goal of an average of fewer than 11% unpopped kernels

i.e.

Hoi u = 11 Ha: p<11 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is te = -1.895. The rejection region for this left-tailed test is R = t:t < -1.895 (3) Test Statistics The t-statistic is computed as follows: X - UO s/ n 7.76 - 11 3.16/8 -2.9 (4) Decision about the null hypothesis Since it is observed that t = -2.9 <te = -1.895, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0115, and since p = 0.0115 <0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 11, at the 0.05 significance level.

Answers: t = -2.9

P-value = 0.0115