Question

Pat an an experiment to test optimum power and time settings for microwave popcorn. His

goal was to deliver popcorn with fewer than 12?% 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.

6.7?,

11.2?,

13.8?,

6.4?,

12.7?,

12.2?,

7.3?,

6.5

Calculate the t statistics

t=?

  

Answer 1

The Mean and standard deviation is as calculated below

#	X	The Mean = 9.6% S = SQRT (variance) = SQRT [Sum (x-x^bar)^2/n-1] = 3.65% Given mu = 12% and n = 8 t = x^bar -mu/s/ squareroot n = 9.67-12/3.165/ squareroot 8 = -2.1448 t = -2.1448	(X- The Mean = 9.6% S = SQRT (variance) = SQRT [Sum (x-x^bar)^2/n-1] = 3.65% Given mu = 12% and n = 8 t = x^bar -mu/s/ squareroot n = 9.67-12/3.165/ squareroot 8 = -2.1448 t = -2.1448 )2
1	6.7	9.6	8.41
2	11.2	9.6	2.56
3	13.8	9.6	17.64
4	6.4	9.6	10.24
5	12.7	9.6	9.61
6	12.2	9.6	6.76
7	7.3	9.6	5.29
8	6.5	9.6	9.61
	76.8	Sum(x- The Mean = 9.6% S = SQRT (variance) = SQRT [Sum (x-x^bar)^2/n-1] = 3.65% Given mu = 12% and n = 8 t = x^bar -mu/s/ squareroot n = 9.67-12/3.165/ squareroot 8 = -2.1448 t = -2.1448 )2	70.12
Mean( The Mean = 9.6% S = SQRT (variance) = SQRT [Sum (x-x^bar)^2/n-1] = 3.65% Given mu = 12% and n = 8 t = x^bar -mu/s/ squareroot n = 9.67-12/3.165/ squareroot 8 = -2.1448 t = -2.1448 )	9.600	Variance	10.0171
		SD	3.1650

The Mean = 9.6%

SD = Sqrt(Variance) = SQRT[Sum(x - )²/n-1] = 3.165%

The Mean = 9.6% S = SQRT (variance) = SQRT [Sum (x-x^bar)^2/n-1] = 3.65% Given mu = 12% and n = 8 t = x^bar -mu/s/ squareroot n = 9.67-12/3.165/ squareroot 8 = -2.1448 t = -2.1448

Given $\mu$ = 12% and n = 8

t = -2.1448