Question

Data on the rate at which a volatile liquid will spread across a surface are in the table. Complete parts athrough

c.

Time(Minutes): 0,2,4,6,8,10,12,14,16,18,20,25,30,45,60

Mass (Pounds): 6.62, 5.96, 5.47, 4.85, 4.38, 4.05, 3.61, 3.09, 2.74 ,2.48, 2.23, 1.56, 0.94, 0.18, 0.00

Find a 98​% confidence interval for the mean mass of all spills with an elapsed time of 56minutes. Interpret the result.

What is the confidence​ interval?

(    ),(   )

​(Round to three decimal places as​ needed.)

Interpret the result. Choose the correct answer below.

A. We are 98​% confident that the interval will contain 56minutes.

B. We are 98​% confident that the interval will contain the mean mass of the spill before 56minutes has passed.

C. We are 98​% confident that the interval will not contain the mean mass of the spill at

56minutes.

D. We are 98​% confident that the interval will contain the mean mass of the spill after 56minutes.

b.Find a 98​% prediction interval for the mass of a single spill with an elapsed time of

56minutes. Interpret the result.

What is the prediction​ interval?

(.   ),(.  )

​(Round to three decimal places as​ needed.)

Interpret the result. Choose the correct answer below.

A.We are 98​% confident that the interval will contain the mass of the spill after 56minutes.

B.We are 98​% confident that the interval will contain 56minutes.

C.We are 98​% confident that the interval will contain the mass of the spill before 56minutes has passed.

D.We are 98​% confident that the interval will not contain the mass of the spill after 56minutes.

c.Compare the​ intervals, parts aand

b.

Which interval is​ wider? Will this always be the​ case? Explain. Fill in the blanks below.

The (prediction/confidence/neither) interval is wider. This (will/will not) always be the case because the error of this interval is the (random error/ sum of two errors/neither)

Answer 1

• a)

  98% confidence interval: (-2.5793,0.3990)

  D We are 98​% confident that the interval will contain the mean mass of the spill after 56 minutes

• b)

  98% prediction interval:(-2.2872,3.0851)

  A.We are 98​% confident that the interval will contain the mass of the spill after 56 minutes

  The prediction interval is wider. This will always be the case because the error of this interval Which interval is​ wider

• option:neither

Calculation below using excel

No	x=Time(Minutes)	y=Mass (Pounds)	(x-xbar)^2	(y-ybar)^2	(x-xbar)*(y-ybar)
1	0	6.62	324	11.623554	-61.368
2	2	5.96	256	7.5588338	-43.98933333
3	4	5.47	196	5.1045871	-31.63066667
4	6	4.85	144	2.6874138	-19.672
5	8	4.38	100	1.3673404	-11.69333333
6	10	4.05	64	0.7044804	-6.714666667
7	12	3.61	36	0.1594671	-2.396
8	14	3.09	16	0.0145604	0.482666667
9	16	2.74	4	0.2215271	0.941333333
10	18	2.48	0	0.5338738	0
11	20	2.23	4	0.9617071	-1.961333333
12	25	1.56	49	2.7247004	-11.55466667
13	30	0.94	144	5.1559271	-27.248
14	45	0.18	729	9.1849404	-81.828
15	60	0	1764	10.30838	-134.848
sum	270	48.16	3830	58.311293	-433.48
mean	18	3.210666667	sxx	syy	sxy

slope=b1=sxy/sxx	-0.1132
intercept=b0=ybar-(slope*xbar)	5.2479
SST	SYY	58.3113
SSR	sxy^2/sxx	49.0613
SSE	syy-sxy^2/sxx	9.2500
r^2	SSR/SST	0.8414
r	sxy/sqrt(sxx*syy)	-0.9173
error variance s^2	SSE/(n-2)	0.7115
S^2b1	s^2/sxx	0.0002
standard error b1=se(b1)	sqrt(s^2b1)	0.0136
test statistics	b1/se(b1)	-8.3037
tcritical=For (1-alpha)% CI value of t	0.98	2.6245

Prediction at xi	56	-1.0902
standard error confidence interval for mean at prediction	s*sqrt(1/n+ (xi-Xbar)^2/Sxx)	0.5619
Lower confidence interval for mean at given point xi	y_pred -tcritical*s*sqrt(1/n+ (xi-Xbar)^2/Sxx)	-2.5793
Upperr confidence interval for mean at given point xi	y_pred +tcritical*s*sqrt(1/n+ (xi-Xbar)^2/Sxx)	0.3990
standard error Prediction interval for mean at given point xi	s*sqrt(1+1/n+ (xi-Xbar)^2/Sxx)	1.0135
Lower Prediction interval for mean at given point xi	y_pred -tcritical*s*sqrt(1+1/n+ (xi-Xbar)^2/Sxx)	-2.2872
Upperr Prediction interval for mean at given point xi	y_pred +tcritical*s*sqrt(1+1/n+ (xi-Xbar)^2/Sxx)	3.0851

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