In: Statistics and Probability
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)
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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