Lead is a poisonous substance, one that poisons water and could cost several lives. The measurementof lead using traditional laboratory methods has been expensive. Researchers have developed a newlaboratory method to measure the level of Lead in water samples. The concentration of lead in wateris measured byTwo-Sample T-Test and CISampleNMeanStDevSE Mean1319.701.100.640.6010.90Estimate for difference: 8.800230.3595% CI for difference: (6.498, 11.102)T-Test of difference= 0 (vs not =):T-Value = 12.16 P-Value = 0.001 DF = 3

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Description: Lead is a poisonous substance, one that poisons water and could cost several lives. The measurementof lead using traditional laboratory methods has been expensive. Researchers have developed a newlaboratory method to measure the level of Lead in wat

When the concentration of led in water is measured by μgL, to check that: H0 : μtraditional method…

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Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.3: Measures Of Spread

Problem 1GP

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Question

Lead is a poisonous substance, one that poisons water and could cost several lives. The measurement
of lead using traditional laboratory methods has been expensive. Researchers have developed a new
laboratory method to measure the level of Lead in water samples. The concentration of lead in water
is measured by
Two-Sample T-Test and CI
Sample
N
Mean
StDev
SE Mean
1
3
19.70
1.10
0.64
0.60
10.90
Estimate for difference: 8.800
2
3
0.35
95% CI for difference: (6.498, 11.102)
T-Test of difference
= 0 (vs not =):
T-Value = 12.16 P-Value = 0.001 DF = 3

Expert Solution

Hypotheses

When the concentration of led in water is measured by $\frac{μ g}{L}$ , to check that:

$H_{0} : μ_{t r a d i t i o n a l m e t h o d} = μ_{n e w m e t h o d} v s H_{1} : μ_{t r a d i t i o n a l m e t h o d} \neq μ_{n e w m e t h o d} t h a t i s H_{0} : μ_{t r a d i t i o n a l m e t h o d} - μ_{n e w m e t h o d} = 0 v s H_{1} : μ_{t r a d i t i o n a l m e t h o d} - μ_{n e w m e t h o d} \neq 0$

Test statistic

The test statistic is obtained as:

$T - v a l u e = \frac{x_{t r a d i t i o n a l m e t h o d} - x_{n e w m e t h o d}}{s_{p} \sqrt{\frac{1}{n_{t r a d i t i o n a l m e t h o d}} + \frac{1}{n_{n e w m e t h o d}}}} = 12.16 w h e r e, s_{p} = \sqrt{\frac{(n_{t r a d i t i o n a l m e t h o d} - 1) {s^{2}}_{t r a d i t i o n a l m e t h o d} + (n_{n e w m e t h o d} - 1) {s^{2}}_{n e w m e t h o d}}{(n_{t r a d i t i o n a l m e t h o d} + n_{n e w m e t h o d} - 2)}} n_{t r a d i t i o n a l m e t h o d :} s a m p l e s i z e f o r t r a d i t i o n a l m e t h o d = 3 n_{n e w m e t h o d} : s a m p l e s i z e f o r n e w m e t h o d = 3 x_{t r a d i t i o n a l m e t h o d} : s a m p l e m e a n f o r t r a d i t i o n a l m e t h o d = 19.70 x_{n e w m e t h o d} : s a m p l e m e a n f o r n e w m e t h o d = 10.90 s_{t r a d i t i o n a l m e t h o d} : s a m p l e s \tan d a r d d e v i a t i o n f o r t r a d i t i o n a l m e t h o d = 1.10 s_{n e w m e t h o d} : s a m p l e s \tan d a r d d e v i a t i o n f o r n e w m e t h o d = 0.60 s_{p} = p o o l e d s \tan d a r d d e v i a t i o n$

P-value

The P-value of the test is obtained as:

$P - v a l u e = 2 \times P (t > | 12.6 |) = 0.001$ $A t 5 % l e v e l o f s i g n i f i c a n c e, a s t h e P - v a l u e < 0.05 w e r e j e c t H_{0} a t α = 0.05$

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