Answer :
This question is incomplete, the complete question is;
Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 18ug/g.
22.5 , 9 , 18.5 , 22 , 16 , 18.5 , 15 , 3 , 8, 6
a) what is the null and alternative hypothesis
b) determine the test statistic
Answer:
a)
the null and alternative hypothesis will be;
Null hypothesis H₀ : Ц = 18 ug/g
Alternative hypothesis H₁ : Ц < 18 ug/g
b) the Test Statistic = -1.47
Step-by-step explanation:
Given data;
x = 22.5 , 9 , 18.5 , 22 , 16 , 18.5 , 15 , 3 , 8, 6
number of observation n = 10
level of significance ∝ = 0.01
a)
the null and alternative hypothesis will be;
Null hypothesis H₀ : Ц = 18 ug/g
Alternative hypothesis H₁ : Ц < 18 ug/g
b)
Test statistics;
using the formula;
t = (x" - Ц) / (s/√n)
x" is the sample mean
s is the standard deviation
Ц is the population mean ( 18 )
n is the number of observation ( 10 )
for first we get our mean x"
mean x" = ∑x / n
mean x" = (22.5 + 9 + 18.5 + 22 + 16 + 18.5 + 15 + 3 + 8 + 6) / 10
mean x" = 138.5 /10
mean x" = 14.85
Next we find the standard deviation s
standard deviation s = √[ (∑(x-x")²) / (n - 1)]
x (x - x") (x - x")²
22.5 7.65 58.5225
9 -5.85 34.2225
18.5 3.65 13.3225
22 7.15 51.1225
16 1.15 1.3225
18.5 3.65 13.3225
15 0.15 0.0225
3 -11.85 140.4225
8 -6.85 46.9225
6 -8.85 78.3225
total 437.525
so we substitute to get our standard deviation
standard deviation s = √[ ( 437.525 / (10 - 1)]
= √[ 437.525 / 9]
= √48.6138
standard deviation s = 6.9723
Finally our we substitute our values into the Test statistics formula
t = (x" - Ц) / (s/√n)
t = (14.85 - 18) / (6.9723/√10)
t = -3.25 / 2.2048
t = -1.47
Therefore the Test Statistic = -1.47