Question

1. Listed below are the lead concentrations innug/g measured jn 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 12 ug/g. Assume that the lead concentrations in traditional medicines are normally distributed. What are rhe null and alternatice hypothesis, test statistic, and P-Value.

22, 12.5, 16.5, 6.5, 4.5, 6, 8.5, 4.5, 14.5, 3.5

Answer 1

Step 1:

Ho: 12

Ha: 12

Null hypotheiss states that the lead concentration for all medicines is greater than or equal to 12.

Step 2: Test statistics

n = 10

mean = sum of all terms/ no of terms = 99/10 = 9.9

sample standard deviation = s

data	data-mean	(data - mean)2
22	12.1	146.41
12.5	2.6	6.76
16.5	6.6	43.56
6.5	-3.4	11.56
4.5	-5.4	29.16
6	-3.9	15.21
8.5	-1.4	1.96
4.5	-5.4	29.16
14.5	4.6	21.16
3.5	-6.4	40.96

As the data is normally distriuted and population sd is not given, we will use t statistics.

t = -1.071

P value = TDIST (t statistics, df, 2) = TDIST (1.071, 9, 1)= 0.156

p value = 0. 156

Step 3:

df = 10- 1 = 9

level of significance = 0.01

t critical = - 2.821

As the t stat (-1.071) does not fall in the rejection area, we fail to reject the Null hyppothesis.

As the p value (0.156) is greater than level of significance (0.01) we fail to reject the Null hypothesis.