Question

Listed below are the lead concentrations in μg/g measured in different traditional medicines. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 18 μg/g. Assume that the lead concentrations in traditional medicines are normally distributed.

4 20 10.5 20.5 10.5 9 9 20.5 12 15

Answer #1

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

H0: µ = 18 versus Ha: µ < 18

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 18

Xbar = 13.65555556

S = 5.763486599

n = 9

df = n – 1 = 8

α = 0.05

Critical value = -1.8595

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (13.65555556 – 18)/[ 5.763486599/sqrt(9)]

t = -2.2614

P-value = 0.0268

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the mean lead concentration for all such medicines is less than 18 μg/g.

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