Question

Listed below are the lead concentrations in pg/g measured in different traditional medicines. Use a 0.10 significance level to test the claim that the mean lead concentration for all such medicines is less than 15 ug/g. Assume that the sample is a simple random sample 21.5 5.5 20.5 7.5 4 11 8.5 3 10.5 13.5 Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? O A. Ho: p= 15 ug/g H 15 ug/g OB. Ho: = 15 pg/g H > 15 pg/g 0 D. Ho- = 15 ugg H:< 15 pg/g OC. Hou > 15 ug/g H: <15 pg/g Determine the test statistic. (Round to two decimal places as needed.) Determine the P-value. | (Round to three decimal places as needed.) State the final conclusion that addresses the original claim. Ho. There is evidence to conclude that the mean lead concentration for all such medicines is 15 pg/g

Answer 1

Given the claim that the mean lead concentration in all medicines is less than $\mu$ = 15, hence based on the claim the hypotheses are:

D).

Ho:μ = 15

$Ha: \mu <15$

Based on the claim it will be a left-tailed test since the population standard deviation is unknown hence t-distribution is applicable for hypothesis testing which takes degree of freedom for P-value calculation which is calculated as df = n-1= 10-1= 9.

Now we need to find the mean and the sample standard deviation for the given n= 10 samples as;

N 1 x Σ Xi N4 i=1

Mean = (3 + 4 + 5.5 + 7.5 + 8.5 + 10.5 + 11 + 13.5 + 20.5 + 21.5)/10
= 105.5/10
Mean = 10.55

and the sample standard deviation as:

N S = 1 N - 1 (xi – T)? i=1

= √(1/10 - 1) x ((3 - 10.55)² + ( 4 - 10.55)² + ( 5.5 - 10.55)² + ( 7.5 - 10.55)² + ( 8.5 - 10.55)² + ( 10.5 - 10.55)² + ( 11 - 10.55)² + ( 13.5 - 10.55)² + ( 20.5 - 10.55)² + ( 21.5 - 10.55)²)
= √(1/9) x ((-7.55)² + (-6.55)² + (-5.05)² + (-3.05)² + (-2.05)² + (-0.050000000000001)² + (0.45)² + (2.95)² + (9.95)² + (10.95)²)
= √(0.1111) x ((57.0025) + (42.9025) + (25.5025) + (9.3025) + (4.2025) + (0.0025000000000001) + (0.2025) + (8.7025) + (99.0025) + (119.9025))
= √(0.1111) x (366.725)
= √(40.7431475)
= 6.3834

now the test statistic is calculated as:

X - μo t= 10.55 - 15 6.3834/10 2.204 8/η

t = 2.20

Rejection region:

Based on the given significance level as 0.10 reject Ho if P-value is less than 0.10.

P-value:

The p-value is computed using the excel formula for t-distribution which takes the degree of freedom and the calculated test statistic as parameters.

The excel formula used is =T.DIST(-2.204, 9, TRUE), thus the P-value is computed as 0.027.

Conclusion:

Since the P-value is less than 0.10 hence we can reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean is less than 15.

Reject Ho, There is sufficient evidence to conclude that the mean lead concentration is less than 15