Solution:
Using statistical software output of this problem is,
One sample T hypothesis test:
A)
μ : Mean of variable
H0 : μ = 16
HA : μ < 16
Hypothesis test results:
Variable | Sample Mean | Std. Err. | DF | T-Stat | P-value |
---|---|---|---|---|---|
var1 | 11.4 | 1.4734691 | 9 | -3.1218843 | 0.0061 |
Test statistic
t = -3.12
P-value = 0.006
The p-value is p = 0.006 < 0.10, it is concluded that the null hypothesis is rejected.
Conclusion:
Reject H0. There is sufficient evidence to conclude that the mean lead concentration for all such medicines is less than 16 mglg.
Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.10...
Listed below are the lead concentrations in ug/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 17 pg/g. Assume that the sample is a simple random sample. 17 8 9 13.5 11 8.5 22.5 15.5 14.5 3.5 Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? O A. Ho = 17 pg/g OB. Ho...
Listed below are the lead concentrations in ug/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. 15.5 16.5 16 4.5 32.5 8.5 13.5 7.5 22 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...
Listed below are the lead concentrations in ug/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 16 ug/g. Assume that the sample is a simple random sample. 22 12.5 6 8.5 12 3 14.5 22 10 4 D Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? O A. Ho: u = 16 ug/g...
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