The mean serum-creatinine level measured in 12 patients 24 hours after they rece

Question

The mean serum-creatinine level measured in 12 patients 24 hours after they received a newly proposed antibiotic was 1.2 mg/d 1. If the mean and standard deviation of serum creatinine in the general population are 1.0 and 0.4 mg/dL, respectively, then, using a significance level of.05, test whether the mean serum-creatinine level in this group is different from that of the general population 2. What is the p-value for the test? 3. Suppose 1.52 and a on one-sumple t test is performed basedo even subjects. What is the . Suppose the sample standard deviation of serum creatinine is now 0.6 mg/d Asthat the two-tailed p-value? standard deviation of serum creatinine is not known. Perform the hypothesis test andreport a p-value 5. Compute a two-sided 95% CI for the true mean serum-creatinine level 6. How docs your answer to part 5 relate to your answer to part 4?

Explanation / Answer

1. So here we have the sample mean X'=1.2 and sample size n=12

while the population mean = 1.0 and population sd =0.4

We have to perform the Z test at 95% conf level or 0.05 Significance so., z score is +/-1.96 (From z table look for Z value at the probability of 0.025 &o.975). so Z score must be between -1.96 & +1.96 for the Null hypothesis to be not rejected.

So here Null hypothesis Ho: There is no difference in the sample and population mean

& so the Alternative hypotheisis H1:There is significant difference in the sample and population mean

now using the formula Z=(X'-mu) /(Sd/sqrt(n))

so., Z= (1.2-1)/(0.4/sqrt(12)) =1.732051 i.e. between -1.96 and +1.96 and so we say that we can not reject the Null hypothesis i.e. There is no difference in the sample and population mean so., the sample and population mean are same at 0.05significance

2) to find the p value we need to find the probability from the z table i.e. by loooking the Z value for 1.73 we get 0.9581849 and so the p value is = 2*(1- 0.9581849 ) = 0.08363028

3) Since for 1 sample t test we get t=-1.52 @ degrees of freedom=6 we get p value = 0.08966101 but we have 2 tail test and so look for the t value in the table at probability of (0.08966101/2) &1-(0.08966101/2) and we get -2.021925 & 2.021925 and so these are the t values for the 2 tail test

4) t=(X'-mu)/(sd/sqrt(n)) = (1.2-1)/(0.6/sqrt(12))=1.154701 and from the t table at df=11 we get t value=+/- 2.200985

while 1.154701 is between +/- 2.200985 and so we can not reject the Ho i.e. mean of the sample and population are same.

Now to find the p value we need to find the area from the tvalue of 1.1547 i.e. 0.8635216 and so the p value is (1-0.8635216)*2 =  0.2729568

Hope the above explaination has helped you in understanding the problem. Please upvote if it has really helped. Good Luck!!!

Due to time crunch have answered first 4 questions. Hope you understand.

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