Questions laseResouces Give Up? Feedback Assignment Score: 42.1% Resume Correct
ID: 2946859 • Letter: Q
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Questions laseResouces Give Up? Feedback Assignment Score: 42.1% Resume Correct 100% Question 7 of 19> Attempt 3 1 Attempt 6. Question Correct 3 Attempts Imagine that a product developer at a pharmaceutical company wants to increase the length of time a drug remains in the bloodstream, so he develops a new version of a currently available drug. The half life of the drug (the time the drug takes to reach half its initial concentration), in hours, is used as a measure of how long it persists. The half-life of the currently available drug follows a normal distribution with a mean 11.670 h and a standard deviation of 2.000 h. 10096 The product developer plans to perform a one-sample z-lest using the known standard deviaton of 2.000 h based on data collected from a simple random sample of 25 test subjects. He forms the null hypothesis Ho: ? 11.670 h, where ? is the mean half-life of the new drug in all patients who take it, to represent the assumption that the half-life of the new drug is no different from the current drug. He forms the alternative hypothesis Hi : ? > 1 1.670 h in order to determine if there is cnough evidence to say that the new drug persists in the bloodstream longer than the current drug. If the mcan half-lifc of the new drug in the sample is 13.036 h, compute the one-sample z-statistic for the mcan half-life of the new drug in all patients who take it. Give your answer to three decimal places. 7. Question In Progress 3 Attempts 096 8. Question Correct 3 Attempts 10096 9. Question Correct 4 Altempts 100% 3.673 10. Question 0% Assignment Info ToolsExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: u < 11.67
Alternative hypothesis: u > 11.67
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE),z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.40
z = (x - u) / SE
z = 3.415
where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 3.415.
Thus the P-value in this analysis is less than 0.001.
Interpret results. Since the P-value (0.0001) is less than the significance level (0.05), we have to reject the null hypothesis.
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