Question

4. Long-lasting drugs: One of the ways in which doctors try to determine how long a single dose of pain reliever will provide relief is to measure the drug's half-life, which is the length of time it takes for one-half of the dose to be eliminated from the body. A report of the National Institutes of Health states that the standard deviation of the half-life of the pain reliever oxycodone is =1.43 hours. Assume that a sample of 25 patients is given the drug, and the sample standard deviation of the half-lives was s=1.5 hours. Assume the population is normally distributed. Can you conclude that the true standard deviation is greater than the value reported by the National Institutes of Health? Use the =0.01 level of significance.

Solve without using excel or MINITAB function. Create table

Answer 1

The provided sample variance is s^2 = 2.25 or sd = 1.5

and the sample size is given by n = 25.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: $\sigma^2 =2.0449\ or\ \sigma =1.43$

Ha: $\sigma^2 > 2.0449\ or\ \sigma > 1.43$

This corresponds to a right-tailed test, for which a Chi-Square test for one population variance will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01,

and the rejection region for this right-tailed test is

R={χ2:χ2>42.98}.

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

2 (n-1)s 02 (25 – 1). 2.25 2.0449 = 26.407

(4) Decision about the null hypothesis

Since it is observed that

χ2=26.407≤χU2​=42.98,

it is then concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected.

Therefore, there is not enough evidence to claim that the population variance σ2 or(standard deviation) is greater than 2.0449, at the 0.01 significance level.

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