Answer to Question #269459 in Statistics and Probability for kate

Question

It has been known that a fixed dose of a certain drug results to an average increase of pulse rate by at least 12 beats per minute with a standard deviation of 5. A group of 20 patients given the same dose showed the following increases: 15, 12, 10, 8, 14, 15, 16, 11, 7, 13, 9, 10, 12, 11, 9, 10, 17, 14, 15, 7. Is there proof to show that this group has a lower average increase of pulse rate than the ones in general? Use 0.05 level of significance.

I.                Formulate hypothesis: (20pts)

H₀:

H₁:

 

II.            What is the given level of significance? (5pts)

III.        What is the appropriate test statistics (formula)? (10pts)

IV.        Write the given and your solution (kindly write the CV and TV) 

Answer 1

$$15, 12, 10, 8, 14, 15, 16, 11, 7, 13,$$

$$9, 10, 12, 11, 9, 10, 17, 14, 15, 7.$$

$$\bar{x}=\dfrac{1}{n}\sum _i x_i=\dfrac{235}{20}=11.75$$

$$s^2=\dfrac{1}{n-1}\sum _i (x_i-\bar{x})^2=\dfrac{173.75}{19}$$

$$s=\sqrt{s}=\sqrt{\dfrac{173.75}{19}}\approx3.0240266$$

I. The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq12$

$H_1:\mu<12$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

II. Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

III. The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{11.75-12}{5/\sqrt{20}}\approx-0.2236$$

IV. The 95% confidence interval is

$$CI=(11.75-1.96\times\dfrac{5}{\sqrt{20}},11.75+1.96\times\dfrac{5}{\sqrt{20}})$$

$$=(9.559, 13.941)$$

Critical value is $z_c=-1.6449$

Test value is $z=-0.2236$

Since it is observed that $z = -0.2236 \geq-1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p=P(Z<-0.2236)=0.411534,$ and since $p=0.411534>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is less than $12,$ at the $\alpha = 0.05$ significance level.

Therefore, there is not enough evidence to claim that this group has a lower average increase of pulse rate than the ones in general, at the $\alpha = 0.05$ significance level.