Question

Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 36 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.7 with sample standard deviation s = 3.4. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood.

(a) What is the level of significance?


State the null and alternate hypotheses.

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

What is the value of the sample test statistic? (Round your answer to three decimal places.)

(c) Estimate the P-value.

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

(e) Interpret your conclusion in the context of the application.

Answer 1

Level of significance  

a-0.05

(b) What sampling distribution will you use?

We use t distribution

To Test :-

H0 :-  

μ = 7.4

H1 :-  $\mu \neq7.4$

Test Statistic :-
$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$

t = 2.294

t = (8.7-7.4)/(3.4/V36)

Test Criteria :-
Reject null hypothesis if $| t |\; > \;t_{\alpha /2, n-1}$

a/2,n-1 to.05/2,36- 2.03


리 > taZn-1= 2.2941 > 2.03

Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 2.2941 ) = 0.0279
Reject null hypothesis if P value <    level of significance
P - value = 0.0279 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

a-0.05

There is sufficient evidence to support the claim that the drug has changed (either way) the mean pH level of the blood.