Let x be a random variable that represents the pH of arteria

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 = 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 31 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.6 with sample standard deviation s = 2.9. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood

    (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) Find the P-value. (Round your answer to four decimal places.)

(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 ?

At the = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

At the = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    

At the = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

At the = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

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

There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood. There is insufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.

Solution

B)

We use a z distrbution because n > 30 here.

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   7.4  
Ha:    u   =/   7.4  
              
As we can see, this is a    two   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    8.6          
uo = hypothesized mean =    7.4          
n = sample size =    31          
s = standard deviation =    2.9          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.303902495 [ANSWER]

*********************

c)          
              
Also, the p value is, as this is two tailed,              
              
p =    0.021228118 [ANSWER]

************************

d)

As P < 0.05, then

OPTION A: At the = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. [ANSWER, A]
************************

e)

There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.