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 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.)

Answer 1

Answer:

2.304

Step-by-step explanation:

Given that x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood).

$n=31\\\bar x = 8.6\\s=2.9\\se = \frac{2.9}{\sqrt{31} } \\=0.5209$

$H_0: \bar x =7.4\\H_a: \bar x \neq 7.4$

(Two tailed test at 5% level)

Mean difference = $8.6-7.4=1.2$

b) Here we can use only t test since population std deviation is not known.

df = n-1 = 30

Test statistic t = mean diff/std error = 2.304

p value = 0.028