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 = 8.8 with sample standard deviation s = 3.3. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood.

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

Answer 1

Solution :

Given that,

Population mean = = 7.4

Sample mean = = 8.8

Sample standard deviation = s = 3.3

Sample size = n = 36

Level of significance = = 0.05

This is a two tailed test.

The null and alternative hypothesis is,

Ho: 7.4

Ha: 7.4

The test statistics,

t = ( - )/ (s/)

= ( 8.8 - 7.4 ) / ( 3.3 / 36 )

= 2.545

p-value = 0.0155 ( Using t distribution probability table)

The p-value is p = 0.0155 < 0.05 it is concluded that the null hypothesis is rejected.

Conclusion:

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the drug has changed

(either way) the mean pH level of the blood 7.4, at 0.05 significance level.