Question

12) 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 (Reference: The Merck Manual, a commonly used reference in medical schools and nursing programs). 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?̅ = 8.1 with a sample standard deviation ? = 1.9. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood.

a) What is the level of significance? State the null and alternative hypotheses.

b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the appropriate sampling distribution of the sample test statistic.

c) Find (or estimate) the P-value.

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

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

Answer 1

Solution:-

12)

a) The level of significance is 0.05.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 7.4
Alternative hypothesis: u 7.4

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

b) We sill use t-distribution and we have assumed that distribution is approximately normal.

Sample size is greater than 30 and population standard deviation is not known, hence we should use t-distribution.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.3413

DF = n - 1

D.F = 30
t = (x - u) / SE

t = 2.05

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesised population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 30 degrees of freedom is less than -2.05 or greater than 2.05.

c)

P-value = P(t < - 2.05) + P(t > 2.05)

Use the calculator to determine the p-values.

P-value = 0.0245 + 0.0245

Thus, the P-value = 0.049

d) Interpret results. Since the P-value (0.049) is less than the significance level (0.05), we have to reject the null hypothesis.

e) From the above test we have sufficient evidence in the favor of the claim that the drug has changed (either way) the mean pH level of the blood.