Question

In: Statistics and Probability

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

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 (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 36 patients with arthritis took the drug for 3 months. Blood tests showed that = 6.5 with a sample standard deviation s = 1.9. x   
Do the data indicate that the mean pH level of the blood of the patients who used the arthritis drug is different from (either way) μ = 7.4? Use a 0.05 level of significance.

a) Which distribution applies: the Standard Normal or the t distribution? Why?

b) What is the value of the level of significance? State the null and alternate hypotheses. Is it a left-tailed, right-tailed, or two-tailed test?

c) Compute the value of the sample test statistic (either t* or z*).

d) Find the critical value(s). Sketch the sampling distribution and show the critical value(s) and region(s).

e) Based on your answers in parts (a) to (d), will you reject or fail to reject the null hypothesis at the given level of significance?

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

Solutions

Expert Solution

Let be the true mean pH level of the blood of the patients who used the arthritis drug. We want to test if the data indicate that the mean pH level of the blood of the patients who used the arthritis drug is different from (either way) μ = 7.4. That is we want ot test if

We have the following sample information

n=36 is the sample size

is the sample mean pH level of the blood

is the sample standard deviation of pH level

We do not know the population standard deviation and hence we will estimate it using the sample.

The estimated standard deviation is

The estimated standard error of mean is

a) Which distribution applies: the Standard Normal or the t distribution? Why?

The sample size n=36 is greater than 30, hence using the central limit theorem, we can say that the sample means have normal distribution.

ans: The standard normal distribution applies. This is due the fact that the sample size is greater than 30 and hence using the central limit theorem, we can say that the sampling distribution of  means is normal.

b) What is the value of the level of significance? State the null and alternate hypotheses. Is it a left-tailed, right-tailed, or two-tailed test?

We have been asked to use a 0.05 level of significance.

ans:

The level of significance is

Let be the true mean pH level of the blood of the patients who used the arthritis drug.

The hypotheses are

This is a two-tailed test. (because the alternative hypothesis has "not equal to")

c) Compute the value of the sample test statistic (either t* or z*).

The hypothesized value of mean pH level is

The test statistic is

ans: The value of the sample test statistic is

d) Find the critical value(s). Sketch the sampling distribution and show the critical value(s) and region(s).

This is a 2 tailed-test. The right tail critical value for is

Using the standard normal table, we can get for z=1.96, P(Z<1.96)=0.975

ans: The critical values are -1.96, +1.96

e) Based on your answers in parts (a) to (d), will you reject or fail to reject the null hypothesis at the given level of significance?

We will reject the null hypothesis, if the test statistic lies in the critical region.

Here, the test statistic is -2.84 and it does not lie with in the interval -1.96 to +1.96. That is, the test statistic lies in the rejection region. Hence we will reject the null hypothesis.

ans: We will reject the null hypothesis, at the given level of significance.

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

ans: There is sufficient evidence to support the claim that the mean pH level of the blood of the patients who used the arthritis drug is significantly different from (either way) μ = 7.4.


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