The Student's t distribution table gives critical
values for the Student's t distribution. Use an
appropriate d.f. as the row header. For a
right-tailed test, the column header is the value of
α found in the one-tail area row. For a
left-tailed test, the column header is the value of
α found in the one-tail area row, but you must
change the sign of the critical value t to −t.
For a two-tailed test, the column header is the value of
α from the two-tail area row. The critical values
are the ±t values shown.
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 = 8.0 with sample standard deviation s = 1.5.
Use a 5% level of significance to test the claim that the drug has
changed (either way) the mean pH level of the blood. Solve the
problem using the critical region method of testing (i.e.,
traditional method). (Round your answers to three decimal
places.)
| test statistic | = | |
| critical value | = ± |
State your conclusion ion the context of the application.
Fail to reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level
.Fail to reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.
Reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.
Reject the null hypothesis, there is insufficient evidence that the drug has changed the mean pH level.
Compare your conclusion with the conclusion obtained by using the
P-value method. Are they the same?
We reject the null hypothesis using the traditional method, but fail to reject using the P-value method.
We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.
The conclusions obtained by using both methods are the same.
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 7.4
Alternative Hypothesis, Ha: μ ≠ 7.4
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (8 - 7.4)/(1.5/sqrt(36))
t = 2.4
Rejection Region
This is two tailed test, for α = 0.05 and df = 35
Critical value of t are -2.03 and 2.03.
Hence reject H0 if t < -2.03 or t > 2.03
Reject the null hypothesis, there is sufficient evidence that the drug has changed the mean pH level.
P-value = 0.0218
As P-value < 0.05, reject the null hypothesis.
The conclusions obtained by using both methods are the same.
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