Download 1. The P-value for a two sided test of the null hypothesis H0: µ = 30

1. The P-value for a two sided test of the null hypothesis H0: µ = 30 is 0.04.
a) Does the 95% confidence interval include the value 30? Why?
No. A 95% confidence interval corresponds to a hypothesis test with significance level of  
.05. A 95% confidence interval contains the hypothesized value only if the p-value is greater
than .05. If the p-value is less than the .05 level of significance, the confidence interval does not
contain the hypothesized value.
b) Does the 90% confidence interval include the value 30? Why?
No. A 90% confidence interval corresponds to a hypothesis test with significance level of  
.10. A 90% confidence interval contains the hypothesized value only if the p-value is greater
than .10. If the p-value is less than the .10 level of significance, the confidence interval does not
contain the hypothesized value.
2. The level of calcium in the blood in healthy young adults varies with mean about 9.5
milligrams per deciliter and standard deviation about σ= 0.4. A clinic in rural Guatemala
measures the blood calcium level of 160 healthy pregnant women at their first visit for
prenatal care. The mean is
= 9.57. Is this an indication that the mean calcium level in
the population from which these women come differs from 9.5?
I am assuming a significance level of 0.05 for this test.
a) State H0 and Ha.
H 0 :   9.50
H A :   9.50
b) Carry out the test and give the P-value, assuming that σ= .4 in this population. Report
your conclusion.
The test statistic:
z
x  0

n

9.57  9.5
 2.21
0.4
160
If the null hypothesis is true, the test statistic is a randomly chosen observation from a standard
normal distribution
The p-value and decision rule
The p-value is the probability of observing a test statistic as extreme as or more extreme than the
one we observed (assuming the null hypothesis is true).
Expressed another way, it’s the probability that a randomly selected observation from a standard
normal distribution would be “as extreme as or more extreme than” the value of the test statistic.
P-value = 2  P  Z  z   2  P( Z  2.21) = 2  ( 0.013 ) = 0.0269
We should reject the null hypothesis and conclude the alternative is true if the p-value is less
than or equal to the level of significance.
Reject if: p-value  
Conclusion
Since the p-value is less than  = .05 we should reject the null hypothesis and conclude the
alternative is true. There sufficient evidence to conclude that the mean calcium level differs
from 9.5 at the .05 level of significance.
c). Give a 95% confidence interval for the mean calcium level µ in this population. We are
confident that µ lies quite close to 9.5. This illustrates the fact that a test based on a large
sample (n=160 here) will often declare even a small deviation from H0 to be statistically
significant.
This is a large sample 95% confidence interval for a population mean.
The formula for the confidence interval is:

x  z / 2  

n

The values of the variables in the formula are:
x  Sample mean = 9.57
 = Population standard deviation = 0.400
n  Sample size = 160
z / 2 = The 97.5th percentile of the standard normal (Z) distribution = 1.96
When we plug these values into the formula for the confidence interval we get:

x  z / 2  

n


9.57  1.96  

n

where  = 0.400
n  160
9.57  0.062
(9.508, 9.632)
We are 95% confident that  , the mean of the population from which our sample was selected, is
between 9.508 and 9.632.
3. The one-sample t statistic for testing
H0: µ=10
Ha: µ > 10
from a sample of n=20 observations has the value t=2.10.
a) What are the degrees of freedom for this statistic?
There are n 1  20 1  19 degrees of freedom
b) Give the two critical values t* from Table D that bracket t.
Tables can vary from text to text. My table shows on the row for 19 degrees of freedom the
following values that are closest to 2.10.
2.093 and 2.504
c) Between what two values does the P-value of the test fall?
For 19 degrees of freedom, 2.093 corresponds to a p-value of 0.025, and 2.504 corresponds to a
p-value of 0.010
d) Is the value t=2.10 significant at the 5% level? Is it significant at the 1% level?
From question c) we know that the p-value is between .010 and .025. Since this range is less
than .05, the value t = 2.10 is significant at the 5% level. It is not significant at the 1% level
because some of the range is greater than .01.
e) If you have software available, find the exact P-value?
Using the TDIST function in Excel, I found the exact p-value to be 0.02466.