# 38. A recent article in The Wall Street Journal reported that the 30

```38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is
now less than 6 percent. A sample of eight small banks in the Midwest revealed the
following 30-year rates (in percent):
4.8
5.3
6.5
4.8
6.1
5.8
6.2
5.6
At the .01 significance level, can we conclude that the 30-year mortgage rate for small
banks is less than 6 percent?
Estimate the p-value.
Solution:
Mean of the sample = X = 5.6375
Standard deviation of the sample = σ = 0.63457
Null Hypothesis: H0: 0 = 6 vs. Ha:a &lt; 6
Rejection region can be defined as
z &lt; -z
Here significance level is 0.01,
So, z0.01 = - 2.998 [critical value](Using Statistical Ratio Calculator
significance)
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
x 

n
Where n is the sample size.
Calculating t-test statistics:
z = (5.6375-6)/(0.63457/√ 8) = -1.616
Calculating p-value:
Degree of freedom = DF = 8-1 = 7
P(z8 &lt; -1.616) = 0.075064 (Using http://www.danielsoper.com/statcalc/calc08.aspx
with t = -1.616 and DF = 7)
The p-value is 7.5% which is greater than significance level of 1%. Hence, we fail to reject the
null hypothesis but we can’t conclude that the rates are less than 6%.
32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being overfilled.
Assume the standard deviation of the process is .03 ounces. The quality control department took a
random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5
percent level of significance, can we conclude that the mean weight is greater than 16 ounces?
Determine the p-value.
Solution:
Null Hypothesis: H0: 0 = 16 vs. Ha:a &gt; 10
Rejection region can be defined as
z &gt; z
Here significance level is 0.05,
So, z0.05 = 1.645 [critical value](Using Statistical Ratio Calculator
significance)
Now,
z = (X - μ) / σx
Where X is a normal random variable, μ is the mean, and σ is the standard deviation.
x 

n
Where n is the sample size.
Calculating t-test statistics:
z = (16.05-16)/(0.03/√ 50) = 11.785
Calculating p-value:
Degree of freedom = DF = 50-1 = 49
P(z49 &gt; 11.785) = 0.000000 (Using http://www.danielsoper.com/statcalc/calc08.aspx
with t = 11.785and DF = 49)
The p-value of 0.00% is less than significant level of 5%. Hence, we can reject the null
hypothesis and conclude that the mean weight is greater than 16 ounces.
```