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1. A machine produces 5-inch nails. A sample of 12 nails was selected and their lengths
determined. The results are as follows: 4.91 4.97 4.80 4.81 4.93 4.82 4.84 4.81 4.97 4.95
4.82 4.86 Assuming that a = 0.05, test the hypothesis that the population mean is equal to
5.
•State the null and alternate hypotheses
Here the null hypothesis is Ho: mu = 5 and the alternative hypothesis is Ha: mu ≠ 5.
•Calculate the mean and standard deviation
The mean, x = 4.8742 and the standard deviation, s = 0.0671
•Determine which test statistic applies, and calculate it
Since the population standard deviation is unknown and the sample size is small (<30)
we use a t-test.
The test statistic for testing Ho is given by
( x  5)
= (4.8742 – 5)/(0.0671/√12) = -6.4986
t
s/ n
•Determine the critical value(s).
Since a = 0.05, from Student’s t distribution with (n-1) = 11 degrees of freedom the
critical value is given by,
Critical value = ±2.201
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if |t| > 2.201
That is, we Reject Ho if t < -2.201 or t > 2.201
Here, |t| = 6.4986 > 2.201
So we reject the null hypothesis Ho. Thus the population mean is not equal to 5.
2. A sample of size n = 20 is selected from a normal population to construct a 95%
confidence interval estimate for a population mean. The interval was computed to be
(4.30 to 5.20). Determine the sample standard deviation.
The 95% confidence interval for a population mean is given by
( x - t / 2 s / n , x + t / 2 s / n ),
where n = 20, t / 2 = 2.861.
Now, the width of the interval is given by
( x + t / 2 s / n ) - ( x - t / 2 s / n ) = 2 t / 2 s / n
Thus, 2 t / 2 s / n = 5.20 – 4.30 = 0.90
That is, 2*2.861*s/√20 = 0.90
That is, s = (0.90*√20)/(2*2.861) = 0.7034
Thus the standard deviation is s = 0.7034
3. A random sample of 41 observations was selected from a normally distributed
population. The sample mean was x ¯ = 81, and the sample variance was s2 = 25.0. Does
the sample show sufficient reason to conclude that the population standard deviation is
not equal to 6 at the 0.05 level of significance? Use the p-value method.
•State the null and alternate hypotheses
Here the null hypothesis is Ho: sigma = 6 and the alternative hypothesis is Ha: sigma ≠ 6.
•Determine which test statistic applies, and calculate it
Here we use a Chi-Square test.
The test statistic for testing Ho is given by,
(n  1) s 2
2
, follows a Chi-square distribution with (n-1) = 40 d.f.
 
62
Here it is given that, n = 41, s2 = 25.0.
Thus the test statistic is given by,
 2 = (41 -1)*25/36 = 27.7778
•Determine the corresponding probability, and compare to a
The p-value is given by
p-value = P[  2 >27.7778] = 0.9279
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if the p-value < 0.05
Here, p-value = 0.9279 > 0.05.
So we fail to reject Ho.
Thus the sample does not show sufficient reason to conclude that the population standard
deviation is not equal to 6 at the 0.05 level of significance.
4. An insurance company states that 85% of its claims are settled within 5 weeks. A
consumer group selected a random sample of 65 of the company’s claims and found 53
of the claims were settled within 5 weeks. Is there enough evidence to support the
consumer group’s claim that fewer than 85% of the claims were settled within 5 weeks?
Test using the traditional approach with a = 0.02.
•State the null and alternate hypotheses
Here the null hypothesis is Ho: p = 0.85 and the alternative hypothesis is Ha: p < 0.85.
•Calculate the sample proportion
The sample proportion, p = 53/65 = 0.8154
•Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
p  0.85
follows a Standard Normal distribution.
z
0.85(1  0.85) / n
Here, n = 65.
Thus the test statistic is given by
z = (0.8154 – 0.85)/Sqrt[0.85*0.15/65] = -0.7816
•Determine the critical value(s).
Since a = 0.02, from Standard Normal distribution the critical value is given by,
Critical value = -2.05
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if z < -2.05
Here, z = -0.7816 > -2.05
So we fail to reject Ho. Thus there is not enough evidence to support the consumer
group’s claim that fewer than 85% of the claims were settled within 5 weeks
5. A teacher wishes to compare two different groups of students with respect to their
mean time to complete a standardized test. The time required is determined for each
group. The data summary is given below. Test the claim at a = 0.10, that there is no
difference in variance. Give the critical region, test statistic value, and conclusion for the
F test. n1 = 40 s1 = 13 n2 = 40 s2 = 14 a = 0.10
•State the null and alternate hypotheses
Here the null hypothesis is Ho:  12   22 and the alternative hypothesis is Ha:  12   22 .
•Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
s22
, follows F distribution with ( n2  1, n1  1) d.f.
s12
Here it is given that, n1 = 40, s1 = 13, n2 = 40, s2 = 14.
Thus the test statistic is given by,
F
F = (142/132) = 1.1598
•Determine the critical region
Since a = 0.10, from F distribution with (39, 39) degrees of freedom the critical value is
obtained as 1.514
Thus the critical region is F > 1.514
•State your decision: Should the null hypothesis be rejected?
Here F = 1.1598 < 1.514
So we fail to reject the null hypothesis Ho. Thus there is no significant difference in
variance.
6. A machine produces 9 inch latex gloves. A sample of 68 gloves is selected, and it is
found that 42 are shorter than they should be. Find the 98% confidence interval on the
proportion of all such gloves that are shorter than 9 inches.
The 98% confidence interval for the proportion is given by,
p  z / 2 p(1  p) / n , p  z / 2 p (1  p ) / n ,


where z / 2 = 2.326.
Here it is given that, n = 68, x = 42
Therefore, the sample proportion, p = x/n = 42/68 = 0.6176
Thus, the 98% confidence interval on the proportion of all such gloves that are shorter
than 9 inches is given by
(0.6176 – 2.326*√(0.6176*0.3824/68), 0.6176 + 2.326*√(0.6176*0.3824/68))
= (0.6176 – 0.1371, 0.6176 + 0.1371)
= (0.4805, 0.7547)
7. The pulse rates below were recorded over a 30-second time period, both before and
after a physical fitness regimen. The data is shown below for 8 randomly selected
participants. Is there sufficient evidence to conclude that a significant amount of
improvement took place? Assume pulse rates are normally distributed. Test using a =
0.10. Before 33 40 34 37 55 41 52 45 After 34 44 43 39 56 47 52 48
•State the null and alternate hypotheses
Let u1 and u2 denote the mean pulse rate before and after a physical fitness regimen
respectively
Here the null hypothesis is Ho: u1 = u2 and the alternative hypothesis is Ha:u1< u2.
•Calculate the mean and standard deviation
The mean and standard deviation of the pulse rate before physical fitness regimen is
Mean = 42.125 and standard deviation = 8.0434
The mean and standard deviation of the pulse rate after physical fitness regimen is
Mean = 45.375 and standard deviation = 7.0089
The mean and standard deviation of the difference in pulse rate before and after physical
fitness regimen is
Mean = -3.25 and standard deviation = 3.0119
•Determine which test statistic applies, and calculate it
For testing Ho, we use the paired t test.
The test statistic for testing Ho is given by,
t = dbar/(sd/√n), follows a student’s t distribution with (n-1) degrees of freedom.
Here, dbar = -3.25, sd = 3.0119and n = 8.
Thus the test statistic is given by,
t = -3.25 /(3.0119/√8) = -3.052
•Determine the critical value(s).
Since a = 0.10, from Student’s t distribution with (n-1) = 7 degrees of freedom the critical
value is given by,
Critical value = -1.415
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if t < -1.415
Here, t = -3.052 < - 1.415
So we reject the null hypothesis Ho.
Thus there is sufficient evidence to conclude that a significant amount of improvement
took place at 5% significance level.
8. You are given the following data. Test the claim that there is a difference in the means
of the two groups. Use a = 0.01. Group A Group B x ¯_1 = 7 x ¯_2 = 10 s_1 = 12 s_2 =
11 n_1 = 66 n_2 = 92
•State the null and alternate hypotheses
Here the null hypothesis is Ho: u1 = u2 and the alternative hypothesis is Ha: u1 ≠ u2.
•Determine which test statistic applies, and calculate it
Since the sample sizes are large we can use the z test for testing Ho.
The test statistic for testing Ho is given by,
t = (xbar1 – xbar2)/Sqrt[(s1^2)/n1 + (s2^2)/n2] follows a Standard Normal
distribution(approximately).
Here it is given that, xbar1= 7, xbar2 = 10, s1=12 , s2 = 11, n1 = 66, n2 = 92.
Thus the test statistic is given by,
t = (7 – 10)/Sqrt[(12^2)/66 + (11^2)/92] = - 1.6042
•Determine the critical value(s).
Since a = 0.01, from Standard Normal distribution the critical value is given by,
Critical value = ±2.58
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if |t| > 2.58
That is, we reject Ho if t < -2.58or t > 2.58.
Here, |t| = 1.6042 < 2.58
So we fail to reject the null hypothesis Ho. Thus there is no difference in the means of the
two groups.