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8. You are given the following data. Test the claim that there is a difference in the means
of the two groups. Use a = 0.05.
Group A Group B
x-bar(sub 1) = 10 x-bar(sub 2) = 2
s(sub 1) = 1.0 s(sub 2) = 0.1
n(sub 1) = 99 n(sub 2) = 52
• State the null and alternate hypotheses
• Determine which test statistic applies, and calculate it
• Determine the critical value(s)
• State your decision, should the null hypothesis be rejected?
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 ≠ μ2.
Since the sample sizes are large we can use the z test for testing Ho.
The test statistic for testing Ho is given by,
z = (xbar1 – xbar2)/Sqrt[(s1^2)/n1 + (s2^2)/n2] follows a Standard Normal
distribution(approximately).
Here it is given that, xbar1= 10, xbar2 = 2, s1= 1.0 , s2 = 0.1, n1 = 99, n2 = 52,
Thus the test statistic is given by,
z = (10 – 2)/Sqrt[(1.0^2)/99 + (0.1^2)/52] = 78.8519
Since a = 0.05, from Standard Normal distribution the critical value is given by,
Critical value = ±1.96
The decision rule is Reject Ho if |z| > 1.96
Here, |z| = 78.8519 > 1.96
So we reject the null hypothesis Ho. Thus there is significant difference in the means of
the two groups.
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
• State the null and alternate hypotheses
• Calculate the mean and standard deviation
• Determine which test statistic applies, and calculate it
• Determine the critical values(s)
• State your decision, Should the null hypothesis be rejected?
Before 33 40 34 37 55 41 52 &n bsp; 45
After 34 44 43 39 56 47 52 &nb sp; 48
Let μ1 and μ2 denote the mean pulse rate before and after a physical fitness regimen
respectively
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 < μ2.
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
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 = 3.0119 and n = 8.
Thus the test statistic is given by,
t = -3.25/(3.0119/√8) = -3.0520
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
The decision rule is Reject Ho if t < -1.415
Here, t = -3.0520 < - 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 10% significance level.
6. A machine produces 9 inch latex gloves. A sample of 69 gloves is selected, and it is
found that 19 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 = 69, x = 19
Therefore, the sample proportion, p = x/n = 19/69 = 0.2754
Thus, the 98% confidence interval on the proportion of all such gloves that are shorter
than 9 inches is given by
(0.2754 – 2.326*√(0.2754*0.7246/69), 0.2754 – 2.326*√(0.2754*0.7246/69))
= (0.2754 – 0.1251, 0.2754 + 0.1251)
= (0.1503, 0.4005)
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 = 60 s1 = 48
n2 = 60 s2 = 65
a = 0.10
• State the null and alternate hypotheses
• Determine which test statistic applies, and calculate it
• Determine the critical region
• State your decision, should the null hypothesis be rejected?
a) Here the null hypothesis is Ho:  12   22 and the alternative hypothesis is Ha:  12   22 .
b) The test statistic for testing Ho is given by,
s2
F  22 , follows F distribution with ( n2  1, n1  1) d.f.
s1
Here it is given that, n1 = 60, s1 = 48, n2 = 60, s2 = 65.
Thus the test statistic is given by,
F = (652/482) = 1.8338
c) Since a = 0.10, from F distribution with (59, 59) degrees of freedom the critical value
is obtained as 1.399
Thus the critical region is F > 1.399
Here F = 1.8338 > 1.399
So we reject the null hypothesis Ho. Thus there is significant difference in variance.
4. An insurance company states that 75% of its claims are settled within 5 weeks. A
consumer group selected a random sample of 45 of the company's claims and found 35 of
the claims were settled within 5 weeks. Is there enough evidence to support the consumer
group's claim that fewer than 75% of the claims were settled within 5 weeks? Test using
the traditional approach with a = 0.05.
• State the null and alternate hypotheses
• Calculate the sample proportion
• Determine which test statistic applies, and calculate it
• Determine the critical value(s)
• State your decision, should the null hypothesis be rejected?
Here the null hypothesis is Ho: p ≥ 0.75 and the alternative hypothesis is Ha: p < 0.75.
The sample proportion, p = 35/45 = 0.7778
Here we use the z-test for proportion.
The test statistic for testing Ho is given by,
p  0.75
follows a Standard Normal distribution.
z
0.75(1  0.75) / n
Here, n = 45.
Thus the test statistic is given by
z = (0.7778 – 0.75)/Sqrt[0.75*0.25/45] = 0.4303
The critical value = -1.645
The decision rule is Reject Ho if z < -1.645
Here, z = 0.4303 > -1.645
So we fail to reject Ho. Thus there is not enough evidence to support the consumer
group’s claim that fewer than 75% of the claims were settled within 5 weeks.
3. A random sample of 41 observations was selected from a normally distributed
population. The sample mean was x-bar = 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
• Determine which test statistic applies, and calculate it
• Determine the corresponding probability, and compare to a
• State your decision, should the null hypothesis be rejected?
Here the null hypothesis is Ho: σ = 6 and the alternative hypothesis is Ha: σ ≠ 6.
We use the Chi-square test for testing Ho.
The test statistic for testing Ho is given by,
(n  1) s 2
, follows a Chi-square distribution with (n-1) = 40 d.f.
2 
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
The p-value is given by
p-value = P[  2 >27.7778] = 0.9279
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.
2. A sample of size n = 20 is selected from a normal population to construct a 90%
confidence interval estimate for a population mean. The interval was computed to be
(8.70 to 9.00). Determine the sample standard deviation.
The 90% confidence interval for a population mean is given by
( x - t / 2 s / n , x + t / 2 s / n ),
where n = 20, t / 2 = 1.729.
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 = 9.00 – 8.70 = 0.30
That is, 2*1.729*s/√20 = 0.30
That is, s = (0.30*√20)/(2*1.729) = 0.3880
Thus the standard deviation is s = 0.3880
1. A machine produces 3-inch nails. A sample of 12 nails was selected and their lengths
determined. The results are as follows:
3.02 2.95 2.96 2.96 2.90 2.92 3.02 2.93 3.07 3.07 3.00 2.91
Assuming that a = 0.05, test the hypothesis that the population mean is equal to 3.
• State the null and alternate hypotheses
• Calculate the mean and standard deviation
• Determine which test statistic applies, and calculate it
• Determine the critical value(s)
• State your decision, Should the null hypothesis be rejected?
Here the null hypothesis is Ho: μ = 3 and the alternative hypothesis is Ha: μ ≠ 3.
The mean, x = 2.9758 and the standard deviation, s = 0.0593
c) Here we use a t-test. The test statistic for testing Ho is given by
( x  3)
= (2.9758 – 3)/(0.0593/√12) = -1.4115
t
s/ n
d) 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
e) The decision rule is Reject Ho if |t| > 2.201
Here, |t| = 1.4115 < 2.201
So we fail to reject the null hypothesis Ho.