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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
Here the null hypothesis is Ho: μ = 3 and the alternative hypothesis is Ha: μ ≠ 3.
•Calculate the mean and standard deviation
Sample mean, xbar = 2.9758
Sample standard deviation, s = 0.0593
•Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
t = (xbar - 3)/(s/√n), follows a Student’s t distribution with (n-1) d.f..
Thus the test statistic
t = (2.9758 - 3)/(0.0593/√12) = -1.4115
•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
Or Reject Ho if t < -2.201 or t > 2.201.
Here, |t| = 1.4115 < 2.201
So we fail to reject the null hypothesis Ho.
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 estimate for a population mean is given by
( x  t / 2 s / n , x  t / 2 s / n ),
where t / 2 = 1.729 (From Student’s t distribution with 19 df).
Here it is given that, x  t / 2 s / n = 8.70 and x  t / 2 s / n = 9.00
Subtracting the first equation from the second we get,
( x  t / 2 s / n ) – ( x  t / 2 s / n ) = 9.00 – 8.70
That is, 2 t / 2 s / n = 0.30
That is, s = 0.30* n /(2 t / 2 ) = 0.30*√20/(2*1.729) = 0.3880
Thus the sample standard deviation is 0.3880
3. A random sample of 41 observations was selected from a normally distributed
population. The sample mean was x bar = 60, and the sample variance was s2 = 21.0.
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: σ = 6 and the alternative hypothesis is Ho: σ ≠ 6
•Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
χ2 = (n-1)s2/62, follows a Chi-square distribution with (n-1)= 40 degrees of
freedom.
Thus the test statistic
χ2 = (41 -1)(21.0)/62 = 40*21/36 = 23.3333
•Determine the corresponding probability, and compare to a
The p-value of the test is given by
p-value = P[χ2 > 23.3333] = 0.9836
Here it is given that, a = 0.05
Thus, p-value > a
[The p-value is obtained using the Excel formula =CHIDIST(23.3333,40)]
•State your decision: Should the null hypothesis be rejected?
The decision rule is reject Ho if the p-value < 0.05
Here, p-value > 0.05
So we fail to reject the null hypothesis 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 85 of the company’s claims and found 26
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, pbar = x/n = 26/85 = 0.3059
•Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
z = (pbar – 0.85)/Sqrt(0.85*0.15/n), follows Standard Normal distribution
Thus the test statistic,
z = (0.3059 – 0.85)/Sqrt(0.85*0.15/85) = -14.0491
•Determine the critical value(s).
Since a = 0.02, the critical value is given by,
Critical value = - 2.054.
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if z < -2.054
Here, z = -14.0491 < -2.054
So we reject the null hypothesis Ho.
Thus there is 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 = 24 s1 = 21 n2 = 40 s2 = 39 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
F  2 , follows F distribution with ( n2  1, n1  1) d.f.
s1
Here it is given that, n1 = 24, s1 = 21, n2 = 40, s2 = 39.
Thus the test statistic is given by,
F = (392/212) = 3.449
•Determine the critical region
Since a = 0.10, from F distribution with (39, 23) degrees of freedom the critical value is
obtained as 1.658
Thus the critical region is F > 1.658
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if F > 1.658
Here F = 3.449 > 1.658
So we reject the null hypothesis Ho.
6. A machine produces 9 inch latex gloves. A sample of 17 gloves is selected, and it is
found that 5 are shorter than they should be. Find the 99% confidence interval on the
proportion of all such gloves that are shorter than 9 inches.
The 99% 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 = 2.576.
Here it is given that, n = 17, x = 5
Therefore, the sample proportion, p = x/n = 5/17 = 0.2941
Thus, the 99% confidence interval on the proportion of all such gloves that are shorter
than 9 inches is given by
(0.2941 – 2.576*√(0.2941*0.7059/17), 0.2941 + 2.576*√(0.2941*0.7059/17))
= (0.2941 – 0.2847, 0.2941 + 0.2847)
= (0.0095, 0.5788)
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 =
0.05.
•State the null and alternate hypotheses
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.
•Calculate the mean and standard deviation
The mean and standard deviation of the pulse rate before physical fitness regimen is
Mean = 42.5 and standard deviation = 8.2635
The mean and standard deviation of the pulse rate after physical fitness regimen is
Mean = 46.25 and standard deviation = 8.8277
The mean and standard deviation of the difference in pulse rate before and after physical
fitness regimen is
Mean = -3.75 and standard deviation = 9.6474
•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 = -3.75, sd = 9.6474 and n = 8.
Thus the test statistic is given by,
t = -3.75/(9.6474 /√8) = -1.0994
•Determine the critical value(s).
Since a = 0.05, from Student’s t distribution with (n-1) = 7 degrees of freedom the critical
value is given by,
Critical value = -1.895
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if t < -1.895
Here, t = -1.0994 > - 1.895
So we fail to reject the null hypothesis Ho.
Thus there is no sufficient evidence to conclude that a significant amount of improvement
took place.
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.
•State the null and alternate hypotheses
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 ≠ μ2.
•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= 10, xbar2 = 2, n1 = 99 s1=1.0 , n2 =52, s2 = 0.1
Thus the test statistic is given by,
t = (10 – 2)/Sqrt[(1.0^2)/99 + (0.1^2)/52] = 78.8519
•Determine the critical value(s).
Since a = 0.05, from Standard Normal distribution the critical value is given by,
Critical value = ±1.96
•State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if |t| > 1.96
Or Reject Ho if t < -1.96 or t > 1.96.
Here, |t| = 78.8519 > 1.96
So we reject the null hypothesis Ho.