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7. Two groups of ten sprinters run 100 meters. The times required by sprinters in the first
group are as follows:
13.7 14.2 13.2 11.3 12.6 10.0 10.1 14.1 13.8 13.2
The times required by sprinters in the second group are as follows:
17.3 10.6 14.0 15.9 15.1 13.4 18.9 17.2 17.6 10.1
Assuming that a = 0.05, test the hypothesis that the means of the two populations are
equal.
· State the null and alternate hypotheses
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 ≠ μ2.
· Calculate the mean and standard deviation for each group
The mean and standard deviation of the first group is
xbar1 = 12.62 and s1 = 1.5943
The mean and standard deviation of the first group is
xbar2 = 15.01and s2 = 2.9786
· Calculate the value of the test statistic.
The test statistic for testing Ho is given by,
t = (xbar1 – xbar2)/{sp*Sqrt[(1/n1)+(1/n2)]} follows a Student’s t distribution with
(n1 + n2 -2) degrees of freedom, where sp = Sqrt{[(n1-1)s1^2+(n2-1)s2^2]/ (n1 + n2 -2)] is
the pooled standard deviation.
Here, sp = Sqrt[(9*1.5943^2 + 9*2.9786^2)/18] = 2.3889
Thus the test statistic is given by,
t = (12.62 – 15.01)/ {2.3889*Sqrt[(1/10 + 1/10)]} = -2.2371
· Determine the critical value(s).
Since a = 0.05, from Student’s t distribution with (n1 + n2 -2) = 18 degrees of freedom the
critical value is given by,
Critical value = ±2.101
· State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if |t| > 2.101
Or Reject Ho if t < -2.101 or t > 2.101.
Here, |t| = 2.2371 > 2.101
So we reject the null hypothesis Ho.
1. A machine produces 5-inch nails. A sample of 12 nails was selected and their lengths
determined. The results are as follows:
5.01 5.00 4.93 4.95 4.98 5.05 5.01 5.05 4.97 5.08 5.03 4.91
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: μ = 5 and the alternative hypothesis is Ha: μ ≠ 5.
· Calculate the mean and standard deviation
Sample mean, xbar = 4.9975
Sample standard deviation, s = 0.0515
· Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
t = (xbar - 5)/(s/√n), follows a Student’s t distribution with (n-1) d.f..
Thus the test statistic
t = (4.9975 - 5)/(0.0515/√12) = -0.1680
· 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| = 0.1680 < 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 98%
confidence interval estimate for a population mean. The interval was computed to be
(8.90 to 12.90). Determine the sample standard deviation
The 98% confidence interval estimate for a population mean is given by
( x  t / 2 s / n , x  t / 2 s / n ),
where t / 2 = 2.539 (From Student’s t distribution with 19 df).
Here it is given that, x  t / 2 s / n = 8.90 and x  t / 2 s / n = 12.90
Subtracting the first equation from the second we get,
( x  t / 2 s / n ) – ( x  t / 2 s / n ) = 12.90 – 8.90
That is, 2 t / 2 s / n = 4
That is, s = 4* n /(2 t / 2 ) = 4*√20/(2*2.539) = 3.5221
Thus the sample standard deviation is 3.5221
3. A random sample of 51 observations was selected from a normally distributed
population. The sample mean was x = 49, and the sample variance was s2 = 34.0. Does
the sample show sufficient reason to conclude that the population standard deviation is
not equal to 7 at the 0.02 level of significance? Use the p-value method.
· State the null and alternate hypotheses
Here the null hypothesis is Ho: σ = 7 and the alternative hypothesis is Ho: σ ≠ 7
· Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
χ2 = (n-1)s2/72, follows a Chi-square distribution with (n-1)= 50 degrees of
freedom.
Thus the test statistic
χ2 = (51 -1)(34.0)/72 = 50*34/49 = 34.6939
· Determine the corresponding probability, and compare to a
The p-value of the test is given by
p-value = P[χ2 > 34.6939] = 0.9509
Here it is given that, a = 0.02
Thus, p-value > a
[The p-value is obtained using the Excel formula =CHIDIST(34.6939,50)]
· State your decision: Should the null hypothesis be rejected?
The decision rule is reject Ho if the p-value < 0.02
Here, p-value > 0.02
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 7 at the 0.02 level of significance.
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 α = 0.05.
· State the null and alternate hypotheses
Here the null hypothesis is Ho: p = 0.75 and the alternative hypothesis is Ha: p < 0.75.
· Calculate the sample proportion
The sample proportion, pbar = x/n = 35/45 = 0.7778
· Determine which test statistic applies, and calculate it
The test statistic for testing Ho is given by,
z = (pbar – 0.75)/Sqrt(0.75*0.25/n), follows Standard Normal distribution
Thus the test statistic,
z = (0.7778 – 0.75)/Sqrt(0.75*0.25/45) = 0.4303
· Determine the critical value(s).
Since a = 0.05, the critical value is given by,
Critical value = - 1.645.
· State your decision: Should the null hypothesis be rejected
The decision rule is Reject Ho if z < -1.645
Here, z = 0.4303 > -1.645
So we fail to reject the null hypothesis Ho.
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 = 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 = 45
n2 = 60 s2 = 83
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,
s2
F  22 , follows F distribution with ( n2  1, n1  1) d.f.
s1
Here it is given that, n1 = 60, s1 = 45, n2 = 60, s2 = 83.
Thus the test statistic is given by,
F = (832/452) = 3.4020
· Determine the critical region
Since a = 0.10, from F distribution with (59, 59) degrees of freedom the critical value is
obtained as 1.395
Thus the critical region is F > 1.395
[The critical value is obtained using the Excel formula =FINV(0.1,60,60)]
· State your decision: Should the null hypothesis be rejected?
The decision rule is Reject Ho if F > 1.395
Here F = 3.4020 > 1.395
So we reject the null hypothesis Ho.
6. A machine produces 9 inch latex gloves. A sample of 73 gloves is selected, and it is
found that 58 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 = 73 x = 58
Therefore, the sample proportion, p = x/n = 58/73 = 0.7945
Thus, the 99% confidence interval on the proportion of all such gloves that are shorter
than 9 inches is given by
(0.7945– 2.576*√(0.7945*0.2055/73), 0.7945 + 2.576*√(0.7945*0.2055/73))
= (0.7945 – 0.1218, 0.7945 + 0.1218)
= (0.6727, 0.9163)