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9.
H0 :   6.5 Vs. Ha :   6.5
10.

11. A machine produces 3-inch nails. A sample of 12 nails was selected and the lengths
determined. The results are as follows:
2.89
2.95
3.00
3.05
2.99
2.96
3.10
3.06
3.00
3.12
3.00
2.95
Use these results to test H0: u = 3 and Ha: u not equal to 3 at alpha = 0.05. Give the critical
region, the computed test statistic, and the conclusion. (6 pts)
Ans: H0 : M’ = 3
Ha : M’ not equal to 3
X bar = 3
s = 0.0667
Test statistic = 3 – 3 / 0.0067 = 0
Since test statistic is 0, this means that sample mean is = to population mean. SO, null
hypothesis is accepted. i.e. M’ = 3
12.
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 (8.20 to
9.80). Determine the sample standard deviation.
(4 pts)
Ans: The formula for confidence intervals can be written as :
Upper Limit = M’ + z*Standard error
Lower limit = M’ – z * Standard error
For 95% confidence interval, we have 0.475 probability in one tail, so z = 1.96
Standard error = s/sqrt( n) where, s = sample sd.
So, 9.80 – 8.20 = M’ + z*standard error – M’ + z*standard error
Or, 1.6 = 2*1.96*s/sqrt(20)
s = 1.825
13.
A random sample of 46 observations was selected from a normally distributed
population. The sample mean was = 81, and the sample variance was s2 = 35.0. Does the
sample show sufficient reason to conclude that the population standard deviation is not equal
to 7 at the 0.05 level of significance? Use the p-value method. (6 pts)
Ans: H0: Pop sd = 7
Ha : Pop sd not equal to 7
Test statistic i.e. chi square = (n-1)sample var/pop var
= (81-1)35/49 = 57.143
I cannot proceed further as chi square tables will not give the p value for such high levels of
statistic.
14.
An insurance company states that 75% of its claims are settled within 5 weeks. A
consumer group selected a random sample of 50 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 alpha = 0.05 (6 pts)
Ans: Ho : P = 0.75
Ha : P < 0.75
Given : n=50
P bar = 35/50 = 0.7 (to be tested)
Significance level = 0.05
So, standard error of p bar = sqrt (P*Q/n) = sqrt (0.75*0.25/50) = 0.0612
Z – value = p bar – P / standard error = 0.7 – 0.75 / 0.0612 = - 0.8165
The critical z value = - 1.64
So, the z value lies in the acceptance region, so we do not reject the null hypothesis.
So, there is not enough evidence to support that fewer than 75% claims are settled within 5
weeks.
15.
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 alpha = 0.05, that there is no difference
in variance. Give the critical region, test statistic value, and conclusion for the F test. (6
points)
n1=60, s1=24
n2=120, s2=28
alpha= 0.05
Ans: Ho : σ1 2 = σ2 2
Ho : σ1 2 ≠ σ2 2
F statistic value = s22 / s12 = 282 / 242 = 1.3611
F (0.05, 119, 59) = this figure is not available in any standard Ftable so I am taking the value
of F(0.05,120,60) = 1.47. So, our critical region is above 1.47. The test statistic lies in the
acceptance region so we do not reject the null hypothesis. i.e. there is no difference in the
variance.
16.
A machine produces 9 inch latex gloves. A sample of 80 gloves is selected, and it is
found that 20 are shorter than they should be. Find the 99% confidence interval on the
proportion of all such gloves that are shorter than 9 inches. (4 pts)
Ans: Sample proportion of gloves less than 9 inches = 20/80 = 0.25
n = 80
standard error = sqrt (0.25*0.75/80) = 0.0484
Critical value for 0.495 in one tail, z = 2.58
Upper limit = p bar + z*standard error = 0.25 + 2.58*0.0484 = 0.3749
Lower limit = p bar - z*standard error = 0.25 - 2.58*0.0484 = 0.1251
The 99% confidence interval on the proportion of gloves shorter than 9 inches are 37.49% to
12.51%
17.
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 10 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 alpha = 0.05. (6 pts)
Before
31
25&nbs p; 30
31
33
25
32
45
After
32
29 &nbs p; 30
37
40
35
34
52
Data is missing
18.
You are given the following data. Test the claim that there is a difference in the means
of the two groups. Use alpha = 0.05. (6 pts)
Group A
Xbar1 = 3
s1 = 0.5
n1=100
Group B
xbar2 = 4
s2 = 4
n2 = 40
Ans: Ho : M’1 = M’2
Ha : M’1 not equal to M’2
X1 bar – x2 bar = 3 – 4 = -1 (to be tested)
Standard error = sqrt ( 0.52/100 + 42/40) = 0.6344
Test statistic = -1 – 0 / 0.6344 = - 1.58
Acceptance region = -1.96 to 1.96
Since the test statistic lies in the acceptance region, we accept the null hypothesis i.e. there is
no difference in means.
19.
a.
Determine the p-value for each of the following hypothesis-testing situations: (4 pts)
H0: p = 0.30 and Ha: p not equal to 0.30; z test value = 1.65
Ans: for z = 1.65 , p value = 0.05
b.
H0: u greater than/equal to 30 and Ha: u < 30; t test value = - 1.7 d.f. = 15
Ans: For t = - 1.7, p value = 0.05