ENGI 3423 Probability and Statistics

Faculty of Engineering and Applied Science
2010 Fall

Additional Exercises

[Topics after Test 2]

[Note: The "Symbol" font is used in several places in these solutions.
Some solutions will not display correctly unless the "Symbol" font is installed on your computer.
Click here to download the "Symbol" font.]

A production process gives components whose strengths are normally distributed with a mean of 400 N and a standard deviation of 11.5 N. A modification is made to the process which cannot reduce but may increase the mean strength. It may also change the variance. The strengths of nine randomly selected components from the modified process, in Newtons, are:

396 402 409 409 414 398 394 436 418

Test, at a five per cent level of significance, the hypothesis that the mean strength has not increased.

Let X = component strength (N)
X ~ N(µ, σ²)

Test : µ = 400 vs. : µ > 400 at α = .05
σ² is unknown for the new process.

Data:
n = 9 åx = 3676 åx²= 1502838 a = 0.05

s^2 = 12566/72 = 174.527... s = 13.21089617

Method 1

c = muo + t*s/sqrt{n} = 400 + 1.85*4.40 = 408.188...

\ Reject in favour of at a = 0.05

Method 2

t = (xBar - muo) / (s/sqrt{n})
= 8.444 / 4.40 = 1.917
tc = 1.859 ; tobs > tc
\ Reject .

Method 3

From t-tables, t_{.05, 8} = 1.859... and t_.025,8 = 2.306...
.025 < P[T>t_obs] < .05 and a = .05
[From appropriate software, P[T>t_obs] = .045 < a]
\ Reject .

Confidence Interval Method

c = xBar - t*s/sqrt{n} = 408.4 - 1.85*4.40
= 400.25...
\ we are “95% sure that µ > 400.25...”.
µ = 400 is not inside this CI.
\ Reject .

At a = 5%, there is sufficient evidence to conclude that

there has been an increase in mean strength.

An Excel spreadsheet file and a Minitab project file are also available to illustrate this solution.

A transport firm is very suspicious of the tyre company’s claim that the average lifetime of its tyres is at least 45,000 kilometres. The transport company decides to check this claim by fitting forty of these tyres to its trucks, the tyres being a random sample.

A mean lifetime of 44,164 km with a sample standard deviation of 2,106 km is observed. What may the transport firm conclude about the claim at a level of significance of one per cent?

The burden of proof is on the transport firm, (to disprove the tyre company’s claim).

Test:
: µ > 45000 (tyre company’s claim.)
vs.
: µ < 45000
at a = .01

Data:
n = 40 s = 2106

Method 1

\ Reject in favour of .

OR

Method 2

$tobs = (xBar - muo) / (s/sqrt{n}) = (44164-45000)/(2106/sqrt{40}) = -2.51... t_.01,39 = 2.42 tobs < -t_.01,39$
\ Reject .

OR

Method 3

t_obs = –2.51...
From t-tables, t_{.01, 39} = 2.425... , P[T < t_obs] < 0.01 (= a)
[From appropriate software, P[T < t_obs] = 0.008... < a]
\ Reject .

OR

Confidence Interval Method

The one-sided 99% confidence interval estimate for µ is µ < 44972 km.
This CI does not include µ_o = 45000
\ Reject .

\ the transport company does have sufficient evidence to dispute the tyre company’s claim.

An Excel spreadsheet file is also available to illustrate this solution.

A garage wants to know if a more expensive type of radial tyre has a tread life significantly more than 10,000 km beyond the tread life of a cheaper bias-ply tyre. Only if this is the case will the garage invest in the more expensive type of tyre. A random sample of forty tyres of each type is tested and the tread lives are measured. The radial tyres have a mean tread life of 36,500 km with a standard deviation of 2,200 km, while the bias ply tyres have a mean tread life of 23,800 km with a standard deviation of 1,500 km.

Based on these data, should the garage invest in the radial tyres?

Let X_R = life of a randomly chosen radial tyre
and X_B = life of a randomly chosen bias-ply tyre
The burden of proof is on µ_R > µ_B + 10000
Free choice of a. I choose a = 0.01

Test:
: µ_R – µ_B = 10000 (or less)
vs.
: µ_R – µ_B > 10000

Data: n_R = 40 s_R = 2200

n_B = 40 s_B = 1500

An unpaired two-sample t-test is appropriate;
measurements are on two different sets of individuals,
(not pairs of measurements on the same one set of individuals).

Assume that
(The values of s_R & s_B lend some support to this assumption)
[The sample sizes are just large enough to use the two sample z test instead, which does not require any assumption about the two population variances. The eventual decision will be the same.]
Pooled sample variance:

Method 1

$c = (muR - muB)o + t_.01,78 * sP * sqrt{1/nB + 1/nR} = 10000 + 2.37... * sqrt{3545000 * (1/40 + 1/40)} = 10999.4...$

[t_{.01, 78} has been approximated by t_{.01, 80} from the tables and is used in this solution.
A reasonable level of precision would remain if t_{.01, 78} were to be approximated by t_{.01, ¥} = z_.01.]

\ Reject in favour of

YES the garage should invest in the radial tyres.

OR

Method 2

$t_obs = (xBarR - xBarB) - (muR - muB)o / (sP * sqrt{1/nR + 1/nB}) = (12700 - 10000) / (sqrt{3545000}*sqrt{2/40}) = 6.41... t_.01,78 approx= 2.37... t_obs > t_.01,78$
\ Reject

OR

Method 3

t_obs = 6.41...
From the t-tables, P[T > t_obs] << .005
[From appropriate software, P[T > t_obs] » 5 × 10^–9]
\ Reject at any reasonable a.

OR

Confidence Interval Method

The one sided CI for µ_R – µ_B has its boundary at:
$c = (xBarR - xBarB) - t_.01,78 * sP * sqrt{1/nR + 1/nB}$
= 12700 – 2.37...×421.0... = 11700.5...
This 99% CI for µ_R – µ_B is µ_R – µ_B > 11701 (to nearest km).
The CI does not include 10000.
\ Reject .

An Excel spreadsheet file is also available to illustrate this solution.

A particular type of motor is known to have an output torque whose range in normal operation follows a normal distribution. Seven motors are chosen at random and are tested with the old and new methods of controlling the range of torque values. The results of the tests are as follows:

Motor: 1 2 3 4 5 6 7

New method: 5.25 3.16 4.43 6.12 5.75 2.21 6.01
Old method: 7.83 6.22 7.46 8.83 8.19 5.64 8.88

Motor:	1	2	3	4	5	6	7
New method:	5.25	3.16	4.43	6.12	5.75	2.21	6.01
Old method:	7.83	6.22	7.46	8.83	8.19	5.64	8.88

Justify your choice of method in (b) below.

The data are pairs of measurements on a single set of individuals (the seven motors).
\ the appropriate method is a

paired two-sample t-test

Conduct an appropriate hypothesis test to determine whether there is sufficient evidence to conclude that the range of torques with the new method is at least 2 units less than with the old method.

Let X = torque (new method)
Y = torque (old method)
and D = X – Y

Test:
: µ_D = –2
vs.
: µ_D < –2

Free choice of a. I choose a = 0.05

Motor:	1	2	3	4	5	6	7	Sum	SSq
x = New Method:	5.25	3.16	4.43	6.12	5.75	2.21	6.01
y = Old Method:	7.83	6.22	7.46	8.83	8.19	5.64	8.88
d = Difference:	-2.58	-3.06	-3.03	-2.71	-2.44	-3.43	-2.87	-20.12	58.5004

Data:	n = 7	åd = –20.12	åd² = 58.5004
Þ			s_D = 0.334108622

Method 1

$c = muDo - t_.01,6 * sD/sqrt{n} = -2 - 3.14*0.126 = -2.396...$

\ Reject in favour of .

Method 2

$t_obs = (xBarD - muDo) / (sD/sqrt{n}) = -6.92... t_.01,6 = 3.14 t_obs < -t_.01,6$
\ Reject .

Method 3

t_obs = –6.92...
Clearly P[T < –6.92] << .01
[From software, P[T < –6.92] » 0.0002]
\ Reject at any reasonable a.

An Excel spreadsheet file is also available to illustrate this solution.

Use the simple linear regression model on these data to find the equation of the line of best fit to these data.

Note that the “line of best fit” depends on which variable is the predictor and which is the response. The roles of X and Y are interchanged if the new method is the response, (which is the more natural assignment).

Summary statistics (response = Y = old method):

n = 7 åx = 32.93 åx²= 168.6941

åxy = 260.7758 åy = 53.05 åy²= 411.3579

Therefore, (correct to 3 s.f. in each coefficient),

y = 0.814 x + 3.75

OR

Summary statistics (response = Y = new method):

n = 7 åx = 53.05 åx²= 411.3579

åxy = 260.7758 åy = 32.93 åy²= 168.6941

n S_xx = 65.2028
n S_xy = 78.4941
n S_yy = 96.4738

Therefore, (correct to 3 s.f. in each coefficient),

y = 1.20 x – 4.42

An Excel spreadsheet file is available to illustrate the first version of this solution.

Find the coefficient of determination R² and use it to comment on your answer to part (a) above.

Irrespective of which of old or new methods is the regressor,

\ R² = 97.9% (correct to 3 s.f.)

Very high correlation Þ cannot use unpaired t-test.
\ Choice in (a) is correct.

A Minitab project file and a Minitab Report Pad RTF file are also available to illustrate this solution.

A designer claims that a new type of hull increases the average sustained speed of a speedboat by more than 2 km/h over the average sustained speed of the existing hull design. Random samples of speedboats of the two designs are tested and their sustained speeds (in km/h) are measured on a test course under identical water conditions:

New design (x_A):

     42    36    38    37    39    36

Old design (x_B):

 
     30    37    33    31    34

Conduct an appropriate hypothesis test at a level of significance of 5%. Is there sufficient evidence to accept the designer’s claim? What assumptions have you made?

Calculation of summary statistics:

New design
speeds
x_A Old design
speeds
x_B x_A² x_B²

42 30 1764 900

36 37 1296 1369

38 33 1444 1089

37 31 1369 961

39 34 1521 1156

36 1296

228 165 8690 5475

New design speeds x_A	Old design speeds x_B	x_A²	x_B²

42	30	1764	900
36	37	1296	1369
38	33	1444	1089
37	31	1369	961
39	34	1521	1156
36		1296

228	165	8690	5475

n_A = 6 n_B = 5

n_A = 5 n_B = 4

_A = 38 _B = 33

s_A² = 5.2 s_B² = 7.5

The designer’s claim is that µ_A – µ_B > 2
(which is where the burden of proof lies).
Test
: µ_A – µ_B = 2
vs.
: µ_A – µ_B > 2

We assume that the two populations are independent, nearly normal and
have a common population variance.

Then and V[XbarA-XbarB] = sA^2/nA - sB^2/nB

σ² is unknown, so use the pooled sample variance

sP^2 = (nuA sA^2 + nuB sB^2) / (nuA + nuB)

The sample standard error is $sqrt{sP^2 (1/nA + 1/nB)}$ = sqrt{6.2 (1/6 + 1/5)} = sqrt(2.2814...)

The total number of degrees of freedom is n = n_A + n_B = 9

Method 1

If is true, then _A – _B ~ N(2, σ²)
where σ² is approximated by 2.2814... .
Upper tailed test.
c = D_o + t_{a, n} × (s.e.) = 2 + t_{.050, 9} Ö 2.2814...
= 2 + 1.833 × 1.5104... = 2 + 2.768...
\ c = 4.768...
_A – _B = 38 – 33 = 5 > c

\ reject H_o in favour of at a = .05 .

Method 2

t = ((xAbar-xBbar)-Do)/(s.e.)
= ((38-33)-2)/sqrt(2.2814) = 1.986...

t_c = t_{a, n} = t_{.050, 9} = 1.833

t_obs > t_c
\ reject H_o.

Method 3

As above, t_obs = 1.986 .
t_{.050, 9} = 1.833 and t_{.025, 9} = 2.262 .
t_{.025, 9} > t_obs > t_{.050, 9}
.025 < p-value < .050
But a = .05

\ reject H_o.

YES, we can accept the designer’s claim.

[Note: if we had taken a = .01, then we would have reached the opposite conclusion,
because the p-value is .039 (to 3 d.p.).]

An Excel spreadsheet file and a Minitab project file are also available to illustrate this solution.

The true mean tensile strength of a new type of lightweight cable is claimed to be more than 20 kN. The distribution of actual strengths is known to be normal to a good approximation, with a standard deviation of 1.4 kN. A random sample of five of the new cables has a mean of 21.5 kN
1. Is there sufficient evidence to support the claim?
  
  Let X = strength of a randomly chosen cable. X ~ N(µ, σ²)
  σ = 1.4 kN n = 5 = 21.5 kN
  
  Test: : µ = 20 vs. : µ > 20
  
  σ is known, therefore use z instead of t.
  Free choice of a. I choose a = 0.01
  
  Method 1
  
  $c = muo + z_.01 * sigma/sqrt{n} = 20 + 2.32...*1.4/sqrt{5} = 21.456...$
  = 21.500 > c (just barely)
  \ Reject in favour of at a = 0.01 (or any higher a)
  
  OR
  
  Method 2
  
  $z_obs = (xBar - muo) / (sigma/sqrt{n}) = (21.5 - 20) / (1.4/sqrt{5}) = 2.39...$
  z_.01 = 2.32...
  z_obs > z_.01
  \ Reject .
  
  OR
  
  Method 3
  
  F(–2.39) » .008 < .01
  \ Reject .
  
  OR
  
  Confidence Interval Method
  
  $c = xBar - z_.01 * sigma/sqrt{n} approx= 20.04$
  The 99% CI for µ is µ > 20.04 , which does not include µ = 20.
  \ Reject .
  
  An Excel spreadsheet file is also available to illustrate this solution.
2. Now suppose that the standard deviation is unknown and that the measured standard deviation of the random sample is s = 1.4 kN. Is there sufficient evidence to support the claim?
  
  Now σ is not known and n = 5, therefore replace z by t.
  
  Method 1
  
  $c = muo + t_.01,4 * s/sqrt{n} = 20 + 3.74...*1.4/sqrt{5} = 22.345...$
  
  \ do NOT reject at a = 0.01 .
  [If a = 0.05 were chosen, then
  Þ opposite conclusion]
  
  OR
  
  Method 2
  
  $t_obs = (xBar - muo) / (s/sqrt{n}) = (21.5 - 20) / (1.4/sqrt{5}) = 2.39...$
  t_{.01, 4} = 3.74...
  t_obs < t_{.01, 4}
  \ do NOT reject at a = 0.01 .
  [t_{.05, 4} = 2.13... t_obs > t_{.05, 4} \ Reject at a = 0.05]
  
  OR
  
  Method 3
  
  t_{.05, 4} = 2.13... t_obs = 2.39... and t_{.01, 4} = 3.74...
  0.01 < P[T > t_obs] < 0.05
  \ do NOT reject at a = 0.01,
  [but do reject at a = 0.05]
  p-value (from software) is P[T > t_obs] = 0.037...
  
  OR
  
  Confidence Interval Method
  
  $c = xBar - t_.01,4 * s/sqrt{n} = 21.5 - 3.74...*1.4/sqrt{5} = 19.154...$
  The 99% one-sided CI for µ is µ > 19.15 , which includes µ = 20.
  \ do NOT reject .
  [The 95% CI for µ is µ > 20.17 , which does not include µ = 20 \ Reject .]
  
  An Excel spreadsheet file is available to illustrate this solution, with a = 1%.
  An Excel spreadsheet file is also available for a = 5%

Random samples are drawn from two independent populations, producing the following summary statistics:

Are these data consistent with the hypothesis that the two population means are equal?

The burden of proof is on the alternative hypothesis that the two population means are not equal. Therefore test
: µ_X – µ_Y = 0 vs. : µ_X – µ_Y ¹ 0

From the question, the random quantities X and Y are independent, but we cannot assume equality of the variances.
[In this course, we have not investigated how to test for equality of variances. See section 9.5 of the Devore textbook. As it happens, the data are consistent with s_X² = s_Y².]

The Central Limit Theorem leads to
Xbar ~ N(mu_X, sigma_X^2 / n_X),
Ybar ~ N(mu_Y, sigma_Y^2 / n_Y)
and
Xbar - Ybar ~ N(mu_X - mu_Y, sigma_X^2/50 + sigma_Y^2/90)

Approximate the unknown s_X² and s_Y² by the point estimates s_X² and s_Y² respectively,
and we may approximate t_{a, n} by z_a.

If H_o is true, then, to an acceptable approximation,
Xbar - Ybar ~ N(0, 26^2/50 + 32^2/90) = N(0, 24.897)

Method 1:

$c = Delta_o + z_alpha/2 * sqrt{sX^2/nX + sY^2/nY}$
= 0 + 1.960*sqrt{24.897} (alpha=5%)
= 0 + 2.576*sqrt{24.897} (alpha=1%)
The boundaries of the rejection region are
symmetric about D_o (= 0).
c_L = – c_U = – c .
The boundaries of the rejection region are at

Therefore do not reject H_o.

Method 2:

$z = ((xBar - yBar) - Delta_o) / sqrt{sX^2/nX + sY^2/nY} = ((643-651)-0) / sqrt{24.897}$
z_obs = –1.603...
a = 5% z_c = z_.025 = 1.960
a = 1% z_c = z_.005 = 2.576

In both cases, | z_obs | < z_c, so
do not reject H_o.

Method 3:

z_obs = –1.603... (as in method 2 above).

The p-value = P[| Z | > | z_obs |] = 2 P[ Z < –1.603...]
» 2 F(–1.60) » 2 × .0548 » .110 > .10

Therefore do not reject H_o at any reasonable value of a.

Answer:

YES, these data ARE consistent with equal population means,

(although the p-value suggests that the agreement is not all that close).

[See also the Excel spreadsheet file for this question.]

[Note: If a common variance had been assumed, then s_X² and s_Y² would have been replaced by the pooled sample variance s_P² = 124260/138.
More precise estimates of the boundaries of the rejection region could have been obtained by using t_{a, n} in place of z_a .
With both changes in place, the boundaries of the rejection region for a = 5% would have moved out from ±9.78 to ±10.50.
t_{.025, 138} » 1.98.
The decision to retain H_o at any reasonable value of a is unchanged.]

A study of company performance in two nearby cities was conducted to test for any significant difference between the companies in those cities. A random sample of ten companies in city A had a sample mean performance index of 74.3 with a standard deviation of 3.2 . A random sample of ten companies in city B had a sample mean performance index of 73.2 with a standard deviation of 2.9 .

Is there a significant difference in company performance index between these two cities? State carefully your assumptions and your hypotheses.

We are dealing with two different sets of individuals, (the two sets of companies), the performances of which can be assumed to be independent of each other. The sample variances are close enough to each other to support the assumption that the population variances are equal. One further assumption (that is needed because of the small sample sizes) is that the performances are normally distributed to a fair approximation.

We will therefore conduct an

unpaired two sample t-test.

X_A ~ N(µ_A, s²) , X_B ~ N(µ_B, s²)

Summary statistics:

n_A = 10 , _A = 74.3 , s_A = 3.2
n_B = 10 , _B = 73.2 , s_A = 2.9
n_A = n_B = n – 1 = 9

Test : µ_A – µ_B = 0 vs. : µ_A – µ_B ¹ 0

Pooled sample variance:
sp^2 = (9*(3.2)^2 + 9*(2.9)^2) / (9 + 9)
= 9.325
Standard error:
$s.e. = sqrt{sp^2 * (1/nA + 1/nB)} = sqrt{9.325 * 2/10} = sqrt{1.865}$
= 1.365 65...

Total number of degrees of freedom = n = 18.

We have a free choice of a.
Even with an unusually large choice for a of 10%:

Method 1

c = D_o ± t_{a/2,
n} × (standard error)
= 0 ± 1.734 × Ö(1.865)
= ± 2.368...
_A – _B = 74.3 – 73.2 = 1.100
| 1.1 | < |c|
Therefore do not reject .

[If a = .05 is chosen, then c = ±2.869
and
if a = .01 is chosen, then c = ±3.930
® same result.]
Two tailed rejection region graph

Method 2

$tobs = ((xBarA - xBarB) - Delta_o) / (s.e.) = (74.3 - 73.2 - 0) / sqrt{1.865} = 1.100 / 1.365... = 0.805...$
t_a/2, n = t_{.05, 18} = 1.73... [or t_{.025, 18} = 2.10... or t_{.005, 18} = 2.87...]
|t| < t_{a/2,
n}
t is nowhere near the rejection region.
[A computer package will show that the p-value exceeds .40 !]
Therefore do not reject .

There is no significant difference in performance index between the two cities.

[See also the associated Excel file.]

A study was conducted to analyze the relationship between advertising expenditure and sales. The following data were recorded:

X Y

Advertising ($) Sales ($)

20 310

24 340

30 400

32 420

35 490

X	Y
Advertising ($)	Sales ($)
20	310
24	340
30	400
32	420
35	490

Assume a simple linear regression between sales Y and advertising X. Calculate the coefficients β₀ and β₁ of the line of best fit to these data and estimate the sales when $28 are spent on advertising.

Is there a significant linear association between Y and X?

Extending the table:

	x	y	x²	x y	y²
	20	310	400	6200	96100
	24	340	576	8160	115600
	30	400	900	12000	160000
	32	420	1024	13440	176400
	35	490	1225	17150	240100
Sum:	141	1960	4125	56950	788200

n S_xx = n å x² – (å x)²
= 5 × 4125 – (141)²
= 20625 – 19881
= 744
n S_xy = n å xy – (å x å y)
= 5 × 56950 – 141 × 1960
= 284750 – 276360
= 8390
n S_yy = n å y² – (å y)²
= 5 × 788200 – (1960)²
= 3941000 – 3841600
= 99400

The regression line (to 4 s.f.) is

y = 11.28 x + 73.99

x = 28 y = 11.277... × 28 + 73.992... = 389.744 623 7...

When $28 is spent on advertising, we predict $390 of sales

(correct to the nearest dollar).

There are many choices of method for an hypothesis test for linear association.
Test : r = 0 vs. : r ¹ 0 ,
or, equivalently,
test : b₁ = 0 vs. : b₁ ¹ 0 .

r^2 = Sxy^2 / (Sxx*Syy) = ... = .951...
just over 95% of all variation in y is explained by the linear regression. The correlation (.975...) is very strong, (which suggests that there is a linear association), but the sample size is very small, so we should proceed with a formal hypothesis test.

$t = r * sqrt{n-2} / sqrt{1-r^2} = 7.700...$

or
$t = sqrt{Sxy^2 * (n-2) / (Sxx*Syy - Sxy^2)} = sqrt{8390^2 * 3 /(744*99400 - 8390^2)} = 7.700...$

or
calculate the entries in the ANOVA table as follows:
SSR = Sum(yHat - yBar)^2 = (nSxy)^2 / (n * nSxx)
= 8390^2 / (5 * 744) = 18 922.607 52
SST = Sum(y - yBar)^2 = Syy = nSyy / n = 19880
SSE = SST – SSR = 957.392 48
MSR = SSR / n_R = SSR = 18 922.607 52
MSE = SSE / n_E = SSE / 3 = 319.130 826 6...
f = MSR / MSE = 59.294...

	d.f.	SS	MS	f
R	1	18 922.607 52...	18 922.607 52...	59.294...
E	3	957. 392 48...	319.130 826 6...
T	4	19 880

t = +Ö f = 7.700...

or

t = (betaHat1 - 0) / sb = 11.27... / sqrt{2.144...}
= 7.700...

Compare the observed t = 7.700... to t_{.005, 3} = 5.841...
t > t_{a/2, 3} for any reasonable choice of a.
[The p-value is less than .0046 .]
Therefore reject in favour of

YES, there is a significant linear association between Y and X.

[See also the associated Excel and Minitab project files.]

[Devore 6^th ed., Ch. 12 p. 551 q. 73 - parts (b) & (c) are bonus questions only]

The accompanying set of data is a subset of the data that appeared in the paper “Radial Tension Strength of Pipe and Other Curved Flexural Members” (J. Amer. Concrete Inst., 1980, pp. 33-39). The variables are age of a pipe specimen (x in days) and load necessary to obtain a first crack (y in 1000 lb/ft).

x 20 20 20 25 25 25 31 31 31

y 11.45 10.42 11.14 10.84 11.17 10.54 9.47 9.19 9.54

x	20	20	20	25	25	25	31	31	31
y	11.45	10.42	11.14	10.84	11.17	10.54	9.47	9.19	9.54

Calculate the equation of the estimated regression line.

Summary statistics:

n = 9 å x = 228 å y = 93.76
å x² = 5 958 å x y = 2 348.15 å y² = 982.293 2

n S_xx = 1 638 n S_xy = –243.93 n S_yy = 49.701 2

where n S_xy = n å x y – å x . å y etc.

The regression line (to 4 s.f.) is

y = 14.19 – 0.1489 x

A Minitab project file is also available to illustrate the solution to this part of this question.

* Suppose that a theoretical model suggests that the expected decrease in load associated with a 1 day increase in age is at most .10 (× 1000 lb/ft). Do the data contradict this assertion? State and test the appropriate hypotheses at significance level .05 .

The slope b₁ = the expected increase in load per day increase in age.
The burden of proof is on showing that the [negative] slope is steeper than –0.10.

Test : b₁ = –0.10 vs. : b₁ < –0.10 at a = .05.

A chain of calculations follows.

$SSE = {(nSxx)(nSyy) - (nSxy)^2} / {n (nSxx)} = 1.486...$

or, directly,
$t = {(nSxy) - beta1o*(nSxx)} * sqrt{(n-2) / [(nSxx)(nSyy) - (nSxy)^2]} = -1.43...$

But t_{.05, 7} = –1.89... .

Therefore do not reject .

NO, the data do not contradict the assertion that b₁ ³ –0.10.

Note that the non-zero value of b₁ in the null hypothesis renders the other versions of the t statistic incorrect. They are based on the true [null] value of the slope being zero.

* For purposes of estimating the slope of the true regression line as accurately as possible, would it have been preferable to make a single observation at each of the ages 20, 21, 22, ..., 30 and 31? Explain.

For the original data,
For the suggested new values of x ({20, 21, ..., 30, 31}),
n = 12, å x = 306, å x² = 7946
n S_xx = 1716 S_xx = 1716/12 = 143 .
143 < 182 and we want to be as large as possible because it appears in the denominator of the expression for the variance of the slope estimator, . A smaller value of s_b leads to more precise estimates of the slope.

[OR, the width of the confidence and prediction intervals contains a term . These intervals are narrower when the denominator is larger.]
Therefore

NO, the original data set is preferable,

despite its smaller sample size.

Calculate an estimate of true average load to first crack when age is 28 days. Your estimate should convey information regarding precision of estimation.

95% Confidence interval for E[Y|28] :

$(b0 + b1*xo) +- t * sqrt{MSE*(1/n + n(xo-xBar)^2 / nSxx} = (14.19 + -0.148*28) +- 2.36 * sqrt{0.212*(1/9 + 9(28-228/9)^2 / 1638}$
= 10.020... ± 0.4223...

Therefore the 95% confidence interval, to 2 d.p., is

9.60 £ E[Y|28] £ 10.44

[See also the associated Excel file.]

[Bonus question, to provide practice in the supplementary topic of type II error probabilities.]

Find the probability of committing a type II error when the true population mean is µ = 104 and an upper-tail hypothesis test is conducted at a level of significance of five per cent with a random sample of size 25 on the null hypothesis that µ = 100. It is known that the population variance is 100.

Repeat your calculation in the case when the level of significance of the hypothesis test is one per cent.

: µ = 100 vs. : µ > 100
s² = 100 s = 10
n = 25 standard error = 2

a = .05

Use method 1.

c = µ_o + z_a × (s / Ö(n))
= 100 + 1.64... × 2
= 103.290

Probability of committing type II error
when µ = 104 and and a = .05 is

= P[Z < (c-104) / s.e.]
= Phi((103.290 - 104) / 2)
= F(-0.355)

Graph for type II error

b(104) = .361 (approx.)

a = .01

Use method 1.
c = µ_o + z_a × (s / Ö(n))
= 100 + 2.32... × 2 = 104.652

Probability of committing type II error when µ = 104 and a = .01

= F(+0.326)

b(104) = .606 (approx.)

As expected, b increases as a decreases.

[Bonus question, to provide practice in the supplementary topic of type II error probabilities.
This question is a modification of Devore, 6^th ed., Ch. 8.3, pp. 343-344, q. 38.]

A university library ordinarily has a complete shelf inventory done once every year. Because of new shelving rules instituted the previous year, the head librarian believes it may be possible to save money by postponing the inventory. The librarian decides to select 800 books at random from the library’s collection and to have them searched in a preliminary manner. If the evidence indicates strongly that the true proportion of misshelved or unlocatable books is less than .02, then the inventory will be postponed.

Among the 800 books searched, 12 were misshelved or unlocatable. Test the relevant hypotheses (at a level of significance of .05) and advise the librarian what to do.

Let p represent the true proportion of books that are misshelved or unlocatable.
Postpone the inventory if and only if there is strong evidence for p < .02 .
Therefore we need to test
: p = .02 vs. : p < .02

n = 800
a = .05 z_a = 1.644...
Observed sample proportion

Method 1

$c = p_o - z_alpha sqrt{p_o q_o / n} = .02 - 1.645 sqrt{.02 x .98 / 800}$

= .02 – 0.008 142 334 5...
= 0.011 857 665... = .0119 (to 4 d.p.)
Graph of p.d.f. when p = .2

Reject if and only if < .0119
But = .0150 > c
Therefore do not reject H_o

In the remaining parts of this question, we need to know the value of c (the boundary of the rejection region when the null hypothesis is just barely true) in order to calculate the probability of rejecting (or not rejecting) the null hypothesis when the true value of p is something other than .02 . For this reason, we choose to use method 1 in part (a) of this question, rather than methods 2 or 3.

[In method 2, the z score is –1.010, which is closer to z = 0 than
the critical value of –1.645 .]
[In method 3, P[Z < –1.010] = .1562 > a.]

Whichever method we use, we decide to advise the librarian to

proceed with the inventory.

If the true proportion of misshelved and lost books is actually .01, what is the probability that the inventory will be [unnecessarily] taken?

Now we know that p = .01

P[inventory taken]
= P[ not rejected]

The boundary c of the rejection region
is calculated on the basis that
p = .02 .
From method 1 in part (a),
c = .011 857 665...
Graph showing beta(.01)

$P[P^ < c] = Phi((c-.01)/sqrt{.01x.99/800}) = Phi(.001857.../.003517...) = Phi(0.52807)$

Without linear interpolation,
F(0.53) = .7019 = .702 (to 3 d.p.)

P[inventory taken] = .298 (3 d.p.)

With linear interpolation,
F(0.528 07) » F(0.52) + .807 × {F(0.53) - F(0.52)}
= .6985 + .807 × (.7019 – .6985)
= .6985 + .807 × .0034 = .6985 + .0027...
= .7012 = .701 (to 3 d.p.)
P[inventory taken] = .299.

If the true proportion is .05, what is the probability that the inventory will be postponed?

Now we know that p = .05 > .02 .

The correct decision now is to proceed with the inventory.

The boundary c of the
rejection region is
calculated on the basis
that p = .02 .
From method 1 in part (a),
c = .011 857 665...

P[inventory postponed]
= P[ rejected]
Graph showing alpha(.05)

= $P[P^ < c] = Phi((c-.05)/sqrt{.05x.95/800}) = Phi(-.038142.../.007705...) = Phi(-4.95) = 0$

It is therefore very unlikely indeed that the inventory will be postponed.

What types of errors are the events described in parts (b) and (c) above?

In (b), we are told that the true value of p is .01 .
This is less than .02, so that the alternative hypothesis is true.
The correct decision is to reject the null hypothesis; that is, to postpone the inventory.

Taking the inventory = failure to reject the null hypothesis.
Failure to reject the null hypothesis when it is false is a

type II error.

In (c), we are told that the true value of p is .05 .
This is greater than .02, so that the null hypothesis is true.
The correct decision is to accept the null hypothesis; that is, to proceed with the inventory.

Postponing the inventory = rejecting the null hypothesis.
Rejecting the null hypothesis when it is true is a

type I error.

[Bonus question only in 2008 Fall]
[Devore 6^th ed., Ch. 9.4 p. 398 q. 55]

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (in ksi) of each specimen was determined and the results are summarized in the frequency distribution tabulated below.

Alloy:	A	B
26 - under 30	6	4
30 - under 34	12	9
34 - under 38	15	19
38 - under 42	7	10
sample sizes	40	42

Compute a 95% confidence interval for the difference between the true proportions of all specimens of alloys A and B that have an ultimate strength of at least 34 ksi.

Let p_A = proportion of alloy A that has ultimate strength ³ 34 ksi
and p_B = proportion of alloy B that has ultimate strength ³ 34 ksi.

Then and

The point estimate of is

A 95% CI for p_A – p_B is
= –0.140 476 19 ± 0.208 129 882
The CI is –.349 < p_A – p_B < +.068 (to 2 d.p.)

Can you conclude that there is a significant difference between these two population proportions?

NO because p_A = p_B is included in the 95% CI.
[See also the Excel spreadsheet file for this question.]

[Devore, ex. 12.4, q. 50 modified]

An experiment to measure the macroscopic magnetic relaxation time in crystals (in microseconds, µs) as a function of the strength of the external biasing magnetic field (in kiloGauss, kG) yielded the following data (“An Optical Faraday Rotation Technique for the Determination of Magnetic Relaxation Times”, IEEE Trans. Magnetics, June 1968: 175-178, with data read from a graph that appeared in the article).

x 11.0 12.5 15.2 17.2 19.0 20.8 22.0 24.2 25.3 27.0 29.0

y 187 225 305 318 367 365 400 435 450 506 558

The summary statistics are:
1. Fit the simple linear regression model to these data.
  
  Therefore the equation of the SLR line, (correct to 4 s.f.), is
  
  y = 18.87 x – 8.779
2. What proportion of the total variation in Y is explained by the linear regression model?
  
  The value of R², (the coefficient of determination), is required.
  
  Therefore, (to 3 s.f.), the proportion of the total variation in Y that is explained by the linear regression model is
  
  98.0%
3. Calculate a 95% confidence interval for the expected relaxation time when the field strength is 18 kG.
  
  At x = 18, the expected value of Y is
  
  The point estimate s² of the unknown ² is the mean square error (MSE):
  
  The 95% CI for E[Y | x = 18] is
  
  or, (correct to 2 d.p.),
  
  [318.71, 343.18] µs
4. Calculate a 95% prediction interval for a future relaxation time when the field strength is 18 kG.
  
  The only change is an additional term under the square root:
  
  or, (correct to 2 d.p.),
  
  (291.40, 370.48] µs
[Also see the associated Excel file.]

[Index of solutions]

[Return to your previous page]

Created 1997 11 27 and most recently modified 2010 07 30 by Dr. G.H. George.

n = 7	åx = 32.93	åx²= 168.6941
åxy = 260.7758	åy = 53.05	åy²= 411.3579

n = 7	åx = 53.05	åx²= 411.3579
åxy = 260.7758	åy = 32.93	åy²= 168.6941

n_A =	6	n_B =	5
n_A =	5	n_B =	4
_A =	38	_B =	33
s_A² =	5.2	s_B² =	7.5

n = 9	å x = 228	å y = 93.76
å x² = 5 958	å x y = 2 348.15	å y² = 982.293 2

x	11.0	12.5	15.2	17.2	19.0	20.8	22.0	24.2	25.3	27.0	29.0
y	187	225	305	318	367	365	400	435	450	506	558

Data:	n_R = 40		s_R = 2200
	n_B = 40		s_B = 1500