CHAPTER 09 - Hypothesis Testing
CHAPTER 9—Hypothesis Testing
9.1 Type I error: reject when is true
= probability of a Type I error
Type II error: do not reject when is false
= probability of a Type II error
LO2
9.2 Because the probability of a Type II error might become large.
LO2
9.3 a. : m ≤ 42 versus : m > 42.
b. Type I: decide the mean is > 42 (customers satisfied) when it is really ≤ 42
Type II: decide the mean is ≤ 42 (customers not satisfied) when it is really > 42
LO1, LO2
9.4 a. : m ³ 6 versus : m < 6.
b. Type I: decide < 6 when is really ³ 6
Type II: decide ³ 6 when is really < 6
LO1, LO2
9.5 a. : m = 3 versus : m ¹ 3, where m = mean diameter.
b. Type I: decide ¹ 3 (assign team) when = 3 (team is not needed)
Type II: decide = 3 (do not assign team) when ¹ 3 (team may be needed)
c. Cost of a Type I error
LO1, LO2
9.6 a. : m = 16 versus : m ¹ 16.
b. Type I: decide ¹ 16 (readjust filler) when = 16 (no adjustment is needed)
Type II: decide = 16 (do not readjust) when ¹ 16 (readjustment may be needed)
LO1, LO2
9.7 a. : m 60 versus : m > 60 where m = mean temperature of waste water.
b. Type I: decide > 60 (shut down) when £ 60 (water is cool enough, no shutdown needed)
Type II: decide £ 60 (do not shut down) when > 60 (water is too warm, shutdown needed)
c. Set to make the probability of a Type II error smaller.
LO1, LO2
9.8 A rejection point is a point on the horizontal axis under the standard normal curve that defines an area equal to a, the probability of making a Type I error.
LO3
9.9 A p-value is the probability that we obtain a sample statistic that contradicts the null hypothesis as much as or more than the observed sample statistic.
LO3
9.10 a.
b.
Since 2.5 > 1.645, Reject with a =.05.
c. z = 2.5. p-value = P(Z > 2.5) = 1 – .9938 = .0062
d. Since .0062 is less than .10, .05, and .01 but not less than .001, reject at a = .10, .05, .01, but not at a = .001.
e. Since p-value = .0062 is less than .01, there is very strong evidence against .
LO3
9.11 a.
b.
Since –2 > –2.326, Fail To Reject with a =.01.
c. z = –2. p-value = P(Z < -2) = .0228
d. Since .0228 is less than .10 and .05, but not less than .01 and 001, reject at a = .10 and .05, but not at a = .01 or .001.
e. Since p-value = .0228 is less than .05, there is strong evidence.
LO3
9.12 a.
b.
Since –2.5 < -1.96, Reject with a =.05.
c. z = –2.5. p-value = 2P(Z > 2.5) = 2(1 – .9938) = .0124
d. Since .0124 is less than .10, .05, but not less than .01 and .001, reject at a = .10, .05, but not at a = .01 or .001.
e. there is strong evidence against .
LO3
9.13 a.
b. z = (42.954 – 42) / (2.64/sqrt(65)) = 2.91
Reject Points
Since 1.28<1.645<2.33<2.91<3.09, reject with a =.10, .05, .01, but not with .001.
c. p-value
Since p-value=.0018 is less than .10, .05, and .01, reject at those levels of a, but not with a =.001.
d. There is very strong evidence.
LO1, LO3
9.14
a. : m ³ 6 versus : m < 6.
b.
Rejection points
Since –2.19 is less than –1.28 and –1.645, reject with a =.10 and .05, but not with a =.01 and .001.
c. p-value
Since p-value=.0143 is less than .10 and .05, reject at those levels of a, but not with a =.01 or .001.
d. There is strong evidence.
LO1, LO3
9.15 a. : m £ 60 versus : m > 60.
b.
Since 2.41 > 1.645 we reject for α = 0.05. p-value = 1 – 0.9920 = 0.0080 and since 0.0080 < 0.05 we draw the same conclusion so the plant should be shut down and the cooling system repaired.
LO1, LO3
9.16 z = 1.31, reject point = 1.645, p-value = 1 – 0.9049 = 0.0951 so we FTR and do not shut down the plant.
LO3
9.17 z = 3.09, reject point = 1.645, p-value = 1 – 0.9990 = 0.001 so we reject and we conclude the plant should be shut down and the cooling system repaired.
LO3
9.18 a. : m = 3 versus : m ¹ 3.
b.
, Since 2.37 > 1.96, reject with a = .05.
p-value = 2(.5 – .4911) = .0178, Since p-value=.0178 is less than .05, reject with a = .05.
The problem-solving team should be assigned.
LO1, LO3
9.19 a. : m = 16 versus : m ¹ 16.
b. For a = .01,
When = 16.05, z = 3, reject ; readjust.
p-value = 2(1 – .9987) = .0026<.01, reject , readjust.
CI: [16.007, 16.543], readjust
When = 15.96, z = –2.4, cannot reject : do not readjust.
p-value = 2(1 – .9918) = .0164, not less than .01, cannot reject , do not readjust.
CI: [15.917, 16.003], do not readjust
When = 16.02, z = 1.2, cannot reject ; do not readjust.
p-value=2(1 – .8849) = .2302, not less than .01, cannot reject , do not readjust.
CI: [15.977, 16.063], do not readjust
When = 15.94, z = -3.6, reject ; readjust.
p-value is approximately 0.000 < .001 reject , readjust.
CI: [15.897, 15.983], readjust
LO1, LO3
9.20 a. : m ³ 60 versus : m < 60.
b. so reject , run the commercial.
p-value = P(Z < -3.29) = 0.0005, since 0.0005 < 0.05 run the commercial.
LO1, LO3
9.21 Small sample (n < 30), unknown , normal population.
LO4
9.22 Is s known or unknown?
LO4
9.23 a.
8 degrees of freedom
t0.05 = 1.860; Since -4.3 < -1.860 reject H0 at α = 0.10
t0.025 = 2.306; Since -4.3 < -2.306 reject H0 at α = 0.05
t0.005 = 3.355; Since -4.3 < -3.355 reject H0 at α = 0.01
t0.0005 = 5.041; Since -4.3 > -5.041 do not reject H0 at α = 0.001
LO4
9.24
: m = 19.5 versus : m < 19.5.
-t0.01 = -2.386
Since -2.834 < -2.386, reject at a = 0.01
Since the p-value = 0.0031 you reject at 0.01 but do not reject at 0.001.
LO4
9.25 : m = 50 versus : m > 50.
t = (50.575 – 50) / (1.6438 / sqrt(40)) = 2.212. t0.05 = 1.685 so reject .
p-value = 0.0164 means we reject at 0.05 but do not reject at 0.01.
LO4
9.26 : m ≥ 8 versus : m < 8.
t = (7.4 – 8) / (1.026 / sqrt(15)) = -2.264. –t0.05 = -1.761 so reject and conclude the mean alert time is less than 8.
Type I: conclude the mean alert time is < 8 seconds when the mean alert time is actually more than 8 seconds. Type II: conclude the mean alert time is ≥ 8 seconds when the mean alert time is actually less than 8 seconds.
Since 0.01 < 0.02 < 0.05 there is strong evidence against H0.
LO1, LO2, LO4
9.27 a. : m £ 3.5% versus : m > 3.5%.
Type I: Conclude the mean bad debt ratio is > 3.5% when it actually is not.
Type II: Conclude the mean bad debt ratio is £ 3.5% when it actually is not.
b. , degrees of freedom = 6,
Since 3.143 < 3.62 reject : m £ 3.5% at a =.01. There is very strong evidence that is false.
p-value = 0.006 means there is very strong evidence that the mean bad debt ration for Ohio banks exceeds 3.5%.
d. Very strong evidence that the mean bad debt ratio for Ohio banks exceeds 3.5%.
LO1, LO2, LO4
9.28 Practical importance might be determined by interviewing Ohio bank executives to determine how big of a difference from the 3.5% will impact their business models.
9.29 a. : m = 42, : m > 42, a = .05
b.
so reject .
The p-value of .0019 means we would reject at a = .1, .05 and .01 but would not reject at a = .001 so we have very strong evidence.
LO1, LO4
9.30 a. : m ³ 6, : m < 6
b.
so reject .
The p-value of .0143 means we would reject at .1 and .05 but do not reject at a = .01 and .001.
LO1, LO4
9.31 a. H0: μ = 750 Ha: μ does not equal 750
b.
Reject a = .01. p-value = 0.002261 so there is very strong evidence against.
LO1, LO4
9.32 : m = 4 versus : m ¹ 4
Since t0.05=1.660 & t0.025=1.984 & t0.005=2.626 & t0.0005=3.392 < 4.78 rejectwith extremely strong evidence.
Since the sample mean is greater than 4 we estimate μ > 4.
LO1, LO4
9.33 Since t = -4.97 and p = 0.000, there is extremely strong evidence that μ < 18.8.
LO4
9.34 p is the true proportion of all units that are defective, where is the fraction of the sample that are defective.
LO1
9.35 np and n(1 – p) are both > 5.
LO5
9.36 : p = .3 versus : p ¹ .3.
a.
Since –2.575 < –2.18, do not reject.
b. p-value = 2P(z > 2.18) = .0292
c. Reject at a = .10 and .05, but not at a = .01 or .001.
LO5
9.37 a. : p £ .5 versus : p > .50.
b.
Since 1.19 < 1.28, do not reject for any of the a-values; little evidence.
c.
Since 2.33 < 2.53 < 3.09, reject at a = .01, but not at a = .001; very strong evidence.
d. based on a much larger sample provides stronger evidence that p is greater than .50.
LO1, LO5
9.38 a. : p £ .25 versus : p > .25.
b.
Since 5.31 > 3.09, reject at a = .10, .05, .01, .001; extremely strong evidence.
c. p-value = P(z > 5.31) is less than .001; reject at all given values of a.
d. Probably, .365 is significantly higher than .25 in terms of shares.
LO1, LO5
9.39 a. : p £ .18 versus : p > .18.
b.
p-value = P(z > 1.84) = .0329
Reject at a = .10 and .05, but not at a = .01 or .001; strong evidence
c. Perhaps, but this is subjective.
LO1, LO5
9.40 H0: p = 0.5 Ha: p < 0.5
z = ((117/300 – 0.5)/sqrt(0.5 * 0.5 / 300) = -3.81
p-value = P(Z < -3.81) = 0.00007
Since -3.81 < -3.090 or since 0.00007 < 0.001, there is extremely strong evidence that p < 0.5.
LO1, LO5
9.41 H0: p = 0.73 Ha: p does not equal 0.73
z = ((141/200 – 0.73)/sqrt(0.73 * 0.27 / 200) = -0.80
p-value = 2*P(Z > 0.80) = 0.4238
Insufficient evidence against H0.
LO1, LO5
9.42 a. : p = .95 versus : p < .95.
b.
Reject at each value of a; extremely strong evidence.
c. Probably, is far below the claimed .95.
LO1, LO5
9.43 When we must take some action on the basis of not rejecting .
LO6
9.44 The probability of a Type II error varies depending on the alternative value of .
LO6
9.45 A serious Type II error occurs when the true average m is far away from m0.
LO6
9.46 Power = 1 – b so want the power near one.
LO6
9.47 a. : m £ 60 versus : m > 60.
n = 100, s = 2, a = .025
β = P(Z < z* - |µ0 - µa| / (σ / sqrt(n))), where z* = zα since one sided test.
m = 60.1
m = 60.2,
m = 60.3,
m = 60.4,
m = 60.5,
m = 60.6,
m = 60.7,
m = 60.8,
m = 60.9,
m = 61.0,
b. No, b = .2946 when m = 60.5. Increase the sample size.
c. Plot is not included in this manual; power increase.
LO6
9.48 a. n = 100, s = .023, a = .05
: m = 3 versus : m ¹ 3.
β = P(Z < z* - |µ0 - µa| / (σ / sqrt(n))), where z* = zα/2 since two sided test.
m = 2.990,
m = 2.995,
m = 3.005,
m = 3.01,