Chapter 9: Hypothesis Testing: Single Population 1
Chapter 9:
Hypothesis Testing: Single Population
9.1
9.2 No change in interest rates is warranted
Reduce interest rates to stimulate the economy
9.3: There is no difference in the percentage of underfilled cereal packages
: Lower percentage after the change
9.4a. European perspective:
Genetically modified food stuffs are not safe
They are safe
- U.S. farmer perspective:
Genetically modified food stuffs are safe
They are not safe
9.5 No difference in the total number of votes between Bush and Gore
Bush with more votes
9.6 A random sample is obtained from a population with a variance of 625 and the sample mean is computed. Test the null hypothesis versus the alternative . Compute the critical value and state the decision rule
a. n = 25. Reject if = 100 +1.645(25)/ = 108.225
b. n = 16. Reject if = 100 +1.645(25)/ = 110.28125
c. n = 44. Reject if = 100 +1.645(25)/ = 106.1998
d. n = 32 Reject if = 100 +1.645(25)/ = 107.26994
9.7A random sample of n = 25 is obtained from a population with a variance and the sample mean is computed. Test the null hypothesis versus the alternative with alpha = .05. Compute the critical value and state the decision rule
a. = 225. Reject if = 100 +1.645(15)/ = 104.935
b. = 900. Reject if = 100 +1.645(30)/ = 109.87
c. = 400. Reject if = 100 +1.645(20)/ = 106.58
d. = 600. Reject if = 100 +1.645(24.4949)/= 108.0588
9.8Using the results from the above two exercises, indicate how the critical value is influenced by sample size. Next indicate how the critical value is influenced by the population variance.
The critical value is farther away from the hypothesized value the smaller the sample size n. This is due to the increase in the standard error with a smaller sample size.
The critical value is farther away from the hypothesized value the larger the population variance. This is due to the increased standard error with a larger population variance.
9.9A random sample is obtained from a population with variance = 400 and the sample mean is computed to be 70. Consider the null hypothesis versus the alternative . Compute the p-value
- n = 25. = = -2.50. = .0062
- n = 16. = = -2.00. = .0228
- n = 44. = = -3.32. = .0004
- n = 32. = = -2.83. = .0023
9.10A random sample of n = 25, variance = and the sample mean is = 70. Consider the null hypothesis versus the alternative . Compute the p-value
- = 225. = = -3.33. = .0004
- = 900. = = -1.67. = .0475
- = 400. = = -2.50. = .0062
- = 600. = = -2.04. = .0207
9.11; ; reject if Z.10 < -1.28.
= -1.6, therefore, Reject at the 10% level.
9.12; ; reject if Z.10 < -1.28
= -1.8, therefore, Reject at the 10% level.
9.13a. ; ; reject if Z.05 > 1.645
= 1.4, therefore, Do Not Reject at the 5% level.
- p-value = 1 – FZ(1.4) = 1 - .9192 = .0808
- the p-value would be higher – the graph should show that the p-value now corresponds to the area in both of the tails of the distribution whereas before it was the area in one of the tails.
- A one-sided alternative is more appropriate since we are not interested in detecting possible low levels of impurity, only high levels of impurity.
9.14Test ; , using n = 25 and alpha = .05
- . Reject if , = 2.00. Since 2.00 is greater than the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.
- . Reject if , = 2.00. Since 2.00 is greater than the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.
- Assuming a one-tailed lower tailed test, . Reject if , = -2.50. Since -2.50 is less than the critical value of -1.711, there is sufficient evidence to reject the null hypothesis.
- Assuming a one-tailed lower test, . Reject if , = -2.22. Since -2.22 is less than the critical value of -1.711, there is sufficient evidence to reject the null hypothesis.
9.15 Test ; , using n = 36 and alpha = .05
a. . Reject if , = 2.40. Since 2.40 is greater than -1.697, there is insufficient evidence to reject the null hypothesis.
b. . Reject if , = 2.40. Since 2.40 is greater than -1.697, there is insufficient evidence to reject the null hypothesis.
c. . Reject if , = -3.00. Since -3.00 is less than the critical value of -1.697, there is sufficient evidence to reject the null hypothesis.
d. . Reject if , = -2.67. Since -2.67 is less than the critical value of -1.697, there is sufficient evidence to reject the null hypothesis.
9.16
= -3.33, p-value is < .005. Reject at any common level of alpha
9.17 reject if –2.576 > t1561,.005 > -2.576
= 8.08, p-value is < .010. Reject at any common level of alpha.
9.18
= 3.38, p-value is < .010. Reject at any common level of alpha
9.19 reject if t171,.01 > 2.326
= 5.81, p-value is < .005. Reject at any common level of alpha.
9.20
= -3.35, p-value is < .005. Reject at any common level of alpha.
9.21 reject if |t15, .05/2 | > 2.131
= 1.017, p-value is > .200. Do not reject at the .05 level.
9.22 a. No, the 95% confidence level provides for 2.5% of the area in either tail. This does not correspond to a one-tailed hypothesis test with an alpha of 5% which has 5% of the area in one of the tails.
b. Yes.
9.23
= -1.554, p-value is between .100 and .050. Do not reject at common levels of alpha.
9.24 reject if |t8, .05/2 | > 2.306
= 1.741, therefore, do not reject at the 5% level
9.25 reject if |t7, .10/2 | > 1.895
= -1.815, therefore, do not reject at the 10% level
9.26 The population values must be assumed to be normally distributed.
reject if t19, .05 < -1.729
= -3.189, therefore, reject at the 5% level
9.27a.
= -1.486, p-value = .0797, therefore, reject at alpha levels greater than 7.97%
- Yes, with a larger sample size, the standard error would be smaller and hence, the calculated value of t would be larger. This would yield a smaller p-value and hence the company’s claim could be rejected at a lower significance level than part.
9.28 A random sample is obtained to test the null hypothesis of the proportion of women who said yes to a new shoe model. . What value of the sample proportion is required to reject the null hypothesis with alpha = .03?
- n = 400. Reject if = .25 +1.88 = .2907
- n = 225. Reject if = .25 +1.88 = .30427
- n = 625. Reject if = .25 +1.88 = .28256
- n = 900. Reject if = .25 +1.88 = .2771
9.29 A random sample is obtained to test the null hypothesis of the proportion of women who would purchase an existing shoe model. . What value of the sample proportion is required to reject the null hypothesis with alpha = .05?
- n = 400. Reject if = .25 – 1.645 = .2144
- n = 225. Reject if = .25 – 1.645 = .2025
- n = 625. Reject if = .25 – 1.645 = .2215
- n = 900. Reject if = .25 – 1.645 = .22626
9.30
= 1.79, p-value = 1 – FZ(1.79) = 1 - .9633 = .0367
Therefore, reject at alpha greater than 3.67%
9.31 reject if z.05 < -1.645
= -5.62, p-value = 1 – FZ(5.62) = 1 – 1.0000 = .0000
Therefore, reject at the 5% alpha level
9.32
= -1.26, p-value = 2[1 – FZ(1.26)] = 2[1 – .8962] = .2076
The probability of finding a random sample with a sample proportion this far or further from .5 if the null hypothesis is really true is .2076
9.33 reject if |z.10/2 | > 1.645
= .64, p-value = 2[1 – FZ(.64)] = 2[1 – .7389] = .5222
Therefore, do not reject at the 10% alpha level. The p-value shows the probability of finding a random sample with a sample proportion this far or farther from .5 if the null hypothesis is really true is .5222
9.34
= .85, p-value = 1 – FZ(.85) = 1 – .8023 = .1977
Therefore, reject at alpha levels in excess of 19.77%
9.35
= -1.94, p-value = 1 – FZ(1.94) = 1 – .9738 = .0262
Therefore, reject at alpha levels in excess of 2.62%
9.36
= -1.87, p-value = 1 – FZ(1.87) = 1 – .9693 = .0307
Therefore, reject at alpha levels in excess of 3.07%
9.37 Compute the probability of Type II error and the power for the following
a. . = =
= P(z ≤ -2.36) = .0091. Power = 1 – .0091 = .9909
b. . = =
= P(z ≤ .44) = .6700. Power = 1 – .6700 = .3300
c. . = =
= P(z ≤ -4.36) = .0000. Power = 1 – .0000 = 1.0000
d. . = =
= P(z ≤ -1.16) = .3770. Power = 1 – .3770 = .6230
9.38What is the probability of Type II error if the actual proportion is
a. . =
= = P(-2.94 ≤ z ≤ .98) = .4984 + .3365 = .8349
b. . =
= P(-5.96 ≤ z ≤ -1.99) = .5000 – .4767 = .0233
c. . =
= P(-3.44 ≤ z ≤ .49) = .4997 + .1879 = .6876
d. . =
= P(-.98 ≤ z ≤ 2.94) = .3365 + .4984 = .8349
e. . =
= P(1.48 ≤ z ≤5.44) = .5000 – .4306 = .0694
9.39a. is rejected when < -1.28 or when < 48.72. Given an = 48.2 hours, the decision is to reject the null hypothesis.
b. The power of the test = 1 - = 1 – P(Z > ) = 1 – P(Z > -.28) = .3897
9.40a. is rejected when > 1.645 or when > 3.082. Since the sample mean is 3.07% which is less than the critical value, the decision is do not reject the null hypothesis.
b. The = P(Z < ) = 1 – FZ(.36) = .3594. Power of the test = 1 - = .6406
9.41a. is rejected when –2.275 < < 2.275 or when 3.914 < < 4.086. Since the sample mean was 4.27, which is greater than the upper critical value, the decision is to reject the null hypothesis.
b. = P( < Z < ) = P(-1.08 > Z > 4.07) = .8599
9.42 is rejected when < -1.28 or when p < .477
The power of the test = 1 - = 1 – P(Z >) = 1-P(Z > 1.54) = .9382
9.43a. is rejected when < -1.645 or when p < .2275. Since the sample proportion is .173 which is less than the critical value, the decision is to reject the null hypothesis.
b. The power of the test = 1 - = 1 – P(Z > ) = 1-P(Z > 2.17) = .9850
9.44 a. is rejected when –1.645 > > 1.645 or when .442 > p > .558. Since the sample proportion is .5226 which is within the critical values. The decision is that there is insufficient evidence to reject the null hypothesis.
b. = P( < Z <) = 1-P(-4.55 < Z < -1.21) = .1131
9.45a. ) = P(Z < -2.4) = 0.0082
b. ) = P(Z < -1.2) =0.1151
c. ) = P(Z > -.4) =0.6554
9.46 a. ) = P(Z > 1.33) = .0918
b. ) = P(Z > 2.67) = .0038. The smaller probability of a Type I error is due to the larger sample size which lowers the standard error of the mean.
c. ) = P(Z < -1.5) = .0668
d. i) lower, ii) higher
9.47 a. = 39.6,
Therefore, reject at the 2.5% level but not at the 1% level of significance.
b. = 46.2,
Therefore, reject at the 2.5% level but not at the 1% level of significance.
c. = 38.16,
Therefore, reject at the 5% level but not at the 2.5% level of significance.
d. = 24.79,
Therefore, do not reject at any common level of significance.
9.48 reject if > 12.02
= 13.0757, Therefore, reject at the 10% level
9.49a. s2 = 5.1556
b. reject if > 16.92
= 20.6224. Reject at the 5% level
9.50
= 46.4, p-value = .0214. Reject at the 5% level
9.51The hypothesis test assumes that the population values are normally distributed
reject if > 30.14
= 26.4556. Do not reject at the 5% level
9.52
= 16.961.
Do not reject at the 10% level since >15.66 =
9.53 a. The null hypothesis is the statement that is assumed to be true unless there is sufficient evidence to suggest that the null hypothesis can be rejected. The alternative hypothesis is that the statement that will be accepted if there is sufficient evidence to reject the null hypothesis
- A simple hypothesis assumes a specific value for the population parameter that is being tested. A composite hypothesis assumes a range of values for the population parameter.
- One sided alternatives can be either a one-tailed upper (> greater than) or a one-tailed lower (< less than) statement about the population parameter. Two sided alternatives are made up of both greater than or less than statements and are written as ( not equal to).
- A Type I error is falsely rejecting the null hypothesis. To make a Type I error, the truth must be that the null hypothesis is really true and yet you conclude to reject the null and accept the alternative. A Type II error is falsely not rejecting the null hypothesis when in fact the null hypothesis is false. To make a Type II error, the null hypothesis must be false (the alternative is true) and yet you conclude to not reject the null hypothesis.
- Significance level is the chosen level of significance that established the probability of a making a Type I error. This is represented by alpha. The power of the test is the ability of the hypothesis test to identify correctly a false null hypothesis and reject it.
9.54 The p-value indicates the likelihood of getting the sample result at least as far away from the hypothesized value as the one that was found, assuming that the distribution is really centered on the null hypothesis. The smaller the p-value, the stronger the evidence against the null hypothesis.
9.55 a.
b. reject if t(9,.05) > 1.833
= 1.50, therefore, do not reject at the 5% level
9.56 a. False. The significance level is the probability of making a Type I error – falsely rejecting the null hypothesis when in fact the null is true.
b. True
- True
- False. The power of the test is the ability of the test to correctly reject a false null hypothesis.
- False. The rejection region is farther away from the hypothesized value at the 1% level than it is at the 5% level. Therefore, it is still possible to reject at the 5% level but not at the 1% level.
f. True
g. False. The p-value tells the strength of the evidence against the null hypothesis.
9.57a. = 6.245
reject if t8,.05 < -1.86
= -1.44, therefore, do not reject at the 5% level
9.58a. ) = P(Z < -2) = .0228
b. = P(Z > 3) = .0014
c. i) smaller ii) smaller
- i) smaller ii) larger
9.59 a. reject if z.05 < -1.645
= -1.90, therefore, reject at the 5% level
b. is rejected when < -1.645 or when p < .2195
i)power = 1 – P(Z > ) = 1 – P(Z > 1.14) = .8729
ii)power = 1 – P(Z > ) = 1 – P(Z > -1.64) = .0505
iii)power = 1 – P(Z > ) = 1 – P(Z > -4.1) = .0000
9.60
= -.39, p-value = 2[1-FZ(.39)] = 2[1-.6517] = .6966
Therefore, reject at levels in excess of 69.66%
9.61
= 1.51, p-value = 1-FZ(1.51) = 1-.9345 = .0655
Therefore, reject at levels in excess of 6.55%
9.62 reject if z.05 > 1.645
= 2.356, therefore, reject at the 5% level
9.63
= 2.22, p-value = 1-FZ(2.22) = 1-.9868 = .0132
Therefore, reject at levels in excess of 1.32%
9.64 Cost Model where W = Total Cost: W = 1,000 + 5X
: W
Using the test statistic criteria: (3050 – 3000)/25 = 2.00 which yields a p-value of .0228, therefore, reject at the .05 level.
Using the sample statistic criteria: , , since , therefore, reject at the .05 level.
9.65
. Probability of = 40 given that is 39 is .1170. Therefore, the Vice President’s claim is not very strong.
9.66 Assume that the population of matched differences are normally distributed
= 1.961.
Reject at the 10% level since 1.961 > 1.796 = t(11, .05)
9.67 The hypothesis test is: reject if Z.05 > 1.645
Note: two zero values can be removed since a loaf of bread cannot weigh zero grams:
Using Minitab:
One-Sample T: Dbread
Test of mu = 100 vs mu < 100
Variable N Mean StDev SE Mean
Dbread 37 101.19 32.79 5.39
Variable 95.0% Upper Bound T P
Dbread 110.29 0.22 0.587
At the .05 level of significance, do not reject
9.68 reject
One-Sample T: Salmon Weight
Test of mu = 40 vs mu > 40
Variable N Mean StDev SE Mean
Salmon Weigh 39 49.73 10.60 1.70
Variable 95.0% Lower Bound T P
Salmon Weigh 46.86 5.73 0.000
At the .05 level of significance we have strong enough evidence to reject Ho that the true mean weight of salmon is no different than 40 in favor of Ha that the true mean weight is significantly greater than 40.
: 40 + 1.686(1.70) = 42.8662
Population mean for = .50 (power=.50): tcrit = 0.0: 42.8662 + 0.0(1.70) = 42.8662
Population mean for = .25 (power=.75): tcrit = .681: 42.8662 + .681(1.70) = 44.0239
Population mean for = .10 (power=.90): tcrit = 1.28: 42.8662 + 1.28(1.70) = 45.0422
Population mean for = .05 (power=.95): tcrit = 1.645: 42.8662 + 1.645(1.70) = 45.6627
9.69 a. reject if |z.05|> 1.645
= 1.20, p-value =2[1-FZ(1.2)]= 2302.
Do not reject at the 10% level
b. reject if
= 44.376. Reject at the 10% level
9.70a. Assume that the population is normally distributed
One-Sample T: Grams:
Test of mu = 5 vs mu not = 5
Variable N Mean StDev SE Mean
Grams:11-34 12 4.9725 0.0936 0.0270
Variable 95.0% CI T P
Grams:11-34 ( 4.9130, 5.0320) -1.02 0.331
, reject if |t(11, .025| > 2.201
= -1.018. Do not reject at the 5% level
b. Assume that the population is normally distributed
reject if
= 154.19. Therefore, reject at the 5% level
9.71= 6.245
reject if
= 8.67. Do not reject at the 10% level