CHAPTER 11
STATISTICAL INFERENCE: TWO POPULATIONS
1.The null and the alternative hypotheses are:
H0 : 1 = 2;
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen α = 0.01.
The samples are chosen independently, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic, .
The populations are approximately normal. Hence, this is approximately a two -tailed t-test.
df == = 87.835
For df ≈ 88, t0.005 = approximately 2.64.
Decision rule : reject H0 in favour of H1 if t < -2.64 or if t > 2.64.
t = = 2.59.
Since 2.59 is between –2.64 and 2.64, there is insufficient evidence, at α = 0.01, to reject H0 in favour of H1.
3. The null and alternate hypotheses are:
H0 : 1 = 2;
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen α = 0.05.
Samples are chosen independently, the population variances are unknown, but we are assuming them to be equal. Hence, we shall use as test statistic
T= .
The population distributions are approximately normal. Hence, this is a two-tailed t-test.
df = n1+n2-2 = 10 + 8 - 2= 16.
For df = 16, tα/2 = t0.025 = 2.12.
Hence, decision rule is: reject H0 in favour of H1 if t > 2.12 or if t 2.12.
Pooled estimate of the population variance is,
Since –1.416 is between –2.12 and +2.12, we do not have sufficient evidence, at α = 0.05, to reject H0.
5.Let μ1 and μ2 be the population means of weight gains in infants using Gabbs’ products and the competitors’ products, respectively, during first three month after birth. Then, the null and alternative hypotheses are:
H0 : 12;
H1: 12
This is a lower-tailed test.
We have chosen α = 0.05.
The samples are chosen independently, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic, .
The populations are approximately normal. Hence, this is approximately a lower-tailed t-test.
df == = 33.784.
For df ≈ 34, tα = t0.05 1.691.
Decision rule : reject H0 in favour of H1 if the computed t-value is less than –1.691.
t = =-0.581.
Since –0.581 > -1.691, we do not have sufficient evidence, at α = 0.05, to reject H0, that is to accept the claim that the babies using the Gibbs brand gain less weight.
α = 1 – 0.99 = 0.01. For df = 34, = t0.005 2.73. A 99% confidence interval estimate for (μ1- μ2) is= -0.2 + (2.73)(0.3442) = (-1.14, 0.74).
7.Let μ1 and μ2 be the population means of turnover rates of oil related stocks and other stocks, respectively. Then, the null and alternative hypotheses are:
H0 : 1 = 2;
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen α = 0.01.
The two samples are independent, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic,
.
The populations are approximately normal. Hence, this is a two-tailed t-test.
df == = 77.048
For df ≈ 77, tα/2 = t0.005 2.641.
Decision rule : reject H0 in favour of H1 if the computed t-value is less than -2.641 or if it is greater than 2.641.
t = = -2.66
Since -2.66 is less than -2.641, there is sufficient evidence, at α = 0.01, to reject H0, that is, to infer that there is a difference in the mean turnover rates.
9.Let the population means of grades of men and women in the examination be μm and μf, respectively. Then, the null and alternative hypotheses are:
H0 : m ≥ f ;
H1: mf
This is a lower-tailed test.
We have chosen α = 0.01.
Samples are chosen independently, and the population variances are unknown, but are known to be equal. Hence, we shall use as test statistic
T= .
The population distributions are approximately normal. Hence, this is a lower-tailed t-test.
df = 9 + 7 - 2 = 14. For df = 14, tα = t0.01 = 2.624.
Hence, decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.624.
For the given sample data,
Since -0.234 is greater than -2.624, we do not have sufficient evidence, at α = 0.01, to reject H0 in favour of H1, that is, to conclude that the mean grade of women is higher.
11.Let be the population mean values of number of defective units on the day shift and the afternoon shift, respectively. Let.
The null and the alternative hypotheses are:
H0 : μD ≤ 0
H1: μD > 0
This is an upper-tailed test.
We have chosen = 0.05.
We shall use T = as the test statistic.
The population distributions are approximately normal. Hence, for D = 0, the distribution of T is approximately student’s t-distribution with df = (n - 1) = 3.
So, this is an upper-tailed t-test.
For df = 3, tα = t0.05 = 2.353.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 2.353.
Day / 10 / 12 / 15 / 19Afternoon / 8 / 9 / 12 / 15
d / 2 / 3 / 3 / 4
Since the computed t-value (=7.35) is greater than 2.353, there is sufficient evidence, at = 0.05, to reject H0 in favour of H1, that is, to conclude that there are less defective units produced in the afternoon shift.
13.Let be the population mean values of student weights (in lbs.) upon arrival on campus, and one year later, respectively. Let .
The null and the alternative hypotheses are :
H0 : μD ≥ 0
H1 : μD < 0
This is a lower-tailed test.
We have chosen = 0.01.
We shall use as test statistic, T = .
The population distributions are approximately normal. Hence, for D = 0, the distribution of T is approximately student’s t-distribution with df = (n - 1) = 10.
So, this is a lower-tailed t-test.
For df = 10, tα = t0.01 = 2.764.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than –2.764.
For the given sample data, the sample values ofD are:
On arrival / 124 / 157 / 98 / 190 / 103 / 135 / 149 / 176 / 200 / 180 / 256After 1 year / 142 / 157 / 96 / 212 / 116 / 134 / 150 / 184 / 209 / 180 / 269
d / -18 / 0 / 2 / -22 / -13 / 1 / -1 / -8 / -9 / 0 / -13
Since the computed t-value (=-2.92) is less than -2.764, there is sufficient evidence, at = 0.01, to reject H0 in favour of H1, that is, to conclude that there is a weight gain in the first year.
α = 1 – 0.95 = 0.05. For df = 10, = t0.025 = 2.228. A 95% confidence interval estimate for mean increase in weight during the first year is = (1.74, 12.99).
15.Let be the population mean values of strength before and after the use of special vitamins, respectively. Let .
The null and the alternative hypotheses are
H0 : μD ≥ 0;
H1 : μD < 0.
This is a lower-tailed test.
We have chosen = 0.01.
We shall use as test statistic, T = .
The population distributions are approximately normal. Hence, for D = 0, the distribution of T is approximately student’s t-distribution with df = (n - 1) = 9.
So, this is a lower-tailed t-test
For df = 9, tα = t0.01 = 2.821.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.821.
For the given sample data, the sample values of D are:
Before / 86 / 115 / 156 / 96 / 52 / 57 / 85 / 53 / 89 / 57After / 89 / 114 / 156 / 97 / 51 / 58 / 87 / 54 / 88 / 56
d / -3 / 1 / 0 / -1 / 1 / -1 / -2 / -1 / 1 / 1
1.43
Since the computed t-value (=-0.885) is greater than -2.821, we do not have sufficient evidence, at = 0.01, to reject H0 in favour of H1, that is, to conclude that there the special vitamin increases the strength.
17.The null and the alternative hypotheses are :
H0 : p1≤ p2
H1 : p1p2
This is an upper-tailed test.
We have selected = 0.05.
The samples are selected independently. Therefore, we shall use as test statistic:
Hence, the sample sizes are large enough for the use of normal approximation. This is an upper-tailed Z-test.
The decision rule is: reject H0 in favour of H1 if the computed z-value is greater than .
The pooled estimate of proportion is .
.
Since the computed z-value (=1.614) is less than 1.645, we do not have sufficient evidence, = 0.05, to reject H0 in favour of H1, that is to conclude that p1 is greater than p2.
A 98 percent confidence interval estimate for (p1-p2) is:
19.Let p1 and p2 be, respectively, population fractions of vines treated with Action and Pernod 5, respectively, that get infested. Then, the null and alternative hypotheses are:
H0 : p1- p2 ≤ 0.02
H1 : p1-p2 > 0.02
This is an upper-tailed test.
We shall use as test statistic
Hence, the sample sizes are large enough for normal approximation. This is an upper-tailed Z-test.
We have chosen α = 0.05; zα = z0.05 = 1.645.
The decision rule is: reject H0 in favour of H1 if the computed z-value is greater than 1.645.
= 1.045
Since the computed z-value (= 1.045) is less than 1.645, we do not have sufficient evidence, at α = 0.05, to conclude that (p1-p2) is greater than 0.02.
21.Let ps and pm be, respectively, population proportions of single and married drivers insured by Cambridge Insurance that had an accident during the three years period. Then, the null and alternative hypotheses are:
H0 : ps - pm = 0
H1 : ps - pm ≠ 0
This is a two-tailed test.
We have selected = 0.05.
We assume that the samples are selected independently, and therefore, we shall use as test statistic:
Hence, the sample sizes are large enough for the use of normal approximation. This is a two-tailed Z-test.
Zα/2 = z0.025 = 1.96.
The decision rule is: reject H0 in favour of H1 if the computed z-value is less than -1.96 or if it is greater than 1.96.
The pooled estimate of proportion is:
The value of the test statistic is
1.745.
Since the computed z-value (= 1.745) is between -1.96 and 1.96, we do not have sufficient evidence, = 0.05, to reject H0 in favour of H1, that is to conclude that the values of the two population proportions are different.
A 95% confidence interval estimate for the difference between the two population proportions is: = (-0.0067, 0.1067).
23. Let 1 and 2 be the population means of useful life times (in months) of Cooper paint and King paint, respectively. Then, the null and alternative hypotheses are:
H0 : 1 = 2
H1 : 1 ≠ 2
This is a two-tailed test.
We have chosen = 0.01.
The samples are chosen independently, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic, .
The population distributions are approximately normal and the sample sizes are large enough. Hence, this is a two-tailed t-test
df == = 72.9988
For df ≈ 73, tα = t0.005 2.651.
Decision rule : reject H0 in favour of H1 if the computed t-value is less than -2.651 or if it is greater than 2.651.
t = = -2.839
Since –2.839 is less than –2.651, there is sufficient evidence, at = 0.01, to reject H0, that is, to infer that the mean useful lives of the two paints are different.
α = 1 – 0.95 = 0.05. For df = 73, tα/2 = t0.025 = approximately 1.996.
A 95 percent confidence interval estimate for the difference is
= (-0.8 + 0.562) = (-1.362, -0.238).
25.Let 1 and 2 be the population means of costs of health benefit packages (in terms of percentage of salary), for employees, offered by large and small firms, respectively. Then, the null and alternative hypotheses are:
H0 : 1 = 2
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen = 0.05.
The samples are chosen independently, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic, .
The populations are approximately normal and the sample sizes are large enough. Hence, this is a two-tailed t-test
df = = 20.6389
For df ≈ 21, = t0.025 = 2.08.
Decision rule is : reject H0 in favour of H1 if the computed t-value is less than -2.08 or if it is greater than 2.08.
t = = 1.2013
Since 1.2013 is between –2.08 and 2.08, we do not have sufficient evidence, at α = 0.05, to conclude that the population means of percent of employee salaries, spent by large and small firms on health benefits, are different.
27.Let 1 and 2 be the population mean values of number of hamburgers sold per day at northside site and the southside site, respectively. Then, the null and alternative hypotheses are:
H0 : 1 = 2;
H1 : 1 ≠ 2.
This is a two-tailed test.
We have chosen = 0.05.
The samples are chosen independently, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic, .
The populations are approximately normally distributed and the sample sizes are large enough. Hence, this is a two-tailed t-test.
df == =19.7541
For df ≈ 20, tα/2 = t0.025 = 2.086.
Decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.086 or if it is greater than 2.086.
t = = 0.8985
Since 0.8985 is between –2.086 and 2.086, we do not have sufficient evidence, at α = 0.05, to conclude that the mean number of hamburgers sold at the two locations are different.
29.Let 1 and 2 be the population means of amounts purchased on impulse per customer at stores on Peach street and Plum street, respectively. Then, the null and alternative hypotheses are:
H0 : 1 = 2
H1 : 1 ≠ 2
This is a two-tailed test.
We have chosen = 0.01.
The two samples are independent, and the population variances are unknown, but they are known to be equal. Hence, we shall use as test statistic, .
The populations are approximately normally distributed. Hence, this is a two-tailed t-test.
For df = 10 + 14 - 2 = 22, tα/2 = t0.005 = 2.819.
Decision rule is : reject Ho if the computed t-value is less than -2.819 or if it is greater than 2.819 .
For the given sample data,
Since –2.374 is between –2.819 and +2.819, we do not have sufficient evidence, at α = 0.01, to reject H0, that is, to infer that the mean amounts, purchased on impulse at the two stores, are different.
31.Let 1 and 2 be the population means of number of units sold at discount and regular prices, respectively. Then, the null and alternative hypotheses are:
H0 : 1 ≤ 2
H1 : 12
This is an upper-tailed test.
We have chosen = 0.01.
The two samples are independent, and the population variances are unknown, but they are known to be equal. Hence, we shall use as test statistic, .
The populations are approximately normally distributed. Hence, this is an upper-tailed t-test.
For df = 8 + 7 - 2 = 13, tα = t0.01 = 2.65.
Decision rule is: reject Ho in favour of H1 if the computed t- value is greater than 2.65.
For the given sample data,
Since the computed t-value (=0.819) is less than 2.65, we do not have sufficient evidence, at α = 0.01, to reject H0, that is, to infer that price reduction resulted in an increase in mean value of sales.
33.Let be the population mean values of student grades under the quarter system and the semester system, respectively. Let.
Then, the null and the alternative hypotheses are:
H0 : μD ≤ 0
H1 : μD > 0
This is an upper-tailed test.
We have chosen = 0.05.
We have chosen a matched paired sample. Hence, we shall use T = as the test statistic.
The population distributions are approximately normal. Hence, for D = 0, the distribution of T is approximately student’s t-distribution with df = (n - 1) = 9.
So, this is an upper-tailed t-test.
For df = 9, t = t0.05 = 1.833.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 1.833.
For the given sample data, the sample values of D are:
last fall / 2.98 / 2.34 / 3.68 / 3.13 / 3.34 / 2.09 / 2.45 / 2.96 / 2.80 / 4.00this fall / 3.17 / 2.04 / 3.62 / 3.19 / 2.90 / 2.08 / 2.88 / 3.15 / 2.49 / 2.98
d / -0.19 / 0.30 / 0.06 / -0.06 / 0.44 / 0.01 / -0.43 / -0.19 / 0.31 / 0.02
Since the computed t-value (=0.321) is less than 1.833, we do not have sufficient evidence, at α = 0.05, to conclude that student grades declined after the conversion.
35.Let be the population mean values of annual family incomes (in thousands of dollars) of potential buyers at the first and second developments, respectively. Let .Then, the null and the alternative hypotheses are:
H0 : 1 = 2
H1: 1 ≠ 2
This is a two-tailed test.
We have chosen = 0.05.
The samples are chosen independently, the population variances are unknown and they are not known to be equal. Hence, we shall use as test statistic,
.
The population distributions are approximately normal. Hence, this is a two-tailed t-test.
df == = 125.5294
For df ≈ 126, tα/2 = t0.025 1.979.
Decision rule is: reject H0 in favour of H1 if the computed t-value is less than -1.979 or if it is greater than 1.979.
t = = -5.587.
Since the computed t-value (= –5.587) is less than -1.979, there is sufficient evidence, at α = 0.05, to reject H0 in favour of H1, that is, to infer that the two population means are different.
37.Let be the population mean values of contamination measurements, before and after the use of new soap, respectively. Let .
The null and the alternative hypotheses are:
H0 : μD ≤ 0
H1 : μD > 0
This is an upper-tailed test.
We have chosen = 0.05.
Since the sample selected is a matched pair sample, we shall use T = as the test statistic.
The population distributions are approximately normal. Hence, for D = 0, the distribution of T is approximately student’s t-distribution with df = (n - 1) = 7.
So, this is an upper-tailed t-test.
For df = 7, t = t0.05 = 1.895.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is greater than 1.895.
For the given sample data, the sample values of D are:
Before / 6.6 / 6.5 / 9.0 / 10.3 / 11.2 / 8.1 / 6.3 / 11.6After / 6.8 / 2.4 / 7.4 / 8.5 / 8.1 / 6.1 / 3.4 / 2.0
d / -0.2 / 4.1 / 1.6 / 1.8 / 3.1 / 2.0 / 2.9 / 9.6
Since the computed t-value (= 3.02) is greater than 1.895, there is sufficient evidence, at α = 0.05, to reject H0 in favour of H1, that is, to infer that the contamination measurements are lower after the use of the new soap.
39.Let p1 and p2 be the population proportions of current Advil users who, when given the new drug and the old drug, respectively, would indicate that the drug given is more effective. Then, the null and alternative hypotheses are:
H0 : p1-p2 ≤ 0
H1 : p1-p2 > 0
This is an upper-tailed test.
We have selected = 0.05.
The samples are selected independently, and therefore, we shall use as test statistic:
Hence, the sample sizes are large enough for the use of normal approximation. This is an upper-tailed Z-test.
The decision rule is: reject H0 in favour of H1 if the computed z-value is greater than z = z0.05 = 1.645.
The pooled estimate of proportion is: .
Since 1.019 < 1.645, we do not have sufficient evidence, at α = 0.05, to reject H0 in favour H1, that is, to infer that the new drug is more effective.
41.Let p1 and p2 be the population proportions of college women and men, respectively, who smoke at least a pack of cigarettes a day.
Then, the null and alternative hypotheses are:
H0 : p1-p2 = 0
H1 : p1-p2 ≠ 0
This is a two-tailed test.
We have selected = 0.05.
The samples are selected independently, and therefore, we shall use as test statistic:
Hence, the sample sizes are large enough for the use of normal approximation. This is a two-tailed Z-test.
The decision rule is: reject H0 in favour of H1 if the computed z-value is less than -z/2 = -z0.025 = -1.96 or if it is greater than 1.96.
The pooled estimate of proportion is:
Since 1.636 is between –1.96 and +1.96, we do not have sufficient evidence, at α = 0.05, to reject H0, that is, to infer that the two population proportions are different.
43.Let be the population mean values of stock prices of the top 100 companies on November 2001 and at present, respectively. Let .
The null and the alternative hypotheses are:
H0 : μD = 0
H1 : μD ≠ 0
This is a two-tailed test.
We have chosen = 0.05.
The sample is a matched-pair sample. Hence, we shall use T = as the test statistic.
The population distributions are approximately normal. Hence, for D = 0, the distribution of T is approximately student’s t-distribution with df = (n - 1) = 9.
So, this is a two-tailed t-test.
For df = 9, t/2 = t0.025 = 2.262.
Our decision rule is: reject H0 in favour of H1 if the computed t-value is less than -2.262 or if it is greater than 2.262.
Collect the sample data, compute and , and draw the conclusion.
45. (a) Let 1 = mean selling price of homes without pool; 2 = mean selling price of homes with pool. Then, the null and the alternative hypotheses are :
H0 : 1 = 2
H1 : 1 ≠ 2
By sorting the data according to the values of the “Pool” variable, separating the data on houses without pool (value of Pool = 0) from the data on houses with pool (value of Pool=1), and using Minitab and Microsoft Excel we get the following outputs.
From both the outputs, we see that the P-value is approximately 0.001. Since the selected value of α (= 0.05) is greater than the P-value, we have sufficient evidence to reject H0, that is, to infer that the population means of selling prices of houses with and without a pool are different.
(b)Let 1 = mean selling price of homes without garage; 2 = mean selling price of homes with garage. Then, the null and the alternative hypotheses are :
H0 : 1 = 2
H1 : 1 ≠ 2
By sorting the data according to the values of the “garage” variable, separating the data on houses without attached garage (value of garage = 0) from the data on houses with attached garage (value of garage =1), and using Minitab and Microsoft Excel we get the following outputs.
From both the outputs, we see that the P-value is almost 0. Since the selected value of α (= 0.05) is greater than the P-value, we have sufficient evidence to reject H0, that is, to infer that the population means of selling prices of houses with and without an attached garage are different.