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get all questions right for 15
Weights of women in one age group are normall
y
distributed with a standard deviation σ of
1
7
lbs. A researcher wants to estimate the mean weight of all women in this age group. Find how large a sample must be drawn in order to be 90% confident that the sample mean will not differ from the population mean by more than
3
.
5
lbs. Hint: Here, 3.5 is the Margin of Error. Determine the required sample size for estimating the population mean as presented in a formula in Chapter
8
.3)
Answer
A.
as large as possible
B.
76
C.
65
D.
91
The principal of a middle school claims that test scores of seventh-graders at her school vary less than test scores of seventh-graders at a neighboring school whose variation is described by a standard deviation of σ =
14
.7. Identify the null hypothesis, H0 and the alternative hypothesis Ha using σ for a claim about variation. Answer
A.
H0: σ > 14.7 H1: σ ≤ 14.7
B.
H0: σ < 14.7 H1: σ > 14.7
C.
H0: σ ≤ 14.7 H1: σ > 14.7
D.
H0: σ ≥ 14.7 H1: σ < 14.7
The manufacturer of a refrigerator system for beer kegs produces refrigerators that are supposed to maintain a true mean temperature, μ, of 46° F, ideal for a certain type of German pilsner. The owner of the brewery does not agree with the refrigerator manufacturer, and claims he can prove the true mean temperature is incorrect. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms. Answer
A.
There is sufficient evidence to support the claim that the mean temperature is different from 46°F.
B.
There is not sufficient evidence to support the claim that the mean temperature is different from 46°F.
C.
There is not sufficient evidence to support the claim that the mean temperature is equal to 46°F.
D.
There is sufficient evidence to support the claim that the mean temperature is equal to 46°F.
An article in a scientific journal states that fewer than
11
in ten thousand male fireflies are unable to produce light due to a genetic mutation. Identify the Type I error for the test.Answer
A.
The error of rejecting the claim that the true proportion is less than 11 in ten thousand when it really is less than 11 in ten thousand.
B.
The error of failing to reject the claim that the true proportion is less than 11 in ten thousand when it is actually more than 11 in ten thousand.
C.
The error of rejecting the claim that the true proportion is at least 11 in ten thousand when it really is at least 11 in ten thousand.
D.
The error of failing to reject the claim that the true proportion is at least 11 in ten thousand when it is actually less than 11 in ten thousand.
A group of 10 patients who had suffered heart attacks were given a pretest to determine their knowledge of nutrition. After a 6-week course in nutrition, they were given a posttest. Posttest scores minus pretest scores were used to determine 10 different scores. The mean difference, ud or d ̅ , equaled 10.5 and the standard deviation of differences equaled 5.5. Find a 90% confidence interval for the mean difference, notation: ud or d ̅ . Hint: This problem uses the paired t-interval procedure 10.6 in Chapter 10 of the OCR.Answer
A.
0 < ud. <
20
.6
B.
8.0 < ud. < 1
2.0
C.
7.3 < ud. < 13.7
D.
10.5 < ud. < 21.0
Two populations are normally distributed with equal standard deviations. The following statistics are obtained from the populations: n1 = 10, s1 = 5, n2 =
15
, s2 =
4.
What is the pooled sample standard deviation? Hint formula is in Section 10.2 of OCR. Answer
A.
3.9
B.
2.0
C.
19.5
D.
4.4
A chi-square test of significance is essentially concerned with: Answer
A.
the distinction between two interval level variables.
B.
only observed frequencies.
C.
the distinction between e
x
pected and observed frequencies.
D.
the distinction between one ordinal and one interval level variable.
A researcher is interested in knowing the effect of sleep deprivation on reading comprehension. She randomly assigns 200 students to two different groups: one group that is told to get exactly 8 hours of sleep, and one group that is told to get no more than 4 hours of sleep. The next morning, she administers a test to both groups. The results of the test are categorized as
Pass
/
Fail
and are illustrated below. The degrees of freedom for the chi-square test is equal to: Hint: df = (r-1)(c-1).
Pass
Fail
Group 1: 8 hours
83
17
Group 2: 4 hours
61
39
Answer
A.
3.
B.
4.
C.
1.
D.
2.
You take a sample and want to compare the results to the population from which it was drawn. The independent variable is “race” and the dependent variable is a yes/no response to whether they favor the death penalty. What test would you use to see if your results are significant? Answer
A.
a parametric test
B.
a chi-square test
C.
a correlation test
D.
a difference between means test
Find the value of the F test statistic to compare the means for brands X, Y, and Z. The claim is that the populations have the same mean. To save time use One-Way ANOVA.xls. Be sure to enable editing and change values to match this problem.
One-Way ANOVA(2).xls
Brand X
Brand Y
Brand Z
n = 10
n = 10
n = 10
x̅ = 2.5
x̅ = 2.5
x̅ = 2.1
s2 = 0.11
s2 = 0.32
s2 = 0.24
Answer
A.
F = 1.114
B.
F = 1.751
C.
F = 2.388
D.
F = 3.025
Use the given data to find the equation of the regression line. Round the final values to three decimal places, if necessary. Hint: You need to find the linear equation of the line: y = mx + b. To use Excel, enter x values and y values in two columns. Click Insert – Scatterplot. Right click any point on the graph. Left click “Add a Trendline” . Check “Display Equation on Chart” box.
x
3
5
7
15
16
y
8
11
7
14
20
Answer
A.
ŷ = 5.07 + 0.753x
B.
ŷ = 4.07 + 0.850x
C.
ŷ = 4.07 + 0.753x
D.
ŷ = 5.07 + 0.850x
AS
F
Data
| Supplier 1 | Supplier 2 | Supplier | 3 | Supplier | 4 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 18 | 5 | 26.3 | 2 | 0 | 25.4 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 24 | 25.3 | 25.2 | 19 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 17.2 | 20.8 | 22.6 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 21.2 | 24.7 | 17.5 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 24.5 | 22.9 | 20.4 |
COMPUTE
| ANOVA | n | c | Calculations | |||||||||||||||
| SUMMARY | ||||||||||||||||||
| Group | Count | Sum | Average | Variance | ||||||||||||||
| 97.6 | 19.52 | 7.237 | ||||||||||||||||
| 121.3 | 24.26 | 3.683 | ||||||||||||||||
| Supplier 3 | 114.2 | 22.84 | 4.553 | |||||||||||||||
| Supplier 4 | 105.8 | 21. | 16 | 8.903 | ||||||||||||||
| Source of Variation | SS | df | MS | P-value | F crit | |||||||||||||
| Between Groups | 63.2855 | 21.0952 | 3.4616 | 0.0414 | 3.2389 | |||||||||||||
| Within Groups | 97.504 | 6.094 | ||||||||||||||||
| Total | 160.7895 | |||||||||||||||||
| Level of significance | 0.05 |
COMPUTE_FORMULAS
| ANOVA: Single Factor | 21.16 |
| P – value |
TK4
| Tukey Kramer Multiple | Comparison | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Sample | Absolute | Std. Error | Critical | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Mean | Size | Difference | of Difference | Range | Results | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 1: Supplier 1 | Group 1 to Group 2 | 4.74 | 1.1040 | 4.4712 | Means are different | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 2: Supplier 2 | Group 1 to Group 3 | 3.32 | Means are not different | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 3: Supplier 3 | Group 1 to Group 4 | 1.64 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 4: Supplier 4 | Group 2 to Group 3 | 1.42 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Group 2 to Group 4 | 3.1 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Other Data | Group 3 to Group 4 | 1.68 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Numerator d.f. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Denominator d.f. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| MSW | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Q Statistic | 4.05 |
TK4_FORMULAS
| Tukey Kramer Multiple Comparisons | |||||||||||
| 1.1039927536 | 4.4711706521 | ||||||||||
TK3
TK5
| Group 1 to Group 5 | 0.0000 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 5: | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Group 2 to Group 5 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Group 3 to Group 5 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Group 4 to Group 5 |
TK6
| Group 1 to Group 6 | |
| 6: | |
| Group 2 to Group 6 | |
| Group 3 to Group 6 | |
| Group 4 to Group 6 | |
| Group 5 to Group 6 |
TK7
| Group 1 to Group 7 |
| 7: |
| Group 2 to Group 7 |
| Group 3 to Group 7 |
| Group 4 to Group 7 |
| Group 5 to Group 7 |
| Group 6 to Group 7 |
