- A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The correct null hypothesis for this problem is μ <= 8000
- A population has a standard deviation of 50. A random sample of 100 items from this population is selected. The sample mean is determined to be 600. At 95% confidence, the margin of error is 9.8
- A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is 63
- A sample of 1400 items had 280 defective items. For the following hypothesis test,
H0: p ≤ .20Ha: p > .20
the test statistic is 0 - A sample of 60 items from population 1 has a sample variance of 8 while a sample of 40 items from population 2 has a sample variance of 10. If we want to test whether the variances of the two populations are equal, the test statistic will have a value of 1.25
- A school’s newspaper reported that the proportion of students majoring in business is at least 30%. You plan on taking a sample to test the newspaper’s claim. The correct set of hypotheses is H0: p >= .30
Ha: p <= .30 - A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information.
Today
Five Years Ago
82
88
σ2
112.5
54
n
45
36
The test statistic for the difference between the two population means is -3.0 - A two-tailed test is performed at the .05 level of significance. The p-value is determined to be .09. The null hypothesis should not be rejected.
- An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had in the previous year. The results are shown below.
Under Age of 18 Over Age of 18 n1 = 500 n2 = 600 Number of accidents = 180 Number of accidents = 150
We are interested in determining if the accident proportions differ between the two age groups.
The pooled estimator of the population proportion is .300 - Consider the following ANOVA table.
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Between Treatments
2073.6
4
Between Blocks
6000
5
1200
Error
20
288
Total
29
The mean square due to treatments equals 518.4 - Consider the following ANOVA table.
Source of Variation
Sum of Squares
Degrees of Freedom
Mean Square
F
Between Treatments
2073.6
4
Between Blocks
6000
5
1200
Error
20
288
Total
29
The test statistic to test the null hypothesis equals 1.8 - Consider the following hypothesis problem.
n = 14
H0: σ2 <= 410
s = 20
Ha: σ2 > 410
The null hypothesis is to be tested at the 5% level of significance. The critical value(s) from the chi-square distribution table is(are) 22.362 - Consider the following hypothesis problem.
n = 30
H0: σ2 = 500
s2 = 625
Ha: σ2 ≠ 500
At the 5% level of significance, the null hypothesis should not be rejected - Consider the following information.
SSTR = 6750 H0: μ1 = μ2 = μ3 = μ4SSE = 8000 Ha: At least one mean is different
The null hypothesis is to be tested at the 5% level of significance. The null hypothesis Should be rejected - For a sample size of 21 at 95% confidence, the chi-square values needed for interval estimation are 9.591 and 34.170
- For an F distribution, the number of degrees of freedom for the numerator can be larger, smaller, or equal to the number of degrees of freedom for the denominator.
- From a population with a variance of 900, a sample of 225 items is selected. At 95% confidence, the margin of error is 3.92
- If the null hypothesis is rejected in hypothesis testing, the alternative hypothesis is true.
- If the probability of a Type I error ( α) is .05, then the probability of a Type II error ( β) must be cannot be computed
- If we are interested in testing whether the proportion of items in population 1 is larger than the proportion of items in population 2, the alternative hypothesis should state p1 – p2 > 0.
- In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations). Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
If, at a 5% level of significance, we want to determine whether or not the means of the five populations are equal, the critical value of F is 2.53 - In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations). Also, the design provided the following information.
SSTR = 200 (Sum of Squares Due to Treatments)
SST = 800 (Total Sum of Squares)
The mean square due to treatments (MSTR) is 50.00 - In an analysis of variance problem involving 3 treatments and 10 observations per treatment, SSE = 399.6. The MSE for this situation is 14.8
- In an analysis of variance, one estimate of σ2 is based upon the differences between the treatment means and the overall sample mean
- In interval estimation, as the sample size becomes larger, the interval estimate becomes narrower
- In order to determine whether or not the means of two populations are equal, either a t test or an analysis of variance can be performed.
- In order to determine whether or not there is a significant difference between the mean hourly wages paid by two companies (of the same industry), the following data have been accumulated.
Company A
Company B
Sample size
80
60
Sample mean
$16.75
$16.25
Population standard deviation
$1.00
$.95
At the 5% level of significance, the null hypothesis should be rejected - In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered.
Downtown Store
North Mall Store
Sample size
25
20
Sample mean
$9
$8
Sample standard deviation
$2
$1
A 95% interval estimate for the difference between the two population means is .071 to 1.929 - Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (lower tail) using α = .1020, z -1.27
- Salary information regarding male and female employees of a large company is shown below.
Male
Female
Sample Size
64
36
Sample Mean Salary (in $1000)
44
41
Population Variance ( )
128
72
The standard error of the difference between the two sample means is 2.0 - Salary information regarding male and female employees of a large company is shown below.
Male
Female
Sample Size
64
36
Sample Mean Salary (in $1000)
44
41
Population Variance (sigma squared)
128
72
If you are interested in testing whether or not the population average salary of males is significantly greater than that of females, at α = .05, the conclusion is that the population average salary of males is greater than females cannot be proved. - The ability of an interval estimate to contain the value of the population parameter is described by the confidence level.
- The average hourly wage of computer programmers with 2 years of experience has been $21.80. Because of high demand for computer programmers, it is believed there has been a significant increase in the average hourly wage of computer programmers. To test whether or not there has been an increase, the correct hypotheses to be tested are H0: μ ≤ 21.80
Ha: μ >= 21.80 - The contents of a sample of 26 cans of apple juice showed a standard deviation of .06 ounces. We are interested in testing whether the variance of the population is significantly more than .003. The p-value for this test is greater than 0.10
- The critical value of F using α = .05 when there is a sample size of 21 for the sample with the smaller variance, and there is a sample size of 9 for the sample with the larger sample variance is 2.45
- The following information was obtained from independent random samples taken of two populations. Assume normally distributed populations with equal variances.
Sample 1
Sample 2
Sample Mean
45
42
Sample Variance
85
90
Sample Size
10
12
The standard error of x1-x2 is 4 - The manager of a laptop computer dealership is considering a new bonus plan in order to increase sales. Currently, the mean sales rate per salesperson is five laptops per week. The correct set of hypotheses for testing the effect of the bonus plan is H0: μ <= 5
Ha: μ > 5. - The manager of the service department of a local car dealership has noted that the service times of a sample of 15 new automobiles has a standard deviation of 4 minutes. A 95% confidence interval estimate for the variance of service times for all their new automobiles is 8.576 to 39.794
- The margin of error in an interval estimate of the population mean is a function of all of the following except sample mean
- The mean of the t distribution is 0
- The random variable for a chi-square distribution may assume any value greater than zero.
- The results of a recent poll on the preference of shoppers regarding two products are shown below.
Product
Shoppers Surveyed
Shoppers FavoringThis Product
A
800
560
B
900
612
The standard error of p1-p2 is .0225 - The results of a recent poll on the preference of shoppers regarding two products are shown below.
Product
Shoppers Surveyed
Shoppers Favoring This Product
A
800
560
B
900
612
The 95% confidence interval estimate for the difference between the populations favoring the products is -.024 to .064 - The sample size that guarantees the estimate of a population proportion satisfying the margin of error requirement is computed using a planning value of p equal to .50
- The t distribution should be used whenever the sample standard deviation is used to estimate the population standard deviation.
- To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below.
Treatment
Observations
A
20
30
25
33
B
22
26
20
28
C
40
30
28
22
The test statistic to test the null hypothesis equals 1.06 - To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below.
Treatment
Observations
A
20
30
25
33
B
22
26
20
28
C
40
30
28
22
The null hypothesis is to be tested at the 1% level of significance. The null hypothesis should not be rejected - Two approaches to drawing a conclusion in a hypothesis test are p-value and critical value
- We are interested in testing whether the variance of a population is significantly more than 625. The null hypothesis for this test is H0: σ2 <= 625
- We can use the normal distribution to make confidence interval estimates for the population proportion, p, when both np >= 5 and n(1 – p) >= 5
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