ST 560 Final Study guide
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 95% confidence interval for the difference between the means of the two populations is: 0 to 6.92 -0.92 to 6.92 -2 to 2 -1.96 to 1.96
-0.92 to 6.92
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 =: a. -1.27. b. -1.64. c. -1.96. d. -1.53.
-1.27 find alpha using z-score chart
Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (lower tail) using α = .0630, z = a. -1.86. b. -1.53. c. -1.96. d. -1.645
-1.53 use z-score chart
The following information was obtained from matched samples taken from two populations. Assume the population of differences is normally distributed. Individual Method 1 Method 2 1 7 5 2 5 9 3 6 8 4 7 7 5 5 6 The 95% confidence interval for the difference between the two population means is: -3.776 to 1.776 -2.776 to 2.776 -1.776 to 1.776 -1.776 to 2.776
-3.776 to 1.776 https://www.chegg.com/homework-help/questions-and-answers/following-information-obtained-matched-samples-taken-two-populations-assume-population-dif-q37635110?trackid=05546cb59079&strackid=8e5c251b897f
The results of a recent poll on the preference of shoppers regarding two products are shown below. Product Shoppers SurveyedShoppers F Favoring This Product A 800 560 B 900 612 The point estimate for the difference between the two population proportions in favor of this product is: .44 .07 .68 .02
.68 Find average of both shoppers nad favored then divide favored shopper/ shoppers 586/850=0.68
The following information was obtained from matched samples taken from two populations.The daily production rates for a sample of workers before and after a training program are shown below. Assume the population of differences is normally distributed. Worker Before After 1 20 22 2 25 23 3 27 27 4 23 20 5 22 25 6 20 19 7 17 18 The point estimate for the difference between the means of the two populations is: 0 -2 -1 1
0 Explanation= find the average of 'before' and 'after' then do 'before' - after = 0
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a 0.95 probability, the sample mean will provide a margin of error of a) 0.196 b) 0.10 c) 1.96 d) 1.64
0.196 alpha = 1-0.95 = 0.05 alpha/2 = 0.025 Find Z score that is 1-0.025 aka 0.975 FROM Z SCORE TABLE ZScore = 1.96 Margin of Error = z*(StandardDeviation/(n)^1/2) MoE = 1.96*(1/(100)^1/2) = 0.196
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 point estimate for the difference between the two population means is 3 2 1 4
1 just subtract the means: 9-8
Read the z statistic from the normal distribution table and circle the correct answer. For a one-tailed test (upper tail) using α = .1230, z = a. 1.645. b. 1.54. c. 1.16. d. 1.96.
1.16
A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%. The test statistic is: a. 1.25. b. .80. c. .05. d. 2.00
1.25
The management of a department store is interested in estimating the difference between the mean credit purchases of customers using the store's credit card versus those customers using a national major credit card. You are given the following information. Store's Card Major Credit Card Sample size 64 49 Sample mean $140 $125 Population standard deviation $10 $8 A 95% confidence interval estimate for the difference between the average purchases of all customers using the two different credit cards is: 11.68 to 18.32 13.04 to 16.96 12.22 to 17.78 13.31 to 16.69
11.68 to 18.32 First find the difference between the means (140-125) = 15 then use margin of Error calculation to find MOE = 3.32 then add and subtract to 15 MOE = +/- 1.96 * sqrt((10^2/64)+(8^2/49))
Consider the following hypothesis problem. n = 14 H0: σ2 < 410 s = 20 Ha: σ2 > 410 The test statistic equals: 12.68 13.68 0.63 13.33
12.68
Consider the following hypothesis problem. n = 30 H0: σ2 = 500 s2 = 625 Ha: σ2 ≠ 500 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): 42.557 16.791 and 46.979 43.773 16.047 to 45.722
16.047 and 45.722
In an interval estimation for a proportion of a population, the value of z at 99.2% confidence is: 2.41 1.96 2.65 1.645
2.65 1-0.992 then use Z score chart
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. The 95% confidence interval for the true average checkout time (in minutes) is: 1 to 5 2.5 to 3.5 2.804 to 3.196 1.36 to 4.64
2.804 to 3.196 alpha: 1-0.95 --> 1.96 95% confidence: 3 +/- 1.96 (1/(sqrt(100)))
Consider the following information. SSTR = 6750 H0: μ1 = μ2 = μ3 = μ4 SSE = 8000 Ha: At least one mean is different The mean square due to treatments (MSTR) equals: 400 2250 1687.5 500
2250 MSTR = SSE - SSTR
The value of F .01 with 9 numerator and 20 denominator degrees of freedom is: 2.39 3.46 2.94 2.91
3.46 Use F = 0.01 tables to find this
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 mean square due to treatments (MSTR) equals: 1.872 5.86 34 36
36 View equation sheet its in module 12 single factor ANOVA
We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor. What is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidence? 100 58 385 200
385 n = (alpha/error)^2 * Prob* (1-prob) n = (1.96/0.05)^2 *0.5*0.5
The use of the normal probability distribution as an approximation of the sampling distribution of is based on the condition that both np and n(1 - p) equal or exceed: .05 30 10 5
5
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five treatments (a total of 65 observations). The following information is provided. SSTR = 200 (Sum of Squares Due to Treatments)SST = 800 (Total Sum of Squares) The test statistic is: 3.75 0.2 5.0 15
5.0 Test Statistic = MSTR/MSE so MSTR = SSTR / (5-1) SSE = SST - SSTR = 600 MSE = SSE/((65-1)-(5-1)) = 10 50/10 = 5
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 40 10 12
50 MSTR = SSTR/(N-1) MSTR = 200/4
A sample of 41 observations yielded a sample standard deviation of 5. If we want to test H 0: σ 2 = 20, the test statistic is: 50 100.75 10 51.25
50 Test Statistic = ((N-1)SD^2)/variance X^2 = ((41-1)*5^2)/20
The following random sample from a population whose values were normally distributed was collected. 10 8 11 11 The 95% confidence interval for μ is: a. 9.25 to 10.75. b. 7.75 to 12.25. c. 8.52 to 11.48. d. 8.00 to 10.00.
7.75 to 12.25 first find Mean: (sum of all numbers)/(number of numbers) Stand Dev: SQRT((10 -Mean) +(8 - mean) +...etc)/(number of numbers[4]) t* == IDK HOW interval is: MEan +/- (t*)*((stand dev)/(sqrt(number of numbers)))
It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is: 75 190 189 74
75 error = (alpha)*(stand dev/sqrt(sample size)) 5 = 1.96*(sqrt(484)/(sqrt(sample size)) n =75
In order to estimate the average electric usage per month, a sample of 81 houses was selected and the electric usage was determined. Assume a population standard deviation of 450 kilowatt-hours. At 95% confidence, the size of the margin of error is: 42 98 1.96 50.00
98 error = (alpha)*(stand dev/sqrt(sample size) error = 1.96 *(450/sqrt(81))
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 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 significantly greater than females average salary of males is significantly lower than females salaries of males and females are equal average salary of males is greater than females cannot be proved
Average salary of males is greater than females cannot be proved
Point estimate for the difference between the means of the two populations is: **didn't type the numbers out but want to know how to solve**
Find the mean of each sample and then subtract the two.
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 p-value is: a. less than .01. b. between .05 to .10. c. between .01 to .025. d. greater than .1.
Greater than 0.1
Regarding inferences about the difference between two population means, the sampling design that uses a pooled sample variance in cases of equal population standard deviations is based on: Independent Samples Conditional Samples Pooled Samples Research Samples
Independent Samples
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 Pop. standard deviation$1.00 $.95 At the 5% level of significance, the null hypothesis: Should be rejected Should not be rejected Should be revised Should not be tested
Should be rejected
A two-tailed test is performed at the .05 level of significance. The p-value is determined to be .09. The null hypothesis: a. has been designed incorrectly. b. could be rejected, depending on the sample size. c. must be rejected. d. should not be rejected.
Should not be rejected
To compute the necessary sample size for an interval estimate of a population proportion, all of the following procedures are recommended when p is unknown except: use the sample proportion from a previous study use judgment or a best guess use the sample proportion from a peliminary sample use 1.0 as an estimate
Use 1.0 as an estimate
If the null hypothesis is not rejected at the 5% level of significance, it: a. will also not be rejected at the 1% level. b. will always be rejected at the 1% level. c. will sometimes be rejected at the 1% level. d. may be rejected or not rejected at the 1% level.
Will also not be rejected at the 1% level
There is a .90 probability of obtaining a value such that
X^2(.95)<X^2<X^2(0.05)
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: xbar: 82 σ^2: 112.5 n: 45 Five years ago: xbar: 88 σ^2: 54 n: 36 The 95% confidence interval for the difference between the two population means is a) -9.92 to -2.08 b) -13.84 to -1.16 c) -3.08 to 3.92 d) -24.77 to 12.23
a) -9.92 to -2.08 Use first equation under Module 10 on equation sheet xbar1-xbar2 = -6 -6+ 3.92 = -2.08 -6 - 3.92 = -9.92
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. Treatment A: 20, 30, 25, 33 Treatment B: 22, 26, 20, 28 Treatment C: 40, 30, 28, 22 The test statistic to test the null hypothesis equals a) 1.06 b) 3.13 c) 0.944 d) 19.231
a) 1.06 https://www.chegg.com/homework-help/questions-and-answers/25-test-whether-difference-treatments-b-c-sample-12-observations-randomly-assigned-3-treat-q28040196
A sample of 51 elements is selected to estimate a 95% confidence interval for the variance of the population. The chi-square values to be used for this interval estimations are a) 32.357 and 71.420 b) 27.99 and 79.49 c) 34.764 and 67.505 d) 12.8786 and 46.9630
a) 32.357 to 71.420 use chi squared table and find value that's alpha/2 and 1-alpha/2 alpha = 0.05
In order to estimate the average electric usage per month a sample of 81 houses was selected and the electric usage was determined. Assume a population standard deviation of 450 kilowatt-hours. The standard error of the mean is a) 50 b) 450 c) 500 d) 81
a) 50 Standard Error = Standard Deviation / sqrt(n) SE = 450 / sqrt(81) = 50
A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. The margin of error at 95% confidence is a) 59.94 b) 50.07 c) 1.998 d) 80
a) 59.94 alpha = 1-.95 = 0.05 alpha/2 = 0.025 Find Z value at 0.975 Z score = 1.96 MoE = z*(standard deviation / sqrt(n)) MoE = 1.96*(240/sqrt(64)) = 58.8 closest answer is 59.94
In completely randomized experimental design involving five treatments, 13 observations were recorded for each of the 5 treatments (a total of 65 observations). Also, the design provided the following information: SSTR: 200 (sum of total squares due to treatments) SST: 800 (total sum of squares) The number of degrees of freedom corresponding to within-treatments is a) 60 b) 5 c) 4 d) 59
a) 60 total observations - number of treatments 65 -5 = 60
A completely randomized design is useful when the experimental units are a) homogenous b) heterogenous c) clustered d) stratified
a) homogenous
Consider the following information: SSTR: 6750, SSE: 8000 H0 = μ1 = μ2 = μ3 = μ4 Ha: at least one mean is different The null hypothesis is to be tested at 5% level of significance. The null hypothesis a) should be rejected b) should not be rejected c) was designed incorrectly d) cannot be tested
a) should be rejected can't find out why tho
In order to determine whether or not there is a significant difference between the mean hourly wages paid by the two companies, the following data have been accumulated Company A: sample size: 80 sample mean (mean1): $16.75 population standard deviation (s1): $1 Company B: sample size: 60 sample mean (mean2): $16.25 population standard deviation (s2): $0.95 at the 5% level of significance, the null hypothesis a) should be rejected b) should not be rejected c) should be revised d) should not be tested
a) should be rejected first: solve for t-stat: mean1-mean2/(sqrt((s1^2/n1)+(s2^2/n)) = 3.013 tcritical: 1.960 you reject bc t-stat>t-critical
The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing whether the variance of the population is significantly more than 0.003. The null hypothesis is a) σ^2 ≤ .003. b) s2 > .003. c) s2 ≤ .003. d) σ2 > .003.
a) σ2 ≤ .003.
The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is: . 26.8 to 124.356. b. 29.555 to 106.529. c. 5.177 to 11.152. d. 5.436 to 10.321.
a. 26.8 to 124.356. b. 29.555 to 106.529. c. 5.177 to 11.152. d. 5.436 to 10.321.
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: a. 8.576 to 39.794. b. 2.144 to 9.948. c. 2.93 to 6.31. d. 9.46 to 34.09.
a. 8.576 to 39.794
The academic planner of a university thinks that at least 35% of the entire student body attends summer school. The correct set of hypotheses to test his belief is: a. H0: p .35 Ha: p < .35. b. H0: p .35 Ha: p > .35. c. H0: p > .35 Ha: p .35. d. H0: p < .35 Ha: p .35.
a. H0: p .35 Ha: p < .35
The probability that the interval estimation procedure will generate an interval that does not contain the actual value of the population parameter being estimated is the: a. same as α. b. margin of error. c. proportion estimate. d. confidence coefficient.
a. same as α.
The probability of making a Type I error is denoted by: a. α. b. 1 - β. c. 1 - α. d. β.
a. α.
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 null hypothesis is: a. σ2 ≤ .003. b. s2 ≤ .003. c. s2 > .003. d. σ2 > .003.
a. σ2 ≤ .003.
The symbol used for the variance of the population is a) σ^2 b) s c) s^2 d) σ
a. σ^2
The results of a recent poll on the preference of shoppers regarding two products are down below Product A: Shoppers Surveyed: 800 Shoppers Favoring This Product: 560 Product B: Shoppers Surveyed: 900 Shoppers Favoring This Product: 612 The 95% confidence interval estimate for the difference between the populations favoring the products is a) 0.046 to 0.064 b) -0.024 to 0.064 c) -0.024 to 0.7 d) 0.6 to 0.7
b) -0.024 to 0.064 p1 = 560/800 = 0.7 p2 = 612/900 = 0.68 Z = 1.96 P^ = (560+612)/(800+900) = 0.6894 Plug into Module 10 confidence interval equation !
Read the t statistic form the t distribution table and circle the correct answer. For a one-tailed test (lower tail) with 22 degrees of freedom at alpha = 0.05 and the value of t = a) 1.383 b) -1.717 c) -1.721 d) -1.383
b) -1.717 t table??
If the null hypothesis to be tested in H0: mean-d = 0. The test statistic is **not all info is here bc can't add in all the numbers) a) 1.77 b) 0 c) -1.96 d) 1.00
b) 0 find the point estimate for the difference between the means of the population -> d = 0 test stat: d / (sd/(sqrt(n)) since d = 0, test stat = 0 https://www.chegg.com/homework-help/questions-and-answers/following-information-obtained-matched-samples-daily-production-rates-sample-workers-after-q233443
Read the t statistic from the t distribution table and circle the correct answer. For a one-tailed test (upper tail), using the sample size of 18, and at the 5% level of significance, t = a) -1.740 b) 1.740 c) 2.12 d) -2.12
b) 1.740 find on t-table DoF = 18-1 = 17 alpha = 0.05 using t-table = 1.740
A sample of 28 elements is selected to estimate a 95% confidence interval for the variance of the population. The chi square values to be used for this interval estimation are a) 14.573 and 43.195 b) 16.151 and 40.113 c) 11.808 and 49.645 d) 15.308 and 44.461
b) 14.573 and 43.195 use chi-square chart with n = 27 and alpha = 0.05 both values of alpha/2 and 1-alpha/2
In hypothesis testing, the critical value is a) the sample as the p-value b) a number that establishes the boundary of the rejection region c) the probability of a Type 1 error d) the probability of a Type 2 error
b) a number that establishes the boundary of the rejection region
The p-value a) can be any positive value b) must be a number between zero and one c) can be any negative value d) must be a number between -1 and 0
b) must be a number between zero and one
When developing an interval estimate for the difference between two population means with the sample sizes of n1 & n2 a) n1 must be equal to n2 b) n1 & n2 can be of different sizes c) n1 must be larger than n2 d) n1 must be smaller than n2
b) n1 & n2 can be of different sizes
Two approaches to drawing a conclusion in a hypothesis test are a) null and alternative b) p-value and critical value c) type 1 and type 2 d) one-tailed and two-tailed
b) p-value and critical value
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 three treatments. You are given the results below. Treatment A: 20, 30, 25, 33 Treatment B: 22, 26, 20, 28 Treatment C: 40, 30, 28, 22 The null hypothesis is to be tested at the 1% level of significance. The null hypothesis a) should be rejected b) should not be rejected c) should be revised d) should not be tested
b) should not be rejected not sure, chegg says it should be rejected https://www.chegg.com/homework-help/questions-and-answers/25-test-whether-difference-treatments-b-c-sample-12-observations-randomly-assigned-3-treat-q28040196
Consider the following hypothesis problem n = 23, s^2 = 60 H0: σ2 ≤ 66 Ha: σ2 > 66 If the test is to be performed at the 0.05 level of significance, the null hypothesis a) should be rejected b) should not rejected c) should be revised d) should not be tested
b) should not be rejected Find Chi square: (23-1)*(60/66) = 20 Chi square critical: use chi square chart and find value at DoF = 22 and alpha = 0.05 critical: 33.92 because chi-square= 20 and critical=33.92, the p value lies between 0.1 and 0.9 so it should be rejected bc p value is greater than 0.1
The probability of committing a Type 1 error when the null hypothesis is true as an equality is a) the confidence level b) the level of significance c) beta d) greater than 1
b) the level of significance
In hypothesis testing, the tentative assumption about the population parameter is a) the alternative hypothesis b) the null hypothesis c) either the null or the alternative d) neither the null nor the alternative
b) the null hypothesis
The following information was obtained from matched samples taken from two populations. Assume the population of differences is normally distributed. If the null hypothesis H0: μd = 0 is tested at the 5% level, a) the null hypothesis should be rejected b) the null hypothesis should not be rejected c) the alternative hypothesis should be revised d) the null hypothesis should be revised
b) the null hypothesis should not be rejected Take the difference between the two samples, (d) and then find the mean t-stat = (sqrt(n)*(d-0)/sd d = -1 sd = 2.236 t-stat = -1 which is larger than t-critical (2.776), we fail to reject .. hard math to show in quizlet https://www.chegg.com/homework-help/questions-and-answers/question-14-following-information-obtained-matched-samples-taken-two-populations-assume-po-q41692924
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 test statistic is: a. 500. b. 30. c. 1.2. d. 31.2.
b. 30.
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 for this ANOVA problem is: a. μ1 = μ2 = μ3 = μ4. b. μ1 = μ2 = μ3. c. μ1 = μ2. d. μ1 = μ2 = ... = μ12.
b. μ1 = μ2 = μ3. If you look it at there are 3 observations so the mean of each being the same is unlikely
Consider the following information. SSTR = 6750 H0: μ1 = μ2 = μ3 = μ4 SSE = 8000 Ha: At least one mean is different The null hypothesis is to be tested at the 5% level of significance. The p-value is: a. less than .01. b. between .025 and .05. c. between .01 and .025. d. greater than .10.
between 0.01 and 0.025
The mean of the t distribution is a) 1 b) problem specific c) 0 d) 0.50
c) 0
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. The standard error of the means equals a) 0.001 b) 0.01 c) .1 d) 1.00
c) 0.1 Standard error = standard deviation / (sqrt(n)) 1/sqrt(100) = 0.1
The critical F value with 6 numerator and 60 denominator degrees of freedom at alpha = 0.05 is a) 2.37 b) 3.74 c) 2.25 d) 1.96
c) 2.25 use F critical values chart
Consider the following hypothesis problem n=23, s^2 = 60 H0 σ2 ≤ 66 Ha σ2 > 66 The test statistic has a value of a) 24 b) 24.20 c) 20 d) 20.91
c) 20 I think? (23-1)*(60/66) = 20
We are interested in conducting a study in order to determine the percentage of voters in a city who would vote for the incumbent mayor. What is the minimum sample size needed to estimate the population proportion with a margin of error not exceeding 4% at 95% confidence? a) 625 b) 600 c) 601 d) 626
c) 601 when the population proportion is unknown (like it is here) assume 0.5 for alpha MoE = z*sqrt((0.5*(1-.5)/n) MoE = 0.04 z = 1.96 n = (0.04/1.96)^2/(.5*.5) n = 600.25 -> 601 !!
The sampling distribution of the ratio of independent sample variances extracted from two normal populations with equal variances is the a) t distribution b) normal distribution c) F distribution d) chi-square distribution
c) F distribution
In the past, 75% of the tourists who visited Chattanooga went to see Rock City. The management of Rock City recently undertook an extensive promotional campaign. They are interested in determining whether the promotional campaign actually increased the proportion of tourist visiting Rock City. The correct set of hypotheses is a) H0: p ≤ 0.75, Ha: p>0.75 b) H0: p>0.75, Ha: ≥ 0.75 c) H0: p≤0.75, Ha: p>0.75 d) H0: p≥0.75, Ha: p<0.75
c) H0: p≤0.75, Ha: p>0.75 idk why this is right, it just is
The producer of a certain bottling equipment claims that the variance of all their filled bottles is 0.027 or less. A sample of 30 bottles showed a standard deviation of 0.2 ounces. The p-value for the test is a) between 0.05 to 0.1 b) 0.05 c) between 0.025 to 0.05 d) 0.025
c) between 0.025 and 0.05 I honestly don't know how to calculate the p value from a chi squared value chi square: (30-1)*(0.2^2/(0.027) = 42.96
The sampling distribution of the quantity (n-1)s^2/σ^2 a) F distribution b) normal distribution c) chi-square distribution d) t distribution
c) chi-square distribution
In factorial designs, the response produced when the treatments of one factor interact with the treatments of another in influencing the response variable is known as a) main effect b) replication c) interaction d) error
c) interaction
The sampling distribution of p1-p2 is approximated by a normal distribution if ___ are all greater than or equal to 5 a) n1p2, p2(1-n2), n2p1, p1(1-n1) b) n1p2, n1(1-p2), n2p1, n2(1-p1) c) n1p1, n1(1-p1), n2p2, n2(1-p2) d) n1p1, p1(1-n1), n2p2, p2(1-n2)
c) n1p1, n1(1-p1), n2p2, n2(1-p2)
The t distribution should be used whenever a) the population standard deviation is known b) the sample size is less than 30 c) the sample standard deviation is used to estimate the population standard deviation d) the population is normally distributed
c) the sample standard deviation is used to estimate the population standard deviation
In ANOVA, which of the following is not affected by whether or not the population means are equal? a) xbar b) between-treatments estimate of σ^2 c) within-treatments estimate of σ^2 d) ratio between and within treatments estimate of σ^2
c) within-treatments estimate of σ^2
In order to test the following hypotheses at an alpha level of significance H0: u≤800 Ha: a>800 the null hypothesis will be rejected, if the test statistic z is a) < -za b) = alpha c) ≥ z(Alpha) d) <za
c) ≥ za
Consider the following ANOVA table. Sourceof Variation Sumof Squares Degreesof Freedom MeanSquare F Between Treatments 2073.6 4 Between Blocks 6000 5 1200 Error 20 288 Total 29 The null hypothesis for this ANOVA problem is: a. μ1 = μ2 = μ3 = μ4 = μ5 = μ6. b. μ1 = μ2 = ... = μ20. c. μ1 = μ2 = μ3 = μ4 = μ5. d. μ1 = μ2 = μ3 = μ4.
c. μ1 = μ2 = μ3 = μ4 = μ5 When looking for null hypothesis look for degrees of freedom to find this so (4+1) = 5 here
If two independent large samples are taken from two populations, the sampling distribution of the difference between the two sample means: can be approximated by a normal distribution can be approximated by a normal distribution will have a mean of 1 will have a variance of 1
can be approximated by a normal distribution
The sample size needed to provide a margin of error of 2 or less with a .95 probability when the population standard deviation equals 11 is a) 10 b) 116 c) 11 d) 117
d) 117 MoE = 2 (or less) MoE = z*(standard deviation/sqrt(n)) Solve for n! z score at 0.95 probability is 1.96 (trust me) n=((1.96*11)/2)^2 = 116.2 so round up! 117
The following information was obtained from independent random samples taken of two populations. Assume normally distributed populations with equal variances Sample 1: sample mean: 45 sample variance: 85 sample size: 10 Sample 2: sample mean: 42 sample variance: 90 sample size: 12 The degrees of freedom for the t distribution are a) 24 b) 21 c) 22 d) 20
d) 20 solve by using the 'v' equation for Case 2 in Module 10 on study guide
In order to estimate the average electric usage per month, a sample of 81 houses was selected and the electric usage was determined. Assume a population standard deviation of 450 kilowatt-hours. At 95% confidence, the size of the margin of error is a) 42 b) 50 c) 1.96 d) 98.00
d) 98.00 95% confidence -> alpha is 0.05 alpha/2 = 0.025 Z score of (1-0.025) = 1.96 MoE = z*(standard deviation/(sqrt(n)) MoE = 1.96*(450/sqrt(81)) = 98
The bottler of a certain soft drinks claims their equipment to be accurate and that the variance of all filled bottles is 0.05 or less. The null hypothesis in a test to confirm the claim would be written as a) H0 σ2 ≥ .05 b) H0 σ2 < .05 c) H0 σ2 > .05. d) H0 σ2 ≤ .05.
d) H0 σ2 ≤ .05
When the p-value is used for hypothesis testing, the null hypothesis is rejected if a) p-value = 1-alpha/2 b) p-value = 1- alpha c) alpha < p-value d) p-value ≤ alpha
d) p-value ≤ alpha
The standard error of x1-x2 is the a) margin of error of x1-x2 b) variance of the sampling distribution of x1-x2 c) pooled estimator of x1-x2 d) the standard deviation of the sampling distribution of x1-x2
d) the standard deviation of the sampling distribution of x1-x2
To compute the necessary sample size for an interval estimate of a population proportion, all of the following procedures are recommended when p is unknown except a) use the sample proportion from a preliminary sample b) use judgement of a best guess c) use the sample proportion from a previous study d) use 1.0 as an estimate
d) use 1.0 as an estimate
When the p-value is used for hypothesis testing, the null hypothesis is rejected if: a. α < p-value. b. p-value = 1 - α/2. c. p-value = 1 - α. d. p-value < α
d. p-value < α
The symbol used for the variance of the sample is: a. s. b. σ. c. σ2. d. s2.
d. s2.
In testing for the equality of k population means, the number of treatments is a) k b) nT c) nT-k d) k-1
k
In general, higher confidence levels provide: a wider confidence intervals unbiased estimates narrower confidence intervals smaller standard error
wider confidence intervals
A sample of 1400 items had 280 defective items. For the following hypothesis test, H0: p ≤ .20 Ha: p > .20 the test statistic is: a. .20. b. zero. c. .14. d. .28.
zero
