• ### Help with Statistics problem?

Best answer: 1. Use z-test only. You may remember that for large n (>30), if the population proportion is θ, then the value (p - θ) / √ {θ (1 - θ) / n } is close to the value of Z. Here n=200, p=85/200=0.425 H0 : θ = 0.50, H1 : θ ≠ 0.50 Under H0, z = (0.425 - 0.50) / √ {0.50 (1 - 0.50) / 200} = -2.121 Two-tailed test, critical region is of... show more
Best answer: 1. Use z-test only. You may remember that for large n (>30), if the population proportion is θ, then the value (p - θ) / √ {θ (1 - θ) / n } is close to the value of Z. Here n=200, p=85/200=0.425 H0 : θ = 0.50, H1 : θ ≠ 0.50 Under H0, z = (0.425 - 0.50) / √ {0.50 (1 - 0.50) / 200} = -2.121 Two-tailed test, critical region is of the form {|Z|>z}. Level of test α= 0.05 P{|Z|>z}= 2*P{Z > 2.121} = 2*P{Z < -2.121} = 2(0.0170) = 0.034< 0.05; significant at 5%. The given data show that the pennies are not fair. At 1%, p-value >0.01, not significant at 1%. The given data unable to show that the pennies are not fair. 2. Testing for difference of 2 means in independent populations. Let µ1 be the population mean sleep hours viewers, µ2 be the population mean for non-viewers. H0: µ2-µ1 = 0 H1: µ2-µ1> 0 Sampling involves n1=80, n2=80, large sample sizes Test Statistics: z-test, z = (435-420)/ √(1100/80 + 900/80) = 3.00 One-tailed test, right, critical region on right tail, of the form {Z>z} p-value, P{Z>3.00} = P{Z<-3.00} = 0.0013 < 0.01 Reject at 1% and hence at 5%. Decision: The claim is supported.
1 answer · Mathematics · 6 years ago
• ### How to Find the Standard Deviation of paired t-test? HELP!!! Due in a few hours!!?

Best answer: Here is the basic for paired t-test. Get d = x - y for the 7 pairs. Look at d as a random variable then find dbar, and sd = std deviation for d. H0: µd = 0, H1: µd > 3. Since n=7 < 30, assume observation as normal, unknown variance, use t-test, here d.f. ν = n - 1 = 6. One tailed test. Critical region is on the right tail,... show more
Best answer: Here is the basic for paired t-test. Get d = x - y for the 7 pairs. Look at d as a random variable then find dbar, and sd = std deviation for d. H0: µd = 0, H1: µd > 3. Since n=7 < 30, assume observation as normal, unknown variance, use t-test, here d.f. ν = n - 1 = 6. One tailed test. Critical region is on the right tail, look at P{T>t}. Calculate test statistic value: t = (dbar - 0)/(sd/√ 7). Find P{T>t}, if < 0.01 significant at this level, else not. Conclude: (Significant) Data shows that ...... (else) Data is unable to show ....
1 answer · Mathematics · 6 years ago
• ### Statistics help! Need help?

Best answer: Take percent as actual observations. Ho: mean=80, Ha: mean<80 n=50, large, use z-test, sigma is estimated by s=6. z=(xbar-mean)/(6/sqrt 50) = (78-80)/0.849 = -2.36. One-tailed left, critical region on the left, {Z<z} p-value= P{Z<-2.36}=0.0091<0.05, significant. This indicates the scheduled flight is unprofitable. show more
Best answer: Take percent as actual observations. Ho: mean=80, Ha: mean<80 n=50, large, use z-test, sigma is estimated by s=6. z=(xbar-mean)/(6/sqrt 50) = (78-80)/0.849 = -2.36. One-tailed left, critical region on the left, {Z<z} p-value= P{Z<-2.36}=0.0091<0.05, significant. This indicates the scheduled flight is unprofitable.
1 answer · Mathematics · 6 years ago
• ### Statistics help urgent!!?

Best answer: One-tailed right H0 : µ = 81, H1: µ > 81 (Note, H0 always take =, irrespective of H1) One-tailed test, right, critical region on the right, of the form {T>t}. Level of test α=0.10. Since n=10< 30, assume population as normal, unknown population variance, use t-test. df = ν = 10-1 = 9. µ=81, xbar=78, s=2 Under H0, value... show more
Best answer: One-tailed right H0 : µ = 81, H1: µ > 81 (Note, H0 always take =, irrespective of H1) One-tailed test, right, critical region on the right, of the form {T>t}. Level of test α=0.10. Since n=10< 30, assume population as normal, unknown population variance, use t-test. df = ν = 10-1 = 9. µ=81, xbar=78, s=2 Under H0, value of test statistic, t = (78-81)/(2/√10) = -4.74< 0. For sure t would fall outside the critical region. Not significant at 0.10. (do not reject H0 at 0.10) Conclusion: Mean > 81 is not supported.
1 answer · Mathematics · 6 years ago
• ### When two dice are rolled, we obtain a sum of 7 or 8 on the dice.?

Best answer: Arrange into a 6x6 form, like (1,1), (1,2).... (1,6) (1,1), (2,2).... Event A={(1,6), (2,5), .. (3,5), (5,3),...} You can also describe the outcome as sums, B={7, 8}. A and B are equivalent events.
Best answer: Arrange into a 6x6 form, like (1,1), (1,2).... (1,6) (1,1), (2,2).... Event A={(1,6), (2,5), .. (3,5), (5,3),...} You can also describe the outcome as sums, B={7, 8}. A and B are equivalent events.
1 answer · Mathematics · 6 years ago
• ### Statistics homework problem help?

Best answer: A. One-tailed right, critical region of the form {T>t}; ν=18, p-value = P{T>2.1} = 0.025 (at t=2.1009) (p-value is slightly larger than 0.025, if you want to be exact) It is advisable to say "p-value is between 0.025 and 0.05", because we want to arrive at, like, p-value <0.05; in testing, exact p=value is not... show more
Best answer: A. One-tailed right, critical region of the form {T>t}; ν=18, p-value = P{T>2.1} = 0.025 (at t=2.1009) (p-value is slightly larger than 0.025, if you want to be exact) It is advisable to say "p-value is between 0.025 and 0.05", because we want to arrive at, like, p-value <0.05; in testing, exact p=value is not important, comparing with 0.05, 0.01 is important. B. Two-tailed, critical region of the form {|T|>t}; ν=18, p-value = 2P{T>2.1} = 0.05 (at t=2.1009) Follow like argument above, p-value is between 0.05 and 0.10. At 0.05, test is not significant.
1 answer · Mathematics · 6 years ago
• ### Stats random variable question?

Best answer: Like X~Binomial n=10, p=0.20, X representing number of defectives; alternatively, Y~Binomial n=10, p=0.80, Y rep no of non-defectives. P{Y= 9 or 10} = ? Kept.
Best answer: Like X~Binomial n=10, p=0.20, X representing number of defectives; alternatively, Y~Binomial n=10, p=0.80, Y rep no of non-defectives. P{Y= 9 or 10} = ? Kept.
1 answer · Mathematics · 8 years ago
• ### What is the logic of variance and standard deviation, and how are they related?

Best answer: You know spread or dispersion of a given data? Some data may be more spread than others. How would you compare? Use measurements like range, inter-quartile range and mean deviation. For mean deviation, you look at the distances each observation from the mean, in absolute values, and take the average of the distances. In similar... show more
Best answer: You know spread or dispersion of a given data? Some data may be more spread than others. How would you compare? Use measurements like range, inter-quartile range and mean deviation. For mean deviation, you look at the distances each observation from the mean, in absolute values, and take the average of the distances. In similar concept, we look at the square of the difference between each observation from the mean; sum all differences and divide the result by the number of observations (or less one). We define the value (i.e. we agree to give its name) as the variance of the data. The drawback is that variance has the unit squared the original measure. Example, cm and cm^2. To overcome, we take the square root of the variance, and call the result as standard deviation. Both are measures of spread or dispersion.
1 answer · Mathematics · 8 years ago
• ### What is the mu of a null hypothesis states that people exercising have a heart rate less than 74?

1 answer · Mathematics · 6 years ago
• ### Statistics Hypothesis Testing Question!?

Best answer: Test of proportion, one population One-tailed right Use z-test only. You may remember that for large n (>30), if the population proportion is θ, then the value (p - θ) / √ {θ (1 - θ) / n } is close to the value of Z. Here n=150. H0 : θ = 0.84, H1 : θ > 0.84 p = 134/150 = 0.0.8933. Under H0, z = (0.8933 - 0.84) / √ {0.84 (1... show more
Best answer: Test of proportion, one population One-tailed right Use z-test only. You may remember that for large n (>30), if the population proportion is θ, then the value (p - θ) / √ {θ (1 - θ) / n } is close to the value of Z. Here n=150. H0 : θ = 0.84, H1 : θ > 0.84 p = 134/150 = 0.0.8933. Under H0, z = (0.8933 - 0.84) / √ {0.84 (1 - 0.84) / 150} = 1.782 One-tailed test, right, critical region is of the form (Z>z}. Level =0.05 P{Z > 1.782} = P{Z < -1.782} = 0.0375 < 0.05; significant at 5%. Santa Clara County drivers are using seat belts more than the 2011 national average of 84%.
1 answer · Mathematics · 6 years ago
• ### What conclusion is appropriate if a chi square test produces a chi square statistic near zero?

Best answer: expected values are close to observed values, tend to favor the null hypothesis; existence of relationship is not supported, or like, departure from expectation is not supported.
Best answer: expected values are close to observed values, tend to favor the null hypothesis; existence of relationship is not supported, or like, departure from expectation is not supported.
1 answer · Mathematics · 6 years ago
• ### Does rejecting the null hypothesis tell us exactly what the relationship between the IV and DV is?

Best answer: It tells the existence of relationship between the variables, with an error of at most like 5%, 1%.
Best answer: It tells the existence of relationship between the variables, with an error of at most like 5%, 1%.
2 answers · Mathematics · 6 years ago
• ### Statistics help! Value of Test Statistic?!?

Best answer: Here is the basic for this particular paired t-test. Get d = x - y for the 7 pairs. Look at d as a random variable then find dbar, and sd = std deviation for d. Test for difference greater than 0. (Effective) H0: µd = 0, H1: µd > 0. Since n=7 < 30, assume observation as normal, unknown variance, use t-test, here d.f. ν = n... show more
Best answer: Here is the basic for this particular paired t-test. Get d = x - y for the 7 pairs. Look at d as a random variable then find dbar, and sd = std deviation for d. Test for difference greater than 0. (Effective) H0: µd = 0, H1: µd > 0. Since n=7 < 30, assume observation as normal, unknown variance, use t-test, here d.f. ν = n - 1 = 6. One tailed test, right. Critical region is on the right tail, of the form {T>t}, look at P{T>t}. Calculate test statistic value: t = (dbar - 0)/(sd/√7). Find p-value, P{T>t}, if < 0.05, test is significant at this level, else not. Conclude: (Significant) Data shows that ...... (else) Data is unable to show .... Alternatively (classical): Get t for P{T>t} = 0.05, t = 1.9432 (T-table, df=6, or use Excel) CR = {T>1.9432}. If calc t inside CR, like t=2.57, reject H0 at 0.05, otherwise, like t=1.22, do not. Conclude: (H0 rejected) Data shows that ...... (else) Data is unable to show ....
1 answer · Mathematics · 6 years ago
• ### Help with hypothesis testing?

Best answer: PAIRED t-test Here is the basic for paired t-test. Get d = x - y for the 9 students. Look at d as a random variable then find dbar, and sd = std deviation for d. H0: µd = 0, H1: µd≠ 0 . Since n=9< 30, assume observation as normal, unknown variance, use t-test, here d.f.ν = 9 - 1 = 8. Two-tailed test. Critical region is on... show more
Best answer: PAIRED t-test Here is the basic for paired t-test. Get d = x - y for the 9 students. Look at d as a random variable then find dbar, and sd = std deviation for d. H0: µd = 0, H1: µd≠ 0 . Since n=9< 30, assume observation as normal, unknown variance, use t-test, here d.f.ν = 9 - 1 = 8. Two-tailed test. Critical region is on both ends, look at P{|T|>t}. Calculate test statistic value: t = (dbar - 0)/(sd/√ 9). From table get P{T>t}, then P{|T|>t} = 2*P{T>t}, if < 0.05 significant at this level, else not. Conclude: (Significant) Data shows that ...... (else) Data is unable to show .... Alternatively (classical): Get t for P{T>t} = 0.025, t = 2.3060 (T-table, df=8, or use Excel) CR = {|T|>2.3060}. If calc t inside CR, like t=3.52, reject H0 at 0.05, otherwise, like t=2.22, do not. Conclude: (H0 rejected) Data shows that ...... (else) Data is unable to show ....
1 answer · Mathematics · 6 years ago

Best answer: This is a case of normal approximation to binomial. n=200 is a large sample. Use z-test. Test of Proportion - one population One-tailed left You may remember that for large n (>30), if the population proportion is θ, then the value (p - θ) / √ {θ (1 - θ) / n } is close to the value of Z. Here n=200. H0 : θ = 0.40, H1 : θ <... show more
Best answer: This is a case of normal approximation to binomial. n=200 is a large sample. Use z-test. Test of Proportion - one population One-tailed left You may remember that for large n (>30), if the population proportion is θ, then the value (p - θ) / √ {θ (1 - θ) / n } is close to the value of Z. Here n=200. H0 : θ = 0.40, H1 : θ < 0.40 p = 74/200 = 0.37. Under H0, z = (0.37 - 0.40) / √ {0.40 (1 - 0.40) / 200} = - 0.866 One-tailed test, left, the critical region is of the form {Z<z}. Level of test α=0.01. P{Z < - 0.866} = 0.1922 > 0.01; not significant at 1%. There is no significance evidence that the % is too large.
1 answer · Mathematics · 6 years ago
• ### Paired samples (statistics help)?

Best answer: Do waiters earn more in tips, one-tailed - P{T>3.7}=0.00027 < 0.05, reject H0. Do waiters or waitresses earn more in tips, two-tailed - P{|T|>3.7}=0.000544 < 0.05, reject H0. Next time I propose you go for z-test, treating d=x-y as new observations, since n=50>30 is large enough. This example may be helpful.... show more
Best answer: Do waiters earn more in tips, one-tailed - P{T>3.7}=0.00027 < 0.05, reject H0. Do waiters or waitresses earn more in tips, two-tailed - P{|T|>3.7}=0.000544 < 0.05, reject H0. Next time I propose you go for z-test, treating d=x-y as new observations, since n=50>30 is large enough. This example may be helpful. PAIRED t-test Here is the basic for paired t-test. Get d = x - y for the n pairs. Look at d as a random variable then find dbar, and sd = std deviation for d. H0: µd = 0, H1: µd>0 . Assume observation as normal, unknown variance, use t-test, here d.f.ν = n - 1 . One tailed test. Critical region is on the right tail, look at P{T>t}. Calculate test statistic value: t = (dbar - 0)/(sd/√n). Find P{T>t}, if < 0.05 significant at this level, else not. Conclude: (Significant) Data shows that ...... (else) Data is unable to show ....
1 answer · Mathematics · 6 years ago
• ### Using the z table, find the critical value (or values) for an = 0.13 right-tailed test.?

Best answer: P{Z>z}=P{Z<-z}=0.13, z=1.126 or 1.13.
Best answer: P{Z>z}=P{Z<-z}=0.13, z=1.126 or 1.13.
1 answer · Mathematics · 6 years ago
• ### Determine the p-value for Ho: μd = 0 and Ha: μd > 0, with n = 25 and t = 1.81.?

Best answer: One-tailed test, right, critical region is of the form {T>t} P-value, df=24, P{T>1.81} = between 0.05 and 0.025; your aim is to determine if this p-value is < or > the given test level, like 0.05 or 0.01. Here < 0.05 (significant at 5%), > 0.01 (not at 1%). If test level is not given, significant because <0.05.... show more
Best answer: One-tailed test, right, critical region is of the form {T>t} P-value, df=24, P{T>1.81} = between 0.05 and 0.025; your aim is to determine if this p-value is < or > the given test level, like 0.05 or 0.01. Here < 0.05 (significant at 5%), > 0.01 (not at 1%). If test level is not given, significant because <0.05. If you want exact p-value go to Excel.
1 answer · Mathematics · 6 years ago
• ### Hypothesis testing question?

Best answer: Hope the following could help you. T-test in single population One-tailed right H0 : µ = 12, H1: µ > 12 (Note, H0 always take =, irrespective of H1) One-tailed test, right, critical region on the right, of the form {T>t}. Since n=25 < 30, assume population as normal, unknown population variance, use t-test. df = ν =... show more
Best answer: Hope the following could help you. T-test in single population One-tailed right H0 : µ = 12, H1: µ > 12 (Note, H0 always take =, irrespective of H1) One-tailed test, right, critical region on the right, of the form {T>t}. Since n=25 < 30, assume population as normal, unknown population variance, use t-test. df = ν = 25-1 = 24. µ=12, xbar=14, s=4.32 Value of test statistic, t = (14-12)/(4.32/√25) = 2.315. Level of test α=0.05. (1) p-value = P{T>2.315} = between 0.01 and 0.025 < 0.05 Significant at 0.05. Conclusion: Mean > 12 is supported. Another way: Using t-table From table t(24): α = 0.05, t=1.7109, calculated t-value=2.315 is inside the critical region. Hence, significant at α = 0.05. Conclusion: same Yours Table t(22): α = 0.10, t=1.3212, calculated t-value >, inside the critical region, < outside. Without sample size, how do you get the degrees of freedom?
1 answer · Mathematics · 6 years ago
• ### I need help with Statistics homework!?

Best answer: Testing the difference of two means in independent populations: t-test Example that may help. Let µ2 be the populations mean length of new bars and µ1 be the population mean length of old bars. H0: µ2-µ1 = 0 H1: µ2-µ1 > 0 Both populations are normal with unknown variances, assume equal variances. Xbar1=2.52 m, xbar2=2.88 m,... show more
Best answer: Testing the difference of two means in independent populations: t-test Example that may help. Let µ2 be the populations mean length of new bars and µ1 be the population mean length of old bars. H0: µ2-µ1 = 0 H1: µ2-µ1 > 0 Both populations are normal with unknown variances, assume equal variances. Xbar1=2.52 m, xbar2=2.88 m, n1=17, n2=12 Sample standard deviations, s1=0.66 m, s2=0.75 m. Pooled variance sp^2=[(n1-1)s1^2 + (n2-1)s2^2]/(n1+n2-2) = 0.4873, sp=0.6981 Test Statistics: t-test, t = (2.88-2.52)/sp*sqrt(1/17 + 1/12) = 1.368, df=17+12-2=27. One-tailed test, right, critical region on right tail. p-value = P{T>1.368} = between 0.05 and 0.10 (table of t-dist), that is > 0.05 (about 0.09 from Excel) Level of Significance: 0.05 Decision Rule: Reject H0 if p-value < 0.05 Decision: Unable to reject H0 at 0.05 since p-value > 0.05. The claim is not supported. For H0: µ2-µ1 = 0 H1: µ2-µ1 not equal 0 Two-tailed, critical region of the form {|T|>t} Say you get t=2.50 for df=27 p-value = P{|T|>2.50} = 2P{T>2.50}=between 0.01 and 0.02 (table of t-dist) See that it is >0.01, <0.02 (and < 0.05 too) Level of Significance: 0.05 Decision Rule: Reject H0 if p-value < 0.05 Decision: Reject H0 at 0.05 since p-value < 0.05. The tips amount differ.
1 answer · Mathematics · 6 years ago