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October 25, 2014

Posts by MathGuru


Total # Posts: 1,445

statistics
Formula: CI95 = mean ± 1.96(sd/√n) You will need to calculate the mean and standard deviation for your data. In the formula, n = sample size. Once you have what you need, substitute into the formula and determine the confidence interval. I hope this will help get ...
July 31, 2014

statistics
Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is found using a z-table for whatever confidence level is used, when no value is stated in the problem p = .50, q = 1 - p, ^2 means squared, * means to multiply, and E = .028. Plug ...
July 28, 2014

stats
I would think either one due to the nature of the tests, but check this to be sure.
July 26, 2014

Introduction to statistics
Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is found using a z-table for 99% confidence, p = .50 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = 0.19. Plug values into the ...
July 17, 2014

statistics
Hint: The mean of the sampling distribution is the population mean.
July 9, 2014

statistics, probability
Formula (if you don't use the binomial probability table): P(x) = (nCx)(p^x)[q^(n-x)] Values: n = 6 p = 0.4 q = 1 - p = 1 - 0.4 = 0.6 i. Find P(5) ii. Find P(4), P(5), and P(6). Add together for your total probability. iii. I'll let you try this one on your own. I hope...
July 9, 2014

statistics
standard error = s/√n Note: The square root of the variance is called the Standard Deviation With your data: 2 = (4.472)/√n Solve for n. I hope this is what you were asking.
July 8, 2014

statistics
CI95 = 51.4 ± 1.96 (16.66) Calculate to determine the interval.
July 7, 2014

Correction
Substitute t for z in my statements. Sorry for any confusion.
July 7, 2014

Statistics
Since your sample size is small, you can use a one-sample t-test formula. With your data: z = (2.8 - 3)/(0.3/√10) Finish the calculation. Next, check a t-table using 9 degrees of freedom (which is n - 1) for a one-tailed test at .05 level of significance. Compare your z-...
July 7, 2014

statistics
Here's one way you might do this: s/[1 + (2.58/√2n)] to s/[1 - (2.58/√2n)] Note: 2.58 represents 99% confidence Find the standard deviation for your data. Use that value for s. Your sample size is 12. I'll let you take it from here.
July 7, 2014

statistics
Here's one way to do this problem: n = 240 p = .03 q = 1 - p = 1 - .03 = .97 For (a), you will need to find P(0) through P(4). Add together, then subtract the total from 1 for the probability. I'll let you try (b). You can use a binomial probability table or calculate ...
July 5, 2014

stats
I'll help with the last two: 9. The alternate hypothesis shows a specific direction in a one-tailed test. 10. 2.262 (degrees of freedom would be n - 1)
July 3, 2014

Statistics (Check Work)
Looks like you have a good handle on this, Jen. Keep up the good work!
June 26, 2014

Statistics (Check Work)
I see you figured this out on your own. Good job!
June 26, 2014

Statistics (Check Work)
Hi Jen! Just an FYI: I'm a "she" instead of a "he" but thanks for the compliments. I'm always glad to help when and where I can. Let's get started. Both questions are the same type of problem. Question 1: Null: pR = pU Alternate: pR > pU ...
June 26, 2014

Statistics
PsyDAG answered a similar post. See below under "Related Questions" for the prior posting.
June 26, 2014

statistics
You might try a formula like this one: s/(1 + 1.96/√2n) to s/(1 - 1.96/√2n) Substitute and calculate. There may be other variations of similar formulas you can use as well.
June 26, 2014

statistics
Try z-scores. In this case, use the following: z = (x - mean)/(sd/√n) With your data: z = (165 - 150)/(90/√25) I'll let you finish the calculation. Next, check a z-table for the probability. Remember the question says "165 friends or more" when you ...
June 24, 2014

statistics
95% confidence interval is equivalent to z = 1.96, so if you round to 2, then your calculations for E are almost correct. I think you missed a 0; I get E = 0.04248. The formula used to determine the margin of error is all you should need to answer the problem.
June 19, 2014

Statistics
Try the binomial probability formula: P(x) = (nCx)(p^x)[q^(n-x)] n = 15 x = 4 p = .20 q = 1 - p = .80 Substitute the values into the formula and calculate your probability.
June 17, 2014

Statistics (Check Answer)
You are welcome! I'm glad the explanation helped.
June 16, 2014

Statistics (Check Answer)
Some notes on finite population correction factor: If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor. For standard error of the mean, use: sd/√n If you need to adjust for the finite ...
June 16, 2014

Statistics
Formula: n = {[(z-value) * sd]/E}^2 ...where n = sample size, sd = standard deviation, E = maximum error, and ^2 means squared. Using the values you have in your problem: n = {[(1.96) * 15]/3}^2 Calculate for sample size. Round your answer.
June 15, 2014

statistics
Formula: n = {[(z-value) * sd]/E}^2 ...where n = sample size, sd = standard deviation, E = maximum error, and ^2 means squared. Using the values you have in your problem: n = {[(1.96) * 15]/3}^2 Calculate for sample size. Round your answer.
June 12, 2014

statistics
Let's try a binomial proportion one-sample z-test. Ho: p = 0.65 Ha: p does not equal 0.65 Test statistic: z = (0.49 - 0.65)/√[(0.65)(0.35)/100] z = -3.35 The null would be rejected at the .05 level for a two-tailed test (p does not equal 0.65). Use a z-table to find ...
June 11, 2014

Statistics
Standard deviation = √npq 1) n = 157 p = .16 (for 16%) q = 1 - p = 1 - .16 = .84 2) n = 209 p = .12 (for 12%) q = 1 - p = 1 - .12 = .88 Substitute and calculate.
June 10, 2014

Statistics
The closer it is to 1 or -1, the stronger the correlation. A correlation of 0.3 is a weaker relationship between variables.
June 10, 2014

Statistics
Try z-scores: z = (x - mean)/sd Your data: 2.33 = (x - 40)/1 Solve for x. Note: 2.33 represents the z-score corresponding to the 1% in your problem.
June 6, 2014

Stats
Formula: P(x) = (nCx)(p^x)[q^(n-x)] Your data: x = 2 p = .05 q = 1 - p = 1 - .05 = .95 n = 20 Substitute and calculate for your probability.
June 5, 2014

Stats
Formula to find sample size: n = [(z-value)^2 * p * q]/E^2 ... where n = sample size, z-value is 1.96 and is found using a z-table for 95% confidence, p = .5 (when no value is stated), q = 1 - p, ^2 means squared, * means to multiply, and E = .02 (or 2%). I'll let you take...
June 5, 2014

STATISTICS
Formula: P(x) = (nCx)(p^x)[q^(n-x)] For (a): x = 5 p = .16 q = 1 - p = 1 - .16 = .84 n = 29 Substitute and calculate for your probability. For (b): x = 0,1,2,3,4 p,q,n stay the same. Add each probability you calculate for the total probability. For (c): Add together the ...
June 5, 2014

Stat
You will need a confidence interval formula for the difference of two population means. Use 1.96 for z (representing 95% confidence). Substitute what you know into the formula and calculate.
June 2, 2014

statistics
Using a Poisson distribution: P(x) = (e^-μ) (μ^x) / x! e = 2.71828 µ = 1.4 x = 4 Substitute and calculate.
May 28, 2014

Statistics
Use a one-sample z-test for proportions. With your data: z = (.43 - .50)/√(.50)(.50)/1000) = -4.358 Reject the null and accept the alternate hypothesis (p < .50).
May 23, 2014

Correction
Correction: Your calculation is correct. z = 5.156 You would still reject the null and accept the alternate hypothesis. Sorry for any confusion.
May 23, 2014

Statistics
Formula: z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size) With your data: z = (26600 - 25000)/(3800/√150) = 4.96 You can reject the null and accept the alternate hypothesis (µ > 25000).
May 23, 2014

statistics
Hypotheses: Ho: µ = 10 -->null hypothesis Ha: µ < 10 -->alternate hypothesis
May 22, 2014

Statistics
z = 2.575 for a 99% confidence. Since your sample sizes are large, I would use a two-sample confidence interval formula for large sample sizes. Critical value at .05 using a z-table for a one-tailed test is z = 1.645 If you use a t-table instead to find critical value, then ...
May 22, 2014

Biology
Here are a few of your questions; maybe you can determine the rest. 14. dermis 15. chemical 16. urine 17. viruses 18. antibodies I'll let you take it from here.
May 22, 2014

Statistics
Using this formula: P(x) = (nCx)(p^x)[q^(n-x)] Note: q = 1 - p We have: P(x) = (3Cx)(.52^x)[.48^(3-x)]
May 22, 2014

biology
(B) T cells
May 22, 2014

Stats
Use z-scores. For this problem: z = (x - mean)/(sd/√n) With your data: z = (8.2 - 8)/(2/√100) = 1.00 z = (8.8 - 8)/(2/√100) = 4.00 Answer: .1586 is the probability (check a z-table between z = 1.00 and z = 4.00)
May 21, 2014

Statistics
a) ± 3.1 b) 56.7 - 3.1 = 53.6 56.7 + 3.1 = 59.8 Interval is 53.6 to 59.8
May 21, 2014

please help Human bio
A. elastin B. collagen C. telomeres Check this!
May 20, 2014

ap stats
50th percentile
May 13, 2014

statistics
Welch's t-test for unequal variances: t = (mean1 - mean2)/√(s^2/n1 + s^2/n2) t = (4.31 - 3.68)/√(0.17^2/10 + 0.22^2/10) t = 0.63/0.08792 = 7.166 (rounded)
May 13, 2014

Statistics
If you do a proportional z-test with this data, it would look something like this: z = (.258 - .20)/√[(.20)(.80)/120] Note: 31/120 = .258 Calculating: z = .058/.0365 = 1.589 The null would not be rejected and the conclusion would be that there is no difference in the ...
May 11, 2014

Statistics
Answer: That Y increases by 20 units for each unit increase in X.
May 8, 2014

Stat college
Mean = np = 9 * .17 = 1.53 Use binomial probability formula (or a binomial probability table). Formula: P(x) = (nCx)(p^x)[q^(n-x)] x = 0,1,2 n = 9 p = .17 q = 1 - p = 1 - .17 = .83 For first part: Find P(2) for probability. For second part: Find P(0) and P(1). Add P(0), P(1), ...
May 6, 2014

STAT
Mean = np = 9 * .17 = 1.53 Use binomial probability formula (or a binomial probability table). Formula: P(x) = (nCx)(p^x)[q^(n-x)] x = 0,1,2 n = 9 p = .17 q = 1 - p = 1 - .17 = .83 For first part: Find P(2) for probability. For second part: Find P(0) and P(1). Add P(0), P(1), ...
May 6, 2014

Statistics (45)
Formula: n = {[(z-value)(sd)]/E}^2 a) n = [(1.96 * 7)/2]^2 b) n = [(2.575 * 7)/2)^2 Calculate. Round to the next highest whole number.
May 6, 2014

Statistics (44)
Formula: n = {[(z-value)(sd)]/E}^2 a) n = [(1.96 * 6.35)/2]^2 b) n = [(1.96 * 6.35)/1)^2 Calculate. Round to the next highest whole number.
May 6, 2014

Statistics (43)
Find mean and standard deviation, then use confidence interval formula. CI95 = mean ± 1.96(sd/√n) n = 20 I'll let you take it from here.
May 6, 2014

Statistics (42)
I'll get you started. a) CI95 = mean ± 1.96 (sd/√n) mean = 96 sd = 16 n = 60 b) Use the same formula, except n = 120 c) Compare the margin of error in a) and b) to answer this question.
May 6, 2014

Statistics (39)
Some notes on finite population correction factor: If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor. For standard error of the mean, use: sd/√n If you need to adjust for the finite ...
May 6, 2014

Statistics (38)
Mean is 4550.7 (add up the numbers and divide by 10) I'll let you determine standard deviation. Use a calculator, or if you do it by hand, use a formula to calculate the standard deviation.
May 6, 2014

Statistics
How about a one-sample t-test? Your sample size is fairly small. Fill in what you know into the formula to calculate the test statistic. population mean = 100 sample mean = 115 standard deviation = 30 sample size = 22 I'll let you take it from here.
May 5, 2014

Statistics
CI98 = p ± 2.33 √(pq/n) p = 42/200 = .21 q = 1 - p = 1 - .21 = .79 n = 200 With your data: CI98 = .21 + 2.33 √[(.21)(.79)/200] I'll let you take it from here to finish.
May 4, 2014

statistics
Formula: n = [(z-value)(p)(q)]/E^2 z-value = 2.33 for 98% confidence p = .5 if no value is stated q = 1 - p E = Maximum error With your data: n = [(2.33)(.5)(.5)]/.03^2 I'll let you take it from here. Round to the next whole number.
May 4, 2014

statistics
Formula: P(x) = (nCx)(p^x)[q^(n-x)] Find P(0), then subtract that value from 1. n = 12 p = .25 q = 1 - p I'll let you take it from here.
May 4, 2014

statistics
How about d?
May 3, 2014

statistics
How about d? You have only one sample with a small sample size.
May 3, 2014

statistics
How about d?
May 3, 2014

statistics
Using an online calculator: Standard Deviation = 9.21 To find variance, square the standard deviation.
April 30, 2014

Statistics
I think you are correct!
April 30, 2014

Statistics
See later post.
April 29, 2014

Statistics
D. All of the above.
April 29, 2014

Statistics
D. Either (b) or (c).
April 29, 2014

Statistics
C. µ1 and µ2 to be unequal.
April 29, 2014

ap statistics
CI95 = p ± 1.96[√(pq/n)] p = 20/75 = .27 q = 1 - p = 1 - .27 = .73 n = 75 Substitute into the formula to find the confidence interval.
April 29, 2014

statistics
Try a one-sample t-test since your sample size is small. You will need to calculate the mean and standard deviation. Use a t-table to find the critical value at .05 level of significance for a one-tailed test. Degrees of freedom is (n - 1). Sample size (n) is 9.
April 29, 2014

Statistics
2.46 A Type I error is rejecting the null when it is true. If the critical value is 2.45, and you have a test statistic of 2.46, you have the least chance among the other choices of rejecting the null and it happens to be true.
April 28, 2014

statistics
If you have equal sample sizes and unequal standard deviations, a Welch's t-test might be appropriate. Check degrees of freedom for this type of test before checking the appropriate table to determine critical value.
April 28, 2014

statistics
True.
April 28, 2014

Statistics
Second choice. Do not reject the null hypothesis because 1.457 lies in the region between -2.326 and 2.326.
April 27, 2014

statistics
Both A and B.
April 27, 2014

ap statistics
Since the probability of a Type II error is equal to beta, I would say D is not correct.
April 24, 2014

ap statistics
B. The probability of rejecting H0 when HA is true.
April 24, 2014

ap statistics
The population mean is equal to the mean of the sampling distribution of the sample means. Standard error of the mean is the standard deviation of the sample means.
April 24, 2014

Statistics
I would think either one due to the nature of the tests, but check this to be sure.
April 22, 2014

statistics
Try this formula: n = [(z-value)^2 * p * (1-p)]/E^2 ...where n = sample size you need, z-value = 1.96 to represent 95% confidence, p and 1-p represent proportions, E = .025 (or 2.5%), and ^ means squared. With your data: n = [(1.96)^2 * .30 * .70]/.025^2 I'll let you ...
April 22, 2014

statistics
Try this formula: n = [(z-value * sd)/E]^2 With your data: n = [(1.96 * 40)/10]^2 I'll let you finish the calculation. Remember to round the answer to the nearest whole number.
April 21, 2014

statistics
Part a) ME = 1.96 * sd/√n With your data: ME = 1.96 * 65/√45 Part b) CI95 = mean ± 1.96(sd/n) With your data: CI95 = 273 ± 1.96(65/√45) I'll let you finish the calculations.
April 21, 2014

statistics
Use a t-table to determine the area. Remember to use the degrees of freedom when looking at the table.
April 21, 2014

statistics
Part a) SE of mean = sd/√n = 5/√40 Part b) ME = 1.96 * 5/√40 I'll let you finish the calculations.
April 21, 2014

Stats
0.1359 is the area under the normal curve between z = -1 and z = -2 Look at a normal distribution table (z-table) to check.
April 16, 2014

Statistics
Type I errors result when you reject the null and it's true. Type II errors result when you accept the null and it's false. If you reject the null hypothesis that the subject is guessing and it's true, you have made a Type I error. If you accept the null that the ...
April 16, 2014

statistics
Standard error of the proportion is: √(pq/n) p = .55 q = 1-p = .45 n = sample size I'll let you take it from here.
April 14, 2014

statistics
Some notes on finite population correction factor: If the population is small and the sample is large (more than 5% of the small population), use the finite population correction factor. For standard error of the mean, use: sd/√n If you need to adjust for the finite ...
April 14, 2014

personal finance
Common stock
April 11, 2014

Statistics
1. Relationship condition (the two variables must be related) 2. Temporal condition (time order) 3. Lack of other explanation condition (not due to confounding variables, etc.)
April 9, 2014

statistics
Regression equation: predicted y = a + bx ...where a represents the y-intercept and b the slope. In your equation, b (slope) = 0.0712
April 9, 2014

Science
If the current is held constant, an increase in voltage will result in an increase in resistance.
April 9, 2014

Statsistics
Here is a hint: It doesn't matter if the correlation is negative or positive, the closer it is to zero, the weaker the correlation.
April 9, 2014

stats
Use a confidence interval formula for the data. Here is an example of one for a 95% confidence interval: CI95 = mean ± 1.96 (sd/√n)
April 9, 2014

statistics
Use a one-sample z-test. Calculate to find the test statistic. Find the critical value using a z-table for .05 significance level for a one-tailed test. The p-value is the actual level of the test statistic and is found using a z-table as well.
April 8, 2014

science
Velocity = Frequency * Wavelength V = 5 hertz * 10 mm V = ? I'll let you finish.
April 8, 2014

Statistics PLEASE HELP ME!
You might try this formula: Pr[T ≤ t] = 1 - e^-ht h = reciprocal of the mean t = time Therefore, the reciprocal of 25 is 1/25 or .04 Substituting into the formula: Pr[T ≤ 20] = 1 - e^-.04(20) = 1 - e^-.8 = 1 - .4493 = .5507 Check these calculations.
April 7, 2014

Statistics
Use confidence interval formulas for proportions. CI95 = p + or - (1.96)(√pq/n) ...where √ = square root, p = x/n, q = 1 - p, and n = sample size. CI90 = p + or - (1.645)(√pq/n) Hint: x = 396, n = 621 Convert all fractions to decimals to work the formulas. I ...
April 7, 2014

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