MathGuru
Popular questions and responses by MathGuru
math
Substitute 1 for x. Simplify. If f(x) = 4x3, find the following. f(1)
asked on February 23, 2007 
math
You have two formulas to use: A = lw >area = length times width P = 2l + 2w >perimeter You know the perimeter, which is 200m of fencing. Let length = x Now let's solve the perimeter equation for w, using what we know: 200 = 2x + 2w 200  2x = 2w (200 
asked on January 17, 2007

Algebra
Fourth root of 80y cubed
posted on December 20, 2014 
Statistics
See your most recent post.
posted on December 18, 2014 
STAT important
Yes, you are correct. Read about the Central Limit Theorem. It might help to clarify the concepts for you.
posted on December 18, 2014 
Math
How about D?
posted on December 18, 2014 
Staatistics
Normal distribution: 99.7% is approximately +/ 3 standard deviations around the mean. 0.3% is outside those limits.
posted on December 14, 2014 
math
2.82843 from an online calculator. If you do this by hand, find the mean, then the variance. Standard deviation is the square root of the variance. Check your work.
posted on December 14, 2014 
algebra 2
I'm assuming your problem is this: (6+i)/(5+i) If so, then you can multiply both the numerator and denominator by an equivalent of 1: (6+i)/(5+i) * (5  i)/(5  i) = (30  5i + 6i  i^2)/(25  i^2) = (30 + i + 1)/(25 + 1) = (31 + i)/26 Check the
posted on December 14, 2014 
statistics
Since n = 200, .6 = 120, .3 = 60, and .1 = 20 Therefore, the expected vale of X is: E(X) = sum x*p = 120(.6) + 60(.3) + 20(.1) = 92 To get the variance, subtract the mean (expected value) from each X. Next, square each of those values. Finally, add the
posted on December 14, 2014 
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
posted on July 31, 2014 
stats
I would think either one due to the nature of the tests, but check this to be sure.
posted on July 28, 2014 
statistics
Formula to find sample size: n = [(zvalue)^2 * p * q]/E^2 ... where n = sample size, zvalue is found using a ztable 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,
posted on July 28, 2014 
Introduction to statistics
Formula to find sample size: n = [(zvalue)^2 * p * q]/E^2 ... where n = sample size, zvalue is found using a ztable 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.
posted on July 18, 2014 
statistics
Hint: The mean of the sampling distribution is the population mean.
posted on July 10, 2014 
statistics, probability
Formula (if you don't use the binomial probability table): P(x) = (nCx)(p^x)[q^(nx)] 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
posted on July 9, 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
posted on July 8, 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.
posted on July 8, 2014 
Statistics
Substitute t for z in my statements. Sorry for any confusion.
posted on July 8, 2014 
Statistics
Since your sample size is small, you can use a onesample ttest formula. With your data: z = (2.8  3)/(0.3/√10) Finish the calculation. Next, check a ttable using 9 degrees of freedom (which is n  1) for a onetailed test at .05 level of
posted on July 8, 2014 
statistics
CI95 = 51.4 ± 1.96 (16.66) Calculate to determine the interval.
posted on July 8, 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.
posted on July 8, 2014 
stats
I'll help with the last two: 9. The alternate hypothesis shows a specific direction in a onetailed test. 10. 2.262 (degrees of freedom would be n  1)
posted on July 3, 2014 
Statistics (Check Work)
Looks like you have a good handle on this, Jen. Keep up the good work!
posted on June 27, 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.
posted on June 26, 2014 
Statistics (Check Work)
I see you figured this out on your own. Good job!
posted on 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 Formula: z = (pR 
posted on June 26, 2014 
Statistics
PsyDAG answered a similar post. See below under "Related Questions" for the prior posting.
posted on June 26, 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
posted on June 24, 2014 
statistics
Try zscores. 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 ztable for the probability. Remember the question says "165 friends or more" when you are
posted on June 24, 2014 
Statistics (Check Answer)
You are welcome! I'm glad the explanation helped.
posted on June 17, 2014 
Statistics
Try the binomial probability formula: P(x) = (nCx)(p^x)[q^(nx)] n = 15 x = 4 p = .20 q = 1  p = .80 Substitute the values into the formula and calculate your probability.
posted on June 17, 2014 
Statistics
Formula: n = {[(zvalue) * 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.
posted on 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
posted on June 16, 2014 
statistics
Let's try a binomial proportion onesample ztest. 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 twotailed test (p does not equal 0.65). Use a
posted on June 13, 2014 
statistics
Formula: n = {[(zvalue) * 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.
posted on June 13, 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.
posted on June 10, 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.
posted on June 10, 2014 
Statistics
Try zscores: z = (x  mean)/sd Your data: 2.33 = (x  40)/1 Solve for x. Note: 2.33 represents the zscore corresponding to the 1% in your problem.
posted on June 6, 2014 
STATISTICS
Formula: P(x) = (nCx)(p^x)[q^(nx)] 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):
posted on June 5, 2014 
Stats
Formula to find sample size: n = [(zvalue)^2 * p * q]/E^2 ... where n = sample size, zvalue is 1.96 and is found using a ztable for 95% confidence, p = .5 (when no value is stated), q = 1  p, ^2 means squared, * means to multiply, and E = .02 (or 2%).
posted on June 5, 2014 
Stats
Formula: P(x) = (nCx)(p^x)[q^(nx)] Your data: x = 2 p = .05 q = 1  p = 1  .05 = .95 n = 20 Substitute and calculate for your probability.
posted on 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.
posted on 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.
posted on May 28, 2014 
Statistics
Correction: Your calculation is correct. z = 5.156 You would still reject the null and accept the alternate hypothesis. Sorry for any confusion.
posted on May 23, 2014 
Statistics
Use a onesample ztest for proportions. With your data: z = (.43  .50)/√(.50)(.50)/1000) = 4.358 Reject the null and accept the alternate hypothesis (p < .50).
posted on 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).
posted on May 23, 2014 
biology
(B) T cells
posted on May 22, 2014 
Statistics
Using this formula: P(x) = (nCx)(p^x)[q^(nx)] Note: q = 1  p We have: P(x) = (3Cx)(.52^x)[.48^(3x)]
posted on May 22, 2014 
Statistics
z = 2.575 for a 99% confidence. Since your sample sizes are large, I would use a twosample confidence interval formula for large sample sizes. Critical value at .05 using a ztable for a onetailed test is z = 1.645 If you use a ttable instead to find
posted on May 22, 2014 
statistics
Hypotheses: Ho: µ = 10 >null hypothesis Ha: µ < 10 >alternate hypothesis
posted on May 22, 2014 
please help Human bio
A. elastin B. collagen C. telomeres Check this!
posted on 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
posted on May 21, 2014 
Stats
Use zscores. 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 ztable between z = 1.00 and z = 4.00)
posted on May 21, 2014 
ap stats
50th percentile
posted on May 15, 2014 
statistics
Welch's ttest 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)
posted on May 15, 2014 
Statistics
If you do a proportional ztest 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
posted on May 13, 2014 
Statistics
Answer: That Y increases by 20 units for each unit increase in X.
posted on May 9, 2014 
Statistics (45)
Formula: n = {[(zvalue)(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.
posted on May 6, 2014 
Statistics (44)
Formula: n = {[(zvalue)(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.
posted on 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.
posted on 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.
posted on 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
posted on 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.
posted on 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^(nx)] 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
posted on May 6, 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^(nx)] 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
posted on May 6, 2014 
statistics
How about d?
posted on May 5, 2014 
statistics
How about d?
posted on May 5, 2014 
statistics
How about d? You have only one sample with a small sample size.
posted on May 5, 2014 
statistics
Formula: P(x) = (nCx)(p^x)[q^(nx)] Find P(0), then subtract that value from 1. n = 12 p = .25 q = 1  p I'll let you take it from here.
posted on May 5, 2014 
statistics
Formula: n = [(zvalue)(p)(q)]/E^2 zvalue = 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.
posted on 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.
posted on May 5, 2014 
Statistics
How about a onesample ttest? 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.
posted on May 5, 2014 
Statistics
See later post.
posted on April 30, 2014 
Statistics
I think you are correct!
posted on April 30, 2014 
statistics
Using an online calculator: Standard Deviation = 9.21 To find variance, square the standard deviation.
posted on April 30, 2014 
statistics
True.
posted on April 29, 2014 
statistics
If you have equal sample sizes and unequal standard deviations, a Welch's ttest might be appropriate. Check degrees of freedom for this type of test before checking the appropriate table to determine critical value.
posted on 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.
posted on April 29, 2014 
statistics
Try a onesample ttest since your sample size is small. You will need to calculate the mean and standard deviation. Use a ttable to find the critical value at .05 level of significance for a onetailed test. Degrees of freedom is (n  1). Sample size (n)
posted on 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.
posted on April 29, 2014 
Statistics
C. µ1 and µ2 to be unequal.
posted on April 29, 2014 
Statistics
D. Either (b) or (c).
posted on April 29, 2014 
Statistics
D. All of the above.
posted on April 29, 2014 
statistics
Both A and B.
posted on 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.
posted on April 28, 2014 
ap statistics
Since the probability of a Type II error is equal to beta, I would say D is not correct.
posted on April 24, 2014 
ap statistics
B. The probability of rejecting H0 when HA is true.
posted on 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.
posted on April 24, 2014 
statistics
Use a ttable to determine the area. Remember to use the degrees of freedom when looking at the table.
posted on April 22, 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.
posted on April 22, 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.
posted on April 22, 2014 
statistics
Try this formula: n = [(zvalue * 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.
posted on April 22, 2014 
statistics
Try this formula: n = [(zvalue)^2 * p * (1p)]/E^2 ...where n = sample size you need, zvalue = 1.96 to represent 95% confidence, p and 1p represent proportions, E = .025 (or 2.5%), and ^ means squared. With your data: n = [(1.96)^2 * .30 * .70]/.025^2
posted on April 22, 2014 
Statistics
I would think either one due to the nature of the tests, but check this to be sure.
posted on April 22, 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
posted on April 16, 2014 
Stats
0.1359 is the area under the normal curve between z = 1 and z = 2 Look at a normal distribution table (ztable) to check.
posted on April 16, 2014 
statistics
Standard error of the proportion is: √(pq/n) p = .55 q = 1p = .45 n = sample size I'll let you take it from here.
posted on 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
posted on April 14, 2014 
personal finance
Common stock
posted on April 11, 2014