# Posts by MathGuru

Total # Posts: 1,453

**Algebra**

Fourth root of 80y cubed

**Math**

How about D?

**STAT- important**

Yes, you are correct. Read about the Central Limit Theorem. It might help to clarify the concepts for you.

**Statistics**

See your most recent post.

**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.

**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 work.

**Staatistics**

Normal distribution: 99.7% is approximately +/- 3 standard deviations around the mean. 0.3% is outside those limits.

**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 squares times their ...

**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 ...

**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 ...

**stats**

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

**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 ...

**statistics**

Hint: The mean of the sampling distribution is the population mean.

**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...

**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.

**statistics**

CI95 = 51.4 ± 1.96 (16.66) Calculate to determine the interval.

**Correction**

Substitute t for z in my statements. Sorry for any confusion.

**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-...

**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.

**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 ...

**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)

**Statistics (Check Work)**

Looks like you have a good handle on this, Jen. Keep up the good work!

**Statistics (Check Work)**

I see you figured this out on your own. Good job!

**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 ...

**Statistics**

PsyDAG answered a similar post. See below under "Related Questions" for the prior posting.

**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.

**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 ...

**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.

**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.

**Statistics (Check Answer)**

You are welcome! I'm glad the explanation helped.

**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 ...

**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.

**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.

**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 ...

**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.

**Statistics**

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

**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.

**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.

**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...

**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 ...

**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.

**statistics**

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

**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).

**Correction**

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

**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).

**statistics**

Hypotheses: Ho: µ = 10 -->null hypothesis Ha: µ < 10 -->alternate hypothesis

**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 ...

**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.

**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)]

**biology**

(B) T cells

**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)

**Statistics**

a) ± 3.1 b) 56.7 - 3.1 = 53.6 56.7 + 3.1 = 59.8 Interval is 53.6 to 59.8

**please help Human bio**

A. elastin B. collagen C. telomeres Check this!

**ap stats**

50th percentile

**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)

**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 ...

**Statistics**

Answer: That Y increases by 20 units for each unit increase in X.

**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), ...

**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), ...

**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.

**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.

**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.

**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.

**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 ...

**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.

**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.

**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.

**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.

**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.

**statistics**

How about d?

**statistics**

How about d? You have only one sample with a small sample size.

**statistics**

How about d?

**statistics**

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

**Statistics**

I think you are correct!

**Statistics**

See later post.

**Statistics**

D. All of the above.

**Statistics**

D. Either (b) or (c).

**Statistics**

C. µ1 and µ2 to be unequal.

**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.

**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.

**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.

**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.

**statistics**

True.

**Statistics**

Second choice. Do not reject the null hypothesis because 1.457 lies in the region between -2.326 and 2.326.

**statistics**

Both A and B.

**ap statistics**

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

**ap statistics**

B. The probability of rejecting H0 when HA is true.

**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.

**Statistics**

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

**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 ...

**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.

**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.

**statistics**

Use a t-table to determine the area. Remember to use the degrees of freedom when looking at the table.

**statistics**

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

**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.

**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 ...

**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.

**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 ...

**personal finance**

Common stock

**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.)