a 12 sided die is rolled find the probability of rolling a number greater than 7 We assume this is a fair die, not loaded. This means each side 1-12 has an equal probability of 1/12 of being rolled. The problem asks, P(Roll > 7) Greater than 7 means our sample space is {8, 9, 10, 11, 12} If each of these 5 faces have an equal probability of being rolled, then we have: P(Roll > 7) = P(Roll = 8) + P(Roll = 9) + P(Roll = 10) + P(Roll = 11) + P(Roll = 12) P(Roll > 7) = 1/12 + 1/12 + 1/12 + 1/12 + 1/12 P(Roll > 7) =[B] 5/12[/B]

A company has 81 employees of whom x are members of a union how many are not in the union You can either be a union member or a non-union member. This is our sample space. If we have 81 employees and x are union members, this means that: Non-Union membes = [B]81 - x[/B]

A firm wants to know with a 98% level of confidence if it can claim that the boxes of detergent it sells contain more than 500g of detergent. From past experience the firm knows that the amount of detergent in the boxes is normally distributed. The firm takes a random sample of n =25 and finds that X = 520 g and s = 75g. What's your final conclusion? (Ho: u = 500; Ha: u > 500) [URL='http://www.mathcelebrity.com/mean_hypothesis.php?xbar=520&n=25&stdev=75&ptype==&mean=500&alpha=0.02&pl=Mean+Hypothesis+Testing']Perform a hypothesis testing of the mean[/URL] [B]Yes, accept null hypothesis[/B]

A random sample of 100 light bulbs has a mean lifetime of 3000 hours. Assume that the population standard deviation of the lifetime is 500 hours. Construct a 95% confidence interval estimate of the mean lifetime. [B]2902 < u < 3098[/B] using our [URL='http://www.mathcelebrity.com/normconf.php?n=100&xbar=3000&stdev=500&conf=95&rdig=4&pl=Large+Sample']confidence interval for the mean calculator[/URL]

A random sample of 144 with a mean of 100 and a standard deviation of 70 is known from a population of 1,000. What is the 95% confidence interval for the unknown population? [URL='http://www.mathcelebrity.com/normconf.php?n=144&xbar=100&stdev=70&conf=95&rdig=4&pl=Large+Sample']Large Sample Confidence Interval Mean Test[/URL] [B]88.5667 < u < 111.4333[/B]

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places) Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get [B]58.89 < u < 63.11[/B]

A random sample of 25 customers was chosen in CCP MiniMart between 3:00 and 4:00 PM on a Friday afternoon. The frequency distribution below shows the distribution for checkout time (in minutes). Checkout Time (in minutes) | Frequency | Relative Frequency 1.0 - 1.9 | 2 | ? 2.0 - 2.9 | 8 | ? 3.0 - 3.9 | ? | ? 4.0 - 5.9 | 5 | ? Total | 25 | ? (a) Complete the frequency table with frequency and relative frequency. (b) What percentage of the checkout times was less than 3 minutes? (c)In what class interval must the median lie? Explain your answer. (d) Assume that the largest observation in this dataset is 5.8. Suppose this observation were incorrectly recorded as 8.5 instead of 5.8. Will the mean increase, decrease, or remain the same? Will the median increase, decrease or remain the same? Why? (a) [B]Checkout Time (in minutes) | Frequency | Relative Frequency 1.0 - 1.9 | 2 | 2/25 2.0 - 2.9 | 8 | 8/25 3.0 - 3.9 | 10 (since 25 - 5 + 8 + 2) = 10 | 10/25 4.0 - 5.9 | 5 | 5/25 Total | 25 | ?[/B] (b) (2 + 8)/25 = 10/25 = [B]40%[/B] c) [B]3.0 - 3.9[/B] since 2 + 8 + 10 + 5 = 25 and 13 is the middle value which occurs in the 3.0 - 3.9 interval (d) [B]Mean increases[/B] since it's a higher value than usual. Median would not change as the median is the most frequent distribution and assuming the 5.8 is only recorded once.

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.44 hours, with a standard deviation of 1.74 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesis (?1 - ?2). Using our confidence interval for [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=+2.31&n2=+40&xbar2=+4.44&stdev2=1.74&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means calculator[/URL], we get: [B]0.0278 < ?1 - ?2 < 1.5322[/B]

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.29 hours, with a standard deviation of 1.58 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1 - u2) What is the interpretation of this confidence interval? A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours B. There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours C. There is 90% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours D. There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours 0.2021 < u1 - u2 < 1.6579 using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=2.31&n2=40&xbar2=4.29&stdev2=1.58&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means confidence interval calculator[/URL] [B]Choice D There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours[/B]

A random sample of n = 10 flash light batteries with a mean operating life X=5 hr. And a sample standard deviation S = 1 hr. is picked from a production line known to produce batteries with normally distributed operating lives. What's the 98% confidence interval for the unknown mean of the working life of the entire population of batteries? [URL='http://www.mathcelebrity.com/normconf.php?n=10&xbar=5&stdev=1&conf=98&rdig=4&pl=Small+Sample']Small Sample Confidence Interval for the Mean test[/URL] [B]4.1078 < u < 5.8922[/B]

A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 120 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 6.0. What's the absolute value of the difference between means? |70 -65| = |5| = 5

A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 100 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 5.0. What is the standard error of the difference in means? [B]0.707106781187[/B] using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+100&xbar1=70&stdev1=5&n2=+100&xbar2=65&stdev2=5&conf=+99&pl=Hypothesis+Test']difference of means calculator[/URL]

A researcher posed a null hypothesis that there was no significant difference between boys and girls on a standard memory test. He randomly sampled 100 girls and 100 boys in a community and gave them the standard memory test. The mean score for girls was 70 and the standard deviation of mean was 5.0. The mean score for boys was 65 and the standard deviation of mean was 5.0. What's the t-value (two-tailed) for the null hypothesis that boys and girls have the same test scores? t = 7.07106781187 using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+100&xbar1=70&stdev1=5&n2=+100&xbar2=65&stdev2=5&conf=+99&pl=Hypothesis+Test']difference of means calculator[/URL]

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

A toy factory makes 5,000 teddy bears per day. The supervisor randomly selects 10 teddy bears from all 5,000 teddy bears and uses this sample to estimate the mean weight of teddy bears and the sample standard deviation. How many degrees of freedom are there in the estimate of the standard deviation? DF = n - 1 DF = 10 - 1 [B]DF = 9[/B]

A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ test, what is the probability that the sample mean scores will be between 85 and 125 points? For x = 125, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+125&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL] Z = 1 P(x < 1) = 0.841345 For x = 85, our z-score and probability is seen [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+85&mean=+105&stdev=+20&n=+1&pl=P%28X+%3C+Z%29']here[/URL] Z = -1 P(x < -1) = 0.158655 So what we want is the probability between these values:
0.841345 - 0.158655 = [B]0.68269[/B]

As the sample size increases, we assume: a. ? increases b. ? increases c. The probability of rejecting a hypothesis increases d. Power increases [B]d. Power increases[/B] [LIST] [*]Power increases if the standard deviation is smaller. [*]If the difference between the means is bigger, the power is bigger. [*]Sample size also increases power [/LIST]

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

Below are data showing the results of six subjects on a memory test. The three scores per subject are their scores on three trials (a, b, and c) of a memory task. Are the subjects getting better each trial? Test the linear effect of trial for the data. A score trial B score trial 2 C Score trial 3 4 6 7 3 7 8 2 8 5 1 4 7 4 6 9 2 4 2 (a) Compute L for each subject using the contrast weights -1, 0, and 1. That is, compute (-1)(a) + (0)(b) + (1)(c) for each subject. (b) Compute a one-sample t-test on this column (with the L values for each subject) you created. Formula t = To computer a one-sample t-test first know the meaning of each letter (a) Each L column value is just -1(Column 1) + 0(Column2) + 1(Column 3) A score trial B score trial 2 C Score trial 3 L = (-1)(a) + (0)(b) + (1)(c) 4 6 7 3 3 7 8 5 2 8 5 3 1 4 7 6 4 6 9 5 2 4 2 0 (b) Mean = (3 + 5 + 3 + 6 + 5 + 0)/6 = 22/6 = 3.666666667 Standard Deviation = 2.160246899 Use 3 as our test mean (3.666667 - 3)/(2.160246899/sqrt(6)) = 0.755928946

Free Confidence Interval for the Mean Calculator - Calculates a (90% - 99%) estimation of confidence interval for the mean given a small sample size using the student-t method with (n - 1) degrees of freedom or a large sample size using the normal distribution Z-score (z value) method including Standard Error of the Mean. confidence interval of the mean

Free Confidence Interval/Hypothesis Testing for the Difference of Means Calculator - Given two large or two small distriutions, this will determine a (90-99)% estimation of confidence interval for the difference of means for small or large sample populations.
Also performs hypothesis testing including standard error calculation.

mean of 106 inches and a standard deviation of 10 inches and for sample of size is 25. Determine the mean and the standard deviation of /x

Do you mean x bar? mean of 106 inches and a standard deviation of 10 inches and for sample of size is 25. Determine the mean and the standard deviation of /x If so, x bar equals the population mean. So it's [B]106[/B]. Sample standard deviation = Population standard deviation / square root of n 10/Sqrt(25) 10/5 [B]2[/B]

one trunk can carry 5068.8 lb. weight of boxes that it carries have a mean of 75lb and a standard deviation of 16 Ib. For Sample size of 64 ,find the mean and standard deviation of /x

one trunk can carry 5068.8 lb. weight of boxes that it carries have a mean of 75lb and a standard deviation of 16 Ib. For Sample size of 64 ,find the mean and standard deviation of /x

Sample Mean = Population Mean Sample Mean = [B]75[/B] Sample Standard Deviation = Population Standard Deviation / Sqrt(n) Sample Standard Deviation = 16/sqrt(64) Sample Standard Deviation = 16 / 8 Sample Standard Deviation = [B]2[/B]

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

Physiologists often use the forced vital capacity as a way to assess a person's ability to move air in and out of their lungs. A researcher wishes to estimate the forced vital capacity of people suffering from asthma. A random sample of 15 asthmatics yields the following data on forced vital capacity, in liters. 5.2 4.9 2.9 5.3 3.0 4.0 5.2 5.2 3.2 4.7 3.2 3.5 4.8 4.0 5.1 Use the data to obtain a point estimate of the mean forced vital capacity for all asthmatics

Physiologists often use the forced vital capacity as a way to assess a person's ability to move air in and out of their lungs. A researcher wishes to estimate the forced vital capacity of people suffering from asthma. A random sample of 15 asthmatics yields the following data on forced vital capacity, in liters. 5.1 4.9 4.7 3.1 4.3 3.7 3.7 4.3 3.5 5.2 3.2 3.5 4.8 4.0 5.1 Use the data to obtain a 95.44% confidence interval for the mean forced vital capacity for all asthmatics. Assume that ? = 0.7.

Perform a one-sample z-test for a population mean. Be sure to state the hypotheses and the significance level, to compute the value of the test statistic, to obtain the P-value, and to state your conclusion. Five years ago, the average math SAT score for students at one school was 475. A teacher wants to perform a hypothesis test to determine whether the mean math SAT score of students at the school has changed. The mean math SAT score for a random sample of 40 students from this school is 469. Do the data provide sufficient evidence to conclude that the mean math SAT score for students at the school has changed from the previous mean of 475? Perform the appropriate hypothesis test using a significance level of 10%. Assume that ? = 73.

A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. x = 20.5, n = 11, ? = 7, H0: µ = 18.7 , Ha: µ ? 18.7 , ? = 0.01

sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. x = 3.7, n = 32, ? = 1.8, H0: µ = 4.2 , Ha: µ ? 4.2 , ? = 0.05

A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn x = 3.26 , S = 0.55, ?N= 9, H0: µ = 2.85, Ha: µ > 2.85 , ? = 0.01

If the probability of winning is X, what is the probability of losing? (Assume there are no ties.) This means you can either win or lose. Since all probabilities in the sample space must add up to 1, then we have: P(Winning) + P(Losing) = 1 P(Losing) = 1 - P(Winning) Since P(Winning) = X, we have: P(Losing) = [B]1 - X[/B]

Imagine that a researcher wanted to know the average weight of 5th grade boys in a high school. He randomly sampled 5 boys from that high school. Their weights were: 120 lbs., 99 lbs, 101 lbs, 87 lbs, 140 lbs. What's the [B][U]standard error of the mean[/U][/B]? 9.29839 using our [URL='http://www.mathcelebrity.com/statbasic.php?num1=120%2C99%2C101%2C87%2C140&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics#standard_error_of_the_mean']statistics calculator[/URL]

Imagine that a researcher wanted to know the average weight of 5th grade boys in a high school. He randomly sampled 5 boys from that high school. Their weights were: 120 lbs., 99 lbs, 101 lbs, 87 lbs, 140 lbs. The researcher posed a null hypothesis that the average weight for boys in that high school should be 100 lbs. What is the [B][U]absolute value[/U][/B] of calculated t that we use for testing the null hypothesis? Mean is 109.4 and Standard Deviation = 20.79182532 using our [URL='http://www.mathcelebrity.com/statbasic.php?num1=120%2C99%2C101%2C87%2C140&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']statistics calculator[/URL] Now use those values and calculate the t-value Abs(t value) = (100 - 109.4)/ 20.79182532/sqrt(5) Abs(tvalue) = [B]1.010928029[/B]

In order to test if there is a difference between means from two populations, which of following assumptions are NOT required? a. The dependent variable scores must be a continuous quantitative variable. b. The scores in the populations are normally distributed. c. Each value is sampled independently from each other value. d. The two populations have similar means [B]a and d [/B] [I]because b and c [U]are[/U] required[/I]

Free Margin of Error from Confidence Interval Calculator - Given a confidence interval, this determines the margin of error and sample mean.

Free Paired Means Difference Calculator - Calculates an estimation of confidence interval for a small or large sample difference of data. Confidence interval for paired means

population MEAN OF ENVIRONMENTAL SPECIALIST SALARY IS $62000.A RANDOM SAMPLE OF 45 SPECIALIST IS DRAWN FROM THE POPULATION. WHAT IS THE LIKELIHOOD THAT THE MEAN SALARY SAMPLE IS $59000. ASSUME SIGMA IS $6000. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=59000&mean=62000&stdev=6000&n=45&pl=P%28X+%3C+Z%29']Z-Score calculator[/URL], we get the probability as [B]0.0004[/B].

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, [U][B]the Type I error is[/B][/U]: a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
d. to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher [B]b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same [/B] [I]A Type I error is when you reject the null hypothesis when it is in fact true[/I]

Free Random Sampling from the Normal Distribution Calculator - This performs hypothesis testing on a sample mean with critical value on a sample mean or calculates a probability that Z <= z or Z >= z using a random sample from a normal distribution.

Free Sample Size Requirement for the Difference of Means Calculator - Given a population standard deviation 1 of σ₁, a population standard deviation 2 of σ₂ a reliability (confidence) value or percentage, and a variation, this will calculate the sample size necessary to make that test valid.

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

[URL]http://www.mathcelebrity.com/marginoferror.php?num=60%2C66&pl=Calculate+Margin+of+Error+and+Sample+Mean[/URL]

Suppose a firm producing light bulbs wants to know if it can claim that its light bulbs it produces last 1,000 burning hours (u). To do this, the firm takes a random sample of 100 bulbs and find its average life time (X=980 hrs) and the sample standard deviation s = 80 hrs. If the firm wants to conduct the test at the 1% of significance, what's you final suggestion? (i..e, Should the producer accept the Ho that its light bulbs have a 1,000 burning hrs. at the 1% level of significance?) Ho: u = 1,000 hours. Ha: u <> 1,000 hours. [URL='http://www.mathcelebrity.com/mean_hypothesis.php?xbar=+980&n=+100&stdev=+80&ptype==&mean=+1000&alpha=+0.01&pl=Mean+Hypothesis+Testing']Perform a hypothesis test of the mean[/URL] [B]Yes, accept null hypothesis[/B]

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. a. If X = average distance in feet for 49 fly balls, then X ~ _______(_______,_______)
b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X. Shade the region corresponding to the probability. Find the probability.
c. Find the 80^th percentile of the distribution of the average of 49 fly balls a. N(250, 50/sqrt(49)) = [B]0.42074[/B] b. Calculate Z-score and probability = 0.08 shown [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+240&mean=+250&stdev=+7.14&n=+1&pl=P%28X+%3C+Z%29']here[/URL] c. Inverse of normal distribution(0.8) = 0.8416. Use NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL] Using the Z-score formula, we have 0.8416 = (x - 250)/50 x = [B]292.08[/B]

The distribution of actual weights of 8 oz chocolate bars produced by a certain machine is normal with µ=8.1 ounces and ?=0.1 ounces. A sample of 5 of these chocolate bars is selected. What is the probability that their average weight is less than 8 ounces? Calculate Z score and probability using [URL='http://www.mathcelebrity.com/probnormdist.php?xone=8&mean=8.1&stdev=0.1&n=5&pl=P%28X+%3C+Z%29']our calculator[/URL]: Z = -2.236 P(X < -2.236) = [B]0.012545[/B]

The hourly wages of employees at Rowan have a mean wage rate of $10 per hour with a standard deviation of $1.20. What is the probability the mean hourly wage of a random sample of 36 employees will be larger than $10.50? Assume the company has a total of 1,000 employees Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=10.5&mean=10&stdev=1.2&n=36&pl=P%28X+>+Z%29']normal distribution calculator[/URL], we get P(x > 10.5) = [B]0.00621[/B]

The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a) What is the probability that a randomly person has an IQ between 85 and 115? b) Find the 90th percentile of the IQ distribution c) If a random sample of 100 people is selected, what is the standard deviation of the sample mean? a) [B]68%[/B] from the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=50&mean=100&stdev=15&n=1&pl=Empirical+Rule']empirical rule calculator[/URL] b) P(z) = 0.90. so z = 1.28152 using Excel NORMSINV(0.9)
(X - 100)/10 = 1.21852 X = [B]113[/B] rounded up c) Sample standard deviation is the population standard deviation divided by the square root of the sample size 15/sqrt(100) = 15/10 =[B] 1.5[/B]

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

First, determine the [URL='http://www.mathcelebrity.com/statbasic.php?num1=87.4%2C86.9%2C89.9%2C78.3%2C75.1%2C70.6&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']mean and standard deviation[/URL] for the [I]sample[/I] Mean = 81.3667 SD = 7.803 Next, use our [URL='http://www.mathcelebrity.com/normconf.php?n=6&xbar=81.3667&stdev=7.803&conf=90&rdig=4&pl=Small+Sample']confidence interval for the mean calculator[/URL] with these values and n = 6 [B]74.9478 < u < 87.7856[/B]

The weight of a 9.5-inch by 6-inch paperback book published by Leaden Publications, Inc., is 16.2 oz. The standard deviation is 2.9 oz. What is the probability that the average weight of a sample of 33 such books is less than 15.89 oz? Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=15.89&mean=16.2&stdev=2.9&n=33&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we get: [B]0.271[/B]

There are 5 red and 4 black balls in a box. If you pick out 2 balls without replacement, what is the probability of getiing at least one red ball? First list out our sample space. At least one means 1 or 2 red balls, so we have 3 possible draws: [LIST=1] [*]Red, Black [*]Black, Red [*]Red, Red [/LIST] List out the probabilities: [LIST=1] [*]Red (5/9) * Black (4/8) = 5/18 [*]Black (4/9) * Red (5/8) = 5/18 [*]Red (5/9) * Red (4/8) = 5/18 [/LIST] Add these up: 3(5)/18 = [B]5/6[/B]

True or False (a) The normal distribution curve is always symmetric to its mean. (b) If the variance from a data set is zero, then all the observations in this data set are identical. (c) P(A AND A^c)=1, where A^c is the complement of A. (d) In a hypothesis testing, if the p-value is less than the significance level ?, we do not have sufficient evidence to reject the null hypothesis. (e) The volume of milk in a jug of milk is 128 oz. The value 128 is from a discrete data set. [B](a) True, it's a bell curve symmetric about the mean (b) True, variance measures how far a set of numbers is spread out. A variance of zero indicates that all the values are identical (c) True. P(A) is the probability of an event and P(Ac) is the complement of the event, or any event that is not A. So either A happens or it does not. It covers all possible events in a sample space. (d) False, we have sufficient evidence to reject H0. (e) False. Volume can be a decimal or fractional. There are multiple values between 127 and 128. So it's continuous.[/B]

What is the sample space for a 10 sided die? Sample space means the set of all possible outcomes. For a 10-sided die, we have: [B]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}[/B]

Which of the following can increase power? a. Increasing standard deviation b. Decreasing standard deviation c. Increasing both means but keeping the difference between the means constant d. Increasing both means but making the difference between the means smaller [B]b. Decreasing standard deviation[/B] [LIST=1] [*]Power increases if the standard deviation is smaller. [*]If the difference between the means is bigger, the power is bigger. [*]Sample size increase also increases power [/LIST]