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critical value - a point on the distribution of the test statistic under the null hypothesis that defines a set of values that call for rejecting the null hypothesis.

A group of students at a school takes a history test. The distribution is normal with a mean of 25,

A group of students at a school takes a history test. The distribution is normal with a mean of 25, and a standard deviation of 4.
(a) Everyone who scores in the top 30% of the distribution gets a certificate.
(b) The top 5% of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state?
(a) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.70&pl=Calculate+Critical+Z+Value']Top 30% is 70% percentile[/URL]
Inverse of normal distribution(0.7) = -0.5244005
Plug into z-score formula, -0.5244005 = (x - 25)/4
[B]x = 22.9024[/B]
(b) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']Top 5% is 95% percentile[/URL]
Inverse of normal distribution(0.95) = 1.644853627
Plug into z-score formula, 1.644853627 = (x - 25)/4
[B]x = 31.57941451[/B]

Analysis of Variance (ANOVA)

Free Analysis of Variance (ANOVA) Calculator - Performs 1 way analysis of variance or 2 way analysis of variance on a set of data with critical value test and conclusion.

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph.
a. The current speed limit is 65 mph. What is the proportion of vehicles less than or equal to the speed limit?
b. What proportion of the vehicles would be going less than 50 mph?
c. A new speed limit will be initiated such that approximately 10% of vehicles will be over the speed limit. What is the new speed limit based on this criterion?
d. In what way do you think the actual distribution of speeds differs from a normal distribution?
a. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=65&mean=71&stdev=8&n=+1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that P(x<65) = [B]22.66%[/B]
b. Using our [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+50&mean=71&stdev=8&n=+1&pl=P%28X+%3C+Z%29']z-score calculator[/URL], we see that P(x<50) = [B]0.4269%[/B]
c. [URL='http://www.mathcelebrity.com/zcritical.php?a=0.9&pl=Calculate+Critical+Z+Value']Inverse of normal for 90% percentile[/URL] = 1.281551566
Plug into z-score formula: (x - 71)/8 = 1.281551566
[B]x = 81.25241252[/B]
d. [B]The shape/ trail differ because the normal distribution is symmetric with relatively more values at the center. Where the actual has a flatter trail and could be expected to occur.[/B]

Chi-Square χ

Free Chi-Square χ^{2} Test Calculator - This calculator determines a χ^{2} chi-square test on a test statistic and determines if it is outside an accepted range with critical value test and conclusion.

Chi-Square Critical Values

Free Chi-Square Critical Values Calculator - Given a probability, this calculates the critical value for the right-tailed and left-tailed tests for the Chi-Square Distribution. CHIINV from Excel is used as well.

Class Frequency Goodness of Fit

Free Class Frequency Goodness of Fit Calculator - Performs a goodness of fit test on a set of data with class boundaries (class boundary) with critical value test and conclusion.

Critical Values for F-test

Free Critical Values for F-test Calculator - Calculate a critical value for the F-Test statistic based on DF1, DF2, and α

Critical Z-values

Free Critical Z-values Calculator - Given a probability from a normal distribution, this will generate the z-score critical value. Uses the NORMSINV Excel function.

Expected Frequency

Free Expected Frequency Calculator - Given a contingency table (two-way table), this will calculate expected frequencies and then determine a conclusion based on a Χ^{2} test with critical value test and conclusion.

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of t

Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
b. Find the 95th percentile, and express it in a sentence.
a. P(X >=0.30), calculate the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+0.30&mean=+0.28&stdev=+0.05&n=+1&pl=P%28X+%3E+Z%29']z-score[/URL] which is:
Z = 0.4
P(x>0.4) = [B]0.344578 or 34.46%[/B]
b. Inverse Normal (0.95) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.95&pl=Calculate+Critical+Z+Value']calculator[/URL] = 1.644853627
Use NORMSINV(0.95) on Excel
0.28 + 0.05(1.644853627) = [B]0.362242681 or 36.22%[/B]

Mcnemar Test

Free Mcnemar Test Calculator - Given a 2 x 2 contingency table and a significance level, this will determine the test statistic, critical value, and hypothesis conclusion using a Mcnemar test.

Random Sampling from the Normal Distribution

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.

Standard Normal Distribution

Free Standard Normal Distribution Calculator - Givena normal distribution z-score critical value, this will generate the probability. Uses the NORMSDIST Excel function.

Student-t Distribution Critical Values

Free Student-t Distribution Critical Values Calculator - Given an α value and degrees of freedom, this calculates the right-tailed test and left-tailed test critical values for the Student-t Distribution

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed wit

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]

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

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed wit

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.
a. If X = distance in feet for a fly ball, then X ~
b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability.
c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.
a. [B]N(250, 50/sqrt(1))[/B]
b. Calculate [URL='http://www.mathcelebrity.com/probnormdist.php?xone=+220&mean=250&stdev=50&n=+1&pl=P%28X+%3C+Z%29']z-score[/URL]
Z = -0.6 and P(Z < -0.6) = [B]0.274253[/B]
c. Inverse of normal distribution(0.8) = 0.8416 using NORMSINV(0.8) [URL='http://www.mathcelebrity.com/zcritical.php?a=0.8&pl=Calculate+Critical+Z+Value']calculator[/URL]
Z-score formula: 0.8416 = (x - 250)/50

x = [B]292.08[/B]

x = [B]292.08[/B]

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered t

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer.
Digit, Probability
1, 0.301
2, 0.176
3, 0.125
4, 0.097
5, 0.079
6, 0.067
7, 0.058
8, 0.051
9, 0.046
[B][U]Fradulent Checks[/U][/B]
Digit, Frequency
1, 36
2, 32
3, 45
4, 20
5, 24
6, 36
7, 15
8, 16
9, 7
Complete parts (a) and (b).
(a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?

Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.

Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.

The margarita is one of the most common tequila-based cocktails, made with tequila, triple sec, and

The margarita is one of the most common tequila-based cocktails, made with tequila, triple sec, and lime
juice, often served with salt on the glass rim. A manager at a local bar is concerned that the bartender is
not using the correct amounts of the three ingredients in more than 50% of margaritas. He secretly
observed the bartender and found that he used the CORRECT amounts in only 9 out of the 39
margaritas in the sample. Use the critical value approach to test if the manager's suspicion is justified
at ? = 0.10. Let p represent the proportion of all margaritas made by the bartender that have
INCORRECT amounts of the three ingredients. Use Table 1.
a. Select the null and the alternative hypotheses.
[B]H0: p ? 0.50; HA: p > 0.50[/B]
[B][/B]
b. Calculate the sample proportion. (Round your answer to 3 decimal places.)
9/39 = [B]0.231
[/B]
c. Calculate the value of test statistic. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Using our [URL='http://www.mathcelebrity.com/proportion_hypothesis.php?x=9&n=39&ptype=%3C&p=+0.5&alpha=+0.10&pl=Proportion+Hypothesis+Testing']proportion hypothesis calculator[/URL], we get:
[B]Test Stat = -3.36[/B]
[B][/B]
d. Calculate the critical value. (Round your answer to 2 decimal places.)
Using the link above, we get a critical value of [B]1.2816
[/B]
e. What is the conclusion?
[B]The manager’s suspicion is not justified since the value of the test statistic does not fall in the rejection region. Do not reject H0[/B]
[B][/B]