Determine the critical values for these tests of a population standard deviation.(a) A right-tailed test with 14 degrees of freedom at the a = 0.01 level of significance(b) A left-tailed test for a sample of size n =25 at the a = 0.1 level of significance(c) A two-tailed test for a sample of size n= 22 at the a = 0.01 level of significanceChi-Square (x²) DistributionArea to the Right of Critical ValueDegrees ofFreedomClick the icon to view a table a critical values for the Chi-Square Distribution.0.9950.990.9750.950.900.100.050.0250.010.005(a) The critical value for this right-tailed test is0.0162.7066.6359.21011.34513.277127.87910.5970.0010.0510.2160.4840.0040.1033.8415.024(Round to three decimal places as needed.)0.0100.0720.2070.4120.2110.5841.0647.3789.34811.1430.0204.6055.991340.1150.2970.5540.3520.7116.2517.7797.8159.48811.07012.83814.86050.8311.1451.6109.23612.83315.08616.7501.63518.54820.27821.9550.6760.8721.2372.20410.64512.59214.44916.8122.1672.7333.3253.9402.8333.49012.01713.36214.06715.50718.47520.09070.9891.2391.6461.69016.0131.3442.18017.5352.0882.5582.7003.2474.1684.86516.91918.30723.58925.1881.73514.68415.98719.02321.666102.15620.48323.209112.6033.0533.8164.5755.57817.27519.67521.92024.72526.7573.5714.4045.0096.3047.0427.7908.54728.30012133.0743.5655.22618.54919.81221.02622.36223.68524.99623.33724.73626.21727.68829.81931.31932.8014.1075.8924.6605.2295.6296.262144.0754.6016.5717.26121.06422.30726.11927.48829.1411530.5785.8126.4087.01526.29627.58732.00033.40934.805165.1426.9087.9629.31223.54228.84534.26724.76925.98927.20428.4127.5648.23110.08517185.6978.6729.39030.19131.52635.71837.15610.86511.65112.4436.26528.869192010.11710.85132.85234.17038.58239.9976.8447.6338.2608.9079.59130.14436.1917.43431.41037.5668.8979.54210.28310.98211.68913.24014.04129.61530.81311.591212235.47936.78138.93240.2898.0348.6439.2609.88610.52032.67133.92435.17241.40112.33842.7962310.19613.09114.84838.07632.00733.19634.38241.63844.181242513.84814.61145.55946.92810.85612.40115.65916.47336.41539.36440.64642.98044.31411.52413.12037.652

Name: Determine the critical values for these tests of a population standar
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Determine the critical values for these tests of a population standard deviation.(a) A right-tailed test with 14 degrees of freedom at the a = 0.01 level of significance(b) A left-tailed test for a sample of size n =25 at the a = 0.1 level of signif

We have to find critical value.

Math

Statistics

Determine the critical values for these tests of a population standard deviation. (a) A right-tailed test with 14 degrees of freedom at the a = 0.01 level of significance (b) A left-tailed test for a sample of size n= 25 at the a = 0.1 level of significance (c) A two-tailed test for a sample of size n= 22 at the a = 0.01 level of significance Degr Freed Click the icon to view a table a critical values for the Chi-Square Distribution. (a) The critical value for this right-tailed test is (Round to three decimal places as needed.) 523 S

Determine the critical values for these tests of a population standard deviation. (a) A right-tailed test with 14 degrees of freedom at the a = 0.01 level of significance (b) A left-tailed test for a sample of size n= 25 at the a = 0.1 level of significance (c) A two-tailed test for a sample of size n= 22 at the a = 0.01 level of significance Degr Freed Click the icon to view a table a critical values for the Chi-Square Distribution. (a) The critical value for this right-tailed test is (Round to three decimal places as needed.) 523 S

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Body mass index (BMI) is a medical screening tool that measures body fat in an individual based on height and weight. The following data sets include four random samples of BMI measurements. Each sample includes the measurements of 7 individuals. Use the range rule of thumb to estimate the standard deviation, then determine which sample has the most variation. Sample 1 25.8,39.2,32.2,18.7,22.7,28.8,25.4 Sample 2 24.5,33.3,31.7,45.9,23.4,28.2,25.8 Sample 3 30.8,19.2,29.9,25.6,28.8,31.0,27.3 Sample 4 31.8,23.9,29.1,52.6,23.6,31.1,20.6
Moviegoers The average "moviegoer" sees 8.5 movies a year. A moviegoer is defined as a person who sees at least one movie in a theater in a 12-month period. A random sample of 42 moviegoers from a large university revealed that the average number of movies seen per person was 10.2. The population standard deviation is 3.2 movies. At the 0.01 level of significance, can it be concluded that this represents a difference from the national average? Part 1 of 5 State the hypotheses and identify the claim with the correct hypothesis. Hotu = 85 Part: 1/5 not claim H₁ 8.5 claim This hypothesis test is a two-tailed ▼test. Part 2 of 5 ▼ ▼ Find the critical value(s). Round the answer to at least two decimal places. If there is more than one critical value, separate them with commas. Critical value(s): X 0.0.... + 8分民图〗b. E
The world is currently being adversely affected by the COFLU-20 pandemic. Persons are required to take certain measures to prevent the spread of this virus; however, persons’ knowledge, attitudes and practices towards COFLU-20 may influence their adherence to these measures. A researcher decided to evaluate the knowledge, attitudes and practices towards COFLU-20 of adults living in the English-speaking Caribbean. To gather the data, the researcher developed an online survey consisting of four (4) sections. Section A gathered demographic data on the respondents including their gender, marital status, highest level of education attained, occupation and country of residence. Section B consisted of 15 questions that gathered data on the respondents’ knowledge of COFLU-20. Section C consisted of 10 questions that gathered data on the respondents’ attitudes towards COFLU-20. The final section, Section D, consisted of 5 questionsthat gathered information on the respondents’ practices against…
Hypothesis Test about a Population Standard Deviation You conducted a research study to analyze the pulse rates of adult men. A simple random sample of 153 adult men from a normally distributed population resulted in a sample standard deviation of pulse rates of 11.33; the sample variance is then 128.40. The given information is then: ? = 153 ? = 11.33 ? 2 = 128.40 Use the sample results with a significance level ? = 0.01 to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
Watching TV: The 2012 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.9. Assume that in a sample of 37 teenagers, the sample standard deviation of daily TV time is 2.2 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 2.9? Use the a=0.10 level of significance.
Cost of Building a Home According to the National Association of Home Builders, the average cost of building a home in the Northeast is $117.91 per square foot. A random sample of 34 new homes indicated that the mean cost was $120.61and the standard deviation was $17. Can it be concluded that the mean cost differs from $117.91, using the 0.10level of significance? Use the P-value method. Use a graphing calculator. (b) Find the P-value. Round the answer to at least four decimal places. P-value=
The ___________ is a weighted average of the Standard Deviations of the conditions of an IV in an independent samples t-test. Pooled estimate of the Standard deviation Standard deviation of the difference Population Standard deviation Standard error of the difference
Does color enhance creativity? Test the indicated claim about the standard deviations or variances of two populations. Subjects are given a creativity exercise on a computer with either a red background or a blue background. The scores are shown in the table. At the 0.05 significance level, test the claim that those tested with red background have creativity scores with a standard deviation equal to the standard deviation for those tested with a blue background. Red Background Blue Background n1 = 39 n2 = 34 xˉx̄1 = 19.6 xˉx̄2 = 20.5 s1 = 0.72 s2 = 0.01 What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)H0: Select an answer x̄₂ x̄₁ μ(red) p₂ p μ₁ μ₂ σ₁ s₁² σ(red) p₁ μ p̂₁ ? ≠ < > ≤ ≥ = Select an answer μ₁ μ₂ p₁ μ(blue) p σ₂ p₂ σ(blue) s₁² x̄₁ x̄₂ μ p̂₁ H1: Select an answer μ σ₁ x̄₁ μ(red) p̂₂ p₁ s₂² p₂ μ₂ μ₁ σ(red) x̄₂ p ? ≤ ≥ ≠ > = < Select an answer μ(blue) σ(blue) p μ μ₁ p₁ x̄₂ s₁² x̄₁ σ₂ p̂₁ p₂ μ₂…
Please answer all sections
Test the claim that the mean GPA of night students is significantly different than 2.6 at the 0.2 significance level. Based on a sample of 50 people, the sample mean GPA was 2.55 with a standard deviation of 0.04The test statistic is: ? (2 decimals)The p-value is: ? (2 decimals)
Detecting gypsy moths The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. Each month, an SRS of 50 traps is inspected, the number of moths in each trap is recorded, and the mean number of moths is calculated. Based on years of data, the distribution of moth counts is discrete and strongly skewed with a mean of 0.5 and a standard deviation of 0.7. a. Explain why it is reasonable to use a Normal distribution to approximate the sampling distribution of for SRSS of size 50. b. Estimate the probability that the mean number of moths in a sample of size 50 is greater than or equal to 0.6. c. In a recent month, the mean number of moths in an SRS of size 50 was = 0.6. Based on this result, is there convincing evidence that the moth population is getting larger in this state? Explain your reasoning.
An experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. The mean tip percentages and standard deviations are given in the accompanying table along with the sample sizes. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. The summarized statistics about giving candy. No candy 21 18.53 1.52 Two candy 21 20.93 2.54 (a) Use a 0.01 significance level to test the claim that giving candy does result in greater tips. (b) Test statistic (Two decimal places. ) (c) P-value (Three decimal places.) (d) Construct confidence interval lower bound (Round to two decimal places.) upper bound (Round to two decimal places.)