**Chi-Square Distribution Analysis for Platelet Counts**To determine the confidence interval for a population standard deviation based on a sample, the following steps and information are essential:1. **Sample Information**: - **Confidence Level**: 90% - **Sample Size (n)**: 40 - **Sample Standard Deviation (s)**: 65.92. **Objective**: - Calculate the degrees of freedom. - Identify the critical values for the Chi-Square distribution. - Estimate the confidence interval for the population standard deviation (σ).3. **Calculations**: - **Degrees of Freedom (df)**: \[ \text{df} = n - 1 \] - Input a whole number for degrees of freedom. - **Chi-Square Critical Values**: - \( \chi^2_L \) (Lower critical value) - \( \chi^2_R \) (Upper critical value) - Round these values to three decimal places.4. **Confidence Interval for σ**: - The interval is represented as: \[ \left( \frac{\sqrt{df} \cdot s}{\sqrt{\chi^2_R}} \right) < \sigma < \left( \frac{\sqrt{df} \cdot s}{\sqrt{\chi^2_L}} \right) \] - Ensure accurate rounding for your final calculation.Use the Chi-Square critical value table to find the necessary values and complete the analysis. This method assumes the sample originates from a normally distributed population.

Answered: Use the given information to find the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Z-value p-value Single Policyholders n₁ = 440 Number making claims = 88 a. Use a = 0.05. Test to determine whether the claim rates differ between single and married male policyholders. (to 2 decimals) (to 4 decimals) X Married Policyholders n2 = 840 Number making claims 126 We can conclude ✔ that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number. X * )
Q scores are normally distributed with μ = 100 and σ = 15. Find the IQ score separatingthe highest 4% of all scores.
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P 25 Click to view page 1 of the table. Click to view page 2 of the table. The z-score that corresponds to P25 is (Round to two decimal places as needed.)
Please answer the question(s) with the same format as the examples. Thank you.
Refer to the figure at the right. At a= 0.10, can you reject the claim that the proportion of 18 to 24-year-olds living in their parents' homes in 2000 was the same for men and women? Assume the survey included a random sample of 270 men and 278 women. Percentage of 18- to 24-year olds living in parents' homes 70, 60.4% 45% 54.6% 50.3% 2000 2012 Men Women Identify the null and alternative hypotheses. Choose the correct answer below. O B. Ho: P, * P2 Ha: P1 = P2 OD. Ho: P1 > P2 YA. Ho: P1 = P2 H3: P, * P2 OC. Ho: P, P2 Ha: P, SP2 Find the critical value(s). The critical value(s) is(are). (Round to two decimal places as needed. Use a comma to separate answers as needed.)
pls answer 1-3 in paper
Suppose baby kittens' weights are normally distributed with a mean of 11.4 and a standard deviation of 2.1. The Z-score tells you how many units above the average (if Z-score is positive) or below the average (if Z- Score is negative) any particular baby kitten's weight is. Find the baby kitten weight that corresponds to the following Z-scores. Use the formula Z where u is the mean, o is the standard deviation, and X is the baby kitten weight. a. Z = = 0.07, X = b. Z = 1.63, X =
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P80
Calculate the z score given the following information: x=20 mean = 15 SD = 5
Two universities give samples of students the same exam. School A records the follwoing sample statistics: ā1=73, 8,=5.5, ni =80. School B records the follwoing sample statistics: z=69, ,=4.4, n2 =90. We wish to build a 95 % cofidence interval to estimate the difference in means between these schools. Define your variable: O Let p be the proprotion School A students that were higher than School B. Let u the mean test score that School A was higher than School B. Let u be the mean test score. O Let u be the mean test score at School A. Create and interpret a 95% Confidence Interval. Round your answers to one decimal place. Select an answer and Select an answer What is the margin of error? Round your answer to one decimal place. Do these results suggest that School A might have higher means on this test? ? Do these results PROVE that School A might have higher means on this test? The campus president at School A made the following statement. "Our students score an average of 5 points…
Find the z-score corresponding to the given value and use the z-score to determine whether the value is significantly low or high. Consider a score to be significantly low if its z-score is less than or equal to -2. Consider a z-score to be significantly high if it is greater than or equal to 2. Round the z-score to the nearest tenth if necessary. A body temperature of 96.7° F given that human body temperatures have a mean of 98.20° F and a standard deviation of 0.62°.
Find the z-scores for which 80% of the distribution's area lies between -z and z. Click to view page 1 of the table. Click to view page 2 of the table. The Z-scores are (Use a comma to separate answers as needed. Round to two decimal places as needed.) C

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