**Analysis of Left-handedness in Males and Females**In a random sample of males, 22 out of 242 were found to write with their left hands. In a random sample of females, 70 out of 522 were left-handed. We want to test the claim that the rate of left-handedness among males is less than that among females using a 0.01 significance level. Below is the detailed analysis:1. **Calculate the z-score:** \[ z = -1.71 \] *(Round to two decimal places as needed.)*2. **Identify the P-value:** \[ \text{P-value} = 0.044 \] *(Round to three decimal places as needed.)*3. **Conclusion Based on Hypothesis Test:** Since the P-value is greater than the significance level ($\alpha = 0.01$), we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.4. **Confidence Interval for the Difference in Proportions:** The 98% confidence interval for $ (p_1 - p_2) $ is: \[ Solution: \, \text{Enter interval values here} \] *(Round to three decimal places as needed.)*This analysis indicates no statistically significant difference in the rates of left-handedness between males and females at the 0.01 significance level.

Given Let sample 1 denote sample of males. Number of males write with left hand, x1=22 Number of…

Answered: In a random sample of males, it was…

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. Single Policyholders Married Policyholders n1=550 n2=1,000 Number making claims = 110 Number making claims = 150 a. Use a=0.05 Test to determine whether the claim rates differ between single and married male policyholders. z-value (to 2 decimals) p-value (to 4 decimals) 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. ( , )
A survey of 700 non fatal accidents showed that 287 involved defective equipment. Find a point estimate for p, the population proportion of accidents that involved defective equipment and round to 4 decimal places if needed.
. Students may be classified as having a math dysfunction—and not have to take statistics—if they score below the 25th percentile on a diagnostic test. The m of the test is 75 and sX 10. Approximately what raw score is the cutoff score needed to avoid taking statistics?
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A study reported a correlation between height and weight r = .5, p < 0.05. What is the proportion of variability in people's weight that can be explained by differences in height? O 0.71 0.5 0.05 0.25
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The average home in the US is expected to cost $240,000. A random sample of 56 homes sold this month showed average price of $232,000. Assume you have access to this data. We are interested in determining cost if the cost of average home has decreased this month. If the test statistic is -1.66 what is the p-value?
In a measurement of B two results were obtained, B1 = (8.3+/- 0.78) mT and B2 = (8.81+/- 0.20) mT . The error on the weighted/arithmetic mean is (Report your answer in mT with 3 significant figures)
Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 33 of the 38 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 94 of the 111 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test? O A. Ho: P1 #P2 Hy:P1 =P2 O B. Ho: P1 =P2 H1: P1 P2 H1: P1 #P2 Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? the null hypothesis. There sufficient evidence to support the claim that echinacea…
Help with to= what is p value =
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.)
The authors of a paper were interested in how the distance a deer mouse will travel for food is related to the distance from the food to the nearest pile of debris. Distances were measured in meters. The data and computer output are given below. Distance from Debris Distance Traveled 6.94 0.00 5.23 6.13 5.21 11.29 7.10 14.35 8.16 12.03 5.50 22.72 9.19 20.11 9.05 26.16 9.36 30.65 Simple Linear Regression Results: Dependent Variable: Traveled Independent Variable: Debris Sample size: 9 R (correlation coefficient) = 0.5657 R-sq = 0.32002088 Estimate of error standard deviation 8.670711 Parameter estimates: Parameter Estimate Std. Err. Alternative DF T-Stat P-Value Intercept -7.6854587 13.332196 ≠ 0 7 -0.5764586 0.5824 Slope 3.2340908 1.7818117 ≠ 0 7 1.8150575 0.1124 a)What is the least squares regression line for the output given above? b) what is the predicted traveled distance given the distance from debris is 6.5 meters?

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