Salary intormation regarding male and female employees of a large company is shown below.Sample meanPopulationstd dev - a(in 1000 TL)Female10026Male15018Calculate the z test statistic and p-value for the above hypothesis testp-value

Given that Sample Mean 1 (\bar X_1)(Xˉ1) = 2626 Population Standard Deviation…

Answered: Calculate the z test statistic and…

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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Only Answer Question 15. Question 15 has parts to it, would love all the parts to be answered please.
A survey of 580 smartphone users showed that 55% of them said that their smartphone is the only thing they could not live without. Use 0.05 significance level to test the claim that more than half of smartphone users identify the smartphone as the only thing they could not live without. Round to four decimal places if necessary. a) Null hypothesis H0 : Alternate hypothesis H1 : b) Type of test: right or left or two c) significance level: α = d) Test statistic: e) Compute p-value of the test statistic.
The accompanying table lists survey results obtained from a random sample of different crime victims. Homicide Robbery Assault Criminal was a stranger 40 72 Crimial was an acquaintance or relative 13 64 Test at the 0.05 level the claim that the type of crime relates to whether the criminal is a stranger. The null hypothesis (H_0) is Select one: O a. p_{criminal was a stranger} = p_{crimial was an acquaintance or relative} O b. p_{homicide} = p_{robbery} = p_{assault} =1/3 O c. The type of crime is dependent on whether the criminal is a stranger O d. None of the other answers in necessary true. O e. The type of crime is independent of whether the criminal is a stranger
Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 12) A supplier of 3.5" disks claims that no more than 1% of the disks are defective. In a random sample of 600 disks, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of significance, test the supplier's claim that no more than 1% are defective. 12)
What is the setup of the null and alternative hypotheses for this test?
in a recent court case it was found that during a. Of 11 years 862 people were selected for grand jury duty and 38% of them were from the same ethnicity of people eligible for grand jury duty 80.4% weirdest ethnicity uses 0.05 significance level to test the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury identify the null hypothesis alternative hypothesis test statistics P value conclusion about the null hypothesis and final conclusion that addresses the original claim used to be Value method in the normal distribution as an approximation to the binomial distribution
In a study of 805 randomly selected medical malpractice lawsuits, it was found that 518 of them were dropped or dismissed. Use a 0.05 significance level to test the claim that most medical malpractice lawsuits are dropped or dismiss Which of the following is the hypothesis test to be conducted? O A. Ho: p+0.5 H1:p= 0.5 О В. Но: р30.5 H,: p#0.5 ОС. Но: р> 0.5 O D. Ho: p= 0.5 H1: p= 0.5 H1: p0.5 What is the test statistic? (Round to two decimal places as needed.) What is the P-value? P-value = (Round to three decimal places as needed.) What is the conclusion about the null hypothesis? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a. O B. Reject the null hypothesis because the P-value is greater than the significance level, a. OC. Fail to reject the null hypothesis because the P-value is greater than the significance level, a. O D. Reject the null hypothesis because the P-value is less than or equal to the significance level,…
For a Lower one-tailed hypothesis test with a sample size of 18 at a = .10, the critical value for the decision would be: O 2.11 O -2.11 O -1.740 O 1.333 O none of these < Previous Next »
2) Two groups of students were surveyed – 86 students were enrolled in a class that required the purchase of lecture notes with mean for this sample being 8.48 and sample variance 0.94. While 35 students were enrolled in another class that did not require the purchase of lecturer notes mean for this sample being 7.80 and sample variance 2.99. The students need to decide whether or not to purchase the lecture notes? a) Hypothesis testing: Do the samples provide sufficient evidence to conclude that there is a difference in the mean responses of the two groups of students? Test using a level of significance = 0.01 (1%). b) Construct a 99% confidence interval for the difference between the two samples and interpret the results. c) Calculate the P-Value and interpret.
A two-tailed hypothesis test for the population mean has two critical values. may have one or two critical values. has only one critical value. O may have more than two critical values.
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Interpret the results presented in the following table. H1: The predictor variables (i.e. Tangibility, Reliability, Responsiveness, Empathy, and Assurance) have an effect on customer satisfaction H2: Tangibility has a positive impact on customer satisfaction. H3: Reliability has a positive impact on customer satisfaction. H4: Responsiveness has a positive impact on customer satisfaction. H5: Empathy has a positive impact on customer satisfaction. H6: Assurance has a negative impact on customer satisfaction.

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Calculate the z test statistic and p-value for the above hypothesis test Pvakue