Topic 4 Review Questions

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Topic 4 Review Questions Chapter 9 9.13A) 1, Given: SE=?; σ= 6 ; n=36 SE=σ/√n =6/√36=6/6=1 B) 2, Given: SE=?; σ= 12 ; n=36 SE=σ/√n =12/√36=12/6=2 C) 5, Given: SE=?; σ= 30 ; n=36 SE=σ/√n=30/√36=30/6=5 D) 100, Given: SE=?; σ= 600 ; n=36 SE=σ/√n=600/√36=600/6=100 Chapter 10 10.10) In this instance, the null hypothesis would be rejected, and it would be determined that the new data sample is not representative of the original population. Given the information n=20 and s=10, T= (30-35)/(10)=5/2.2361=-2.236. Therefore, the degrees of freedom =n-1=19 making 2 tailed p- value=0.0375. I can deduce that the population mean is not 35 since p-value is less than 0.05, which allows me to reject the null hypothesis at the 0.05 level of significance. N-20, X=30, S=10, ?=0.05 ? ℴ :?=35 ? ℴ :?≠35 ?= ?−? ? √ ? ?= 30−35 10 √ 20 ?=−2.236 Chapter 11 11.13a) The alternative hypothesis (H1) is salary discrimination towards female members; if the investigator is primarily concerned with salary discrimination towards female individuals, the null hypothesis (H0) is that there is no salary discrimination towards female members. In other words: H0: Salary discrimination against female members does not exist. H1: Salary discrimination against female employees exists. 11.13b) This is because the probability of producing a Type I mistake is higher at the.01 level of significance than it is at the.05 level of significance (rejection of the null hypothesis is true) is less significant at the.01 level of analysis than it is at the.05 level. It is crucial to reduce the possibility of making a Type I error because of the possible repercussions of an expensive class-action lawsuit. A costly class-action lawsuit could result from a Type I error, thus it is imperative to minimize the risk of making one. Η?:?=$82,500 Η?:?≠$82,500 Two tails The sample means ? ¿$80,100 The population standard deviation, ?=$6,000 The sample size, n=100 Test statistics; Z= ? −? ? √ ? Z= 82,500−80,100 6,000 √ 100 = 4 Decision rule: Reject the null≤? p-value= 2P(?>4 ) =0.000063 Decision; When? =0.05 p-value ≈0.000063<?=0.0500 Or when? =0.01 p- value ≈0.000063<? =0.0100 11.19a) Type I error: rejects the hypothesis before the tests are finished, a smaller significance level must be chosen to minimize errors, which lowers the alpha level's value.

11.19b) Type II error: accepts a false hypothesis though if a bigger sample size would be selected in order to reduce mistakes. As a result, the testing will be less variable. and raising the significance threshold to a higher number from the conventional 5%. Chapter 12 12.8a) sample size n=36, population sd is unknown or Z, population mean is unknown sample mean is unknown Z/ 2= Z0.025= 1.96, Confidence Interval: ±0.1, Range for the true population mean: 32.99 to 33.19 95% =33.09± 0.10or[32.99,33.19 ] Use the z score to generate the confidential interval because this is a sample distribution with a known standard deviation. For a two-tailed confidence interval at 95%, the Z statistic is 1.96. Z=(1+0.95) /2) = 1.96 ?95%= 33.09± 1.96 0.30 √ 36 ?95% = 33.09±0.01 or[32.99,33.19 ] The population mean can be regarded as falling within the 95% confidence interval of 32.99 to 33.19 based on how the confidence interval is interpreted. Given that the interval's lower bound is greater than 32 ounces, there is enough data to draw the conclusion that the maker succeeded in exceeding 32 ounces. 12.8b) It is certain that the cartons weigh more than 32 ounces because 32.992>32.

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