Homework Answers
Determine (a) the test statistic, (b) the degrees of freedom, (c) the P-alue, and (d) test...
Determine (a) the test statistic, (b) the degrees of freedom, (c) the P-alue, and (d) test the hypothesis at the a-0.05 level of significance. Outcome Obsenved Expected 25 25 25 25 A B C Do 22 28 24 26 H,: At least one of the proportions is different from the others Click the icon to wew the chi-square distribution table Round to two decimal places as needed.) (b) (True False) The number of degrees of freedom is oneless than the...
Determine (a) the χ2 test statistic, (b) test degrees of freedom, (c) the critical value using...
Determine (a) the χ2 test statistic, (b) test degrees of freedom, (c) the critical value using α=0.05, and (d) test the hypothesis at the α=0.05 level of significance. Outcome Observed Expected A 18 20 B 18 20 C 22 20 D 22 20 H0:Pa=Pb=Pc=Pd=1/4 H1: at least one of the proportions is different from the others. (a)test statistics is 0.8 (b) There are 3 degrees of freedom (c) The critical value is X ?? (round three decimal places as needed)...
01 2 3 4 Delermine (a) the y2 lest atalistic () Uhe dearees of freedom, fe)...
01 2 3 4 Delermine (a) the y2 lest atalistic () Uhe dearees of freedom, fe) the P-value, and (d) test e hyoothesis at the -005 Jeve srifcarce Expected 6 25.3 162 14s 40S e random vanable X is bnemial wth n 4, p Ho. cick the lan to vlew e ch-scuero dstdbuticn aNo IRound to two decimal places as needed Thera andagraas of freedom. The Pvalua is What is the conclusion of the test at the 0.05 level of...
Column variable Row variable Nursing Psychology Total Nervous 16 3 19 Excited 4 17 21 Total...
Column variable Row variable Nursing Psychology Total Nervous 16 3 19 Excited 4 17 21 Total 20 20 40 Expected Frequencies Column variable Row variable Nursing Psychology Total Nervous 9.5 9.5 19 Excited 10.5 10.5 21 Total 20 20 40 Data Level of Significance 0.05 Number of Rows 2 Number of Columns 2 Degrees of Freedom 1 Results Critical Value 3.841459149 Chi-Square Test Statistic 16.94235589 p-Value 0.0000385 Reject the null hypothesis This is the data set that I need to...
Consider the observed frequency distribution for the set of random variables. a. Perform a chi-square test...
Consider the observed frequency distribution for the set of random variables. a. Perform a chi-square test using alpha=0.05 to determine if the observed frequencies follow the binomial probability distribution when p=0.50 and n=4. b. Determine the p-value and interpret its meaning. Random Variable, X Frequency, Fo 0 29 1 96 2 151 3 96 4 28 Total 400 The chi-square test statistic is chi squared, χ2=______ p-value=______
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations....
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations. Sample 1 Sample 2 x¯1=20.92 x¯2=26.80 s21=2.89 s22=3.81 n1=19 n2=15 Test the null hypothesis H0:μ1=μ2against the alternative hypothesis HA:μ1<μ2. a) Calculate the test statistic for the Welch Approximate t procedure. Round your response to at least 3 decimal places. b) The Welch-Satterthwaite approximation to the degrees of freedom is given by df = 27.983055. Using this information, determine the range in which the p-value...
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations....
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations. Sample 1 Sample 2 x¯1=20.08 x¯2=24.51 s21=2.05 s22=3.20 n1=19 n2=16 Test the null hypothesis H0:μ1=μ2against the alternative hypothesis HA:μ1<μ2. a) Calculate the test statistic for the Welch Approximate t procedure. Round your response to at least 3 decimal places. b) The Welch-Satterthwaite approximation to the degrees of freedom is given by df = 28.610808. Using this information, determine the range in which the p-value...
Given are five observations for two variables, 7 and y. 2 1 3 4 S M...
Given are five observations for two variables, 7 and y. 2 1 3 4 S M 3 7 5 11 14 The estimated regression equation is y = 0.2 +2.67 a. Compute the mean square arror using the following equation to 3 decimals), - MSE SSE - 2 b. Compute the standard error of the estimate using the following equation (to 3 decimals). 5 VMSE SSE N-2 c. Compute the estimated standard deviation by using the following equation (to 3...
Pick 3 correct inferences about the following hypothesis test to determine whether the mean thickness of...
Pick 3 correct inferences about the following hypothesis test to determine whether the mean thickness of aluminum sheets equals 0.05. Assume a 5% level of significance was used. A. The null hypothesis is mean 0.05". B. This is a one-tailed test. C. The chance of a Type 1 error is 1.4%. D, 68% of sheets should be +-0.01" of Hypothesis Test (One-Sample) Sample Size Sample Mean Sample Std Dev Hypothesized Mean Data Set #1 100 0.04739 0.01043 0.05 0.05 0.00104...
4. An exponential random variable X has p.d.f (x,07/», x>0, c.df. F(x:0) 1-exp(-x/8) for x >...
4. An exponential random variable X has p.d.f (x,07/», x>0, c.df. F(x:0) 1-exp(-x/8) for x > 0, and mean θ. A single observation of an exponential random variable X is used to test H0 : θ-2 against H1 : θ-5. The null hypothesis is accepted if and only if the observed value of the random variable is less than 3. (a) What is the probability of committing a Type I error? (b) What is the probability of committing a Type...
Determine (a) the test statistic, (b) the degrees of freedom, (c) the P-alue, and (d) test the hypothesis at the a-0.05 level of significance. Outcome Obsenved Expected 25 25 25 25 A B C Do 22 28 24 26 H,: At least one of the proportions is different from the others Click the icon to wew the chi-square distribution table Round to two decimal places as needed.) (b) (True False) The number of degrees of freedom is oneless than the...
01 2 3 4 Delermine (a) the y2 lest atalistic () Uhe dearees of freedom, fe) the P-value, and (d) test e hyoothesis at the -005 Jeve srifcarce Expected 6 25.3 162 14s 40S e random vanable X is bnemial wth n 4, p Ho. cick the lan to vlew e ch-scuero dstdbuticn aNo IRound to two decimal places as needed Thera andagraas of freedom. The Pvalua is What is the conclusion of the test at the 0.05 level of...
Given are five observations for two variables, 7 and y. 2 1 3 4 S M 3 7 5 11 14 The estimated regression equation is y = 0.2 +2.67 a. Compute the mean square arror using the following equation to 3 decimals), - MSE SSE - 2 b. Compute the standard error of the estimate using the following equation (to 3 decimals). 5 VMSE SSE N-2 c. Compute the estimated standard deviation by using the following equation (to 3...
Pick 3 correct inferences about the following hypothesis test to determine whether the mean thickness of aluminum sheets equals 0.05. Assume a 5% level of significance was used. A. The null hypothesis is mean 0.05". B. This is a one-tailed test. C. The chance of a Type 1 error is 1.4%. D, 68% of sheets should be +-0.01" of Hypothesis Test (One-Sample) Sample Size Sample Mean Sample Std Dev Hypothesized Mean Data Set #1 100 0.04739 0.01043 0.05 0.05 0.00104...
4. An exponential random variable X has p.d.f (x,07/», x>0, c.df. F(x:0) 1-exp(-x/8) for x > 0, and mean θ. A single observation of an exponential random variable X is used to test H0 : θ-2 against H1 : θ-5. The null hypothesis is accepted if and only if the observed value of the random variable is less than 3. (a) What is the probability of committing a Type I error? (b) What is the probability of committing a Type...
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