Answered: Why is the null hypothesis for a…

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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Let X equal the number of female children in a three-child family. We shall use a chi-square goodness-of-fit statistic to test the null hypothesis that the distribution of X is b(3, 0.5) (a) Define the test statistic and critical region, using an a = 0.05 significance level. (b) Among students who were taking statistics, 52 came from families with three children. For these families, x = 0, 1, 2, and 3 for 5, 17, 24, and 6 families, respectively. Calculate the value of the test statistic and state your conclusion, considering how the sample was selected.
A test of H₁: μ=7 versus H₁: u<7 is performed using a significance level of a=0.01. The P-value is 0.20. Part: 0/3 Part 1 of 3 (a) Is H rejected? Since P ≤ Part: 1 / 3 Part 2 of 3 (b) If the true value of u is 1, is the result a Type I error, a Type II error, or a correct decision? a, we do not reject Ho at the a= 0.01 level. The result is a Type I error Part: 2/3 Part 3 of 3 X 3 (c) If the true value of u is 7, is the result a Type I error, a Type II error, or a correct decision? The result is a [Choose one) ▾ Español
The mean ±1 sd of ln [calcium intake (mg)] among 25 females, 12 to 14 years of age, below the poverty level is 6.56 ± 0.64. Similarly, the mean ± 1 sd of ln [calcium intake (mg)] among 40 females, 12 to 14 years of age, above the poverty level is 6.80 ± 0.76. 8.2 Test for a significant difference between the variances of the two groups. Find the P value. Do not use excell. Solve the step by step
A repeated-measures study with a sample of n = 16 participants produces a mean difference of MD = 4 with a variance s2 = 64. Use a two-tailed hypothesis test with significance level = .05 to determine whether it is likely that this sample came from a population with MD = 0. State your hypotheses state the critical t-value calculate the t-value for the sample mean difference state your conclusion about the null hypothesis Show all calculations.
Which of the following is a Null hypothesis for a paired-samples t-test? μd = 0 μd = X μ1 ≠ μ2 μ1 = μ2
0.6.3 Question 3 Is the cubic effect significant? How about quadratic and linear effects? Analysis of Variance Table Response: EX Df Sum Sq Mean Sq F value Pr(>F) MET 1 332 332 0.1246 0.7257742 I(MET^2) 1 37504 37504 14.0572 0.0005141 *** I(MET^3) 1 7245 7245 2.7154 0.1065086 Residuals 44 117390 2668 > anova(quadModel) Analysis of Variance Table Response: EX Df Sum Sq Mean Sq F value Pr(>F) MET 1 332 332 0.120 0.7306134 I(MET^2) 1 37504 37504 13.541 0.0006216 *** Residuals 45 124635 2770 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. Use μ1 for the mean number of families with no children. At α=0.10, is there a difference between the means? Use the P-value method and tables. (a) State the hypothesis and identify the claim with the correct hypothesis. H0: ▼(Choose one) H1: ▼(Choose one) This hypothesis test is a ▼(Choose one) test. (b)Find the critical value(s). Round the answer to 3decimal places, if necessary. If there ismore than one critical value, Critical values: (c)Compute the test value. Round the answer to at least 3 decimal places, if necessary. z= Compute the P -value. Round the answer to four decimal places. P-value= (d)Make the decision. What's the null hypothesis? (e)Summarize the…
What is the name of this hypothesis test? Do male and female servers at Applebee's work the same number of hours? A random sample of 45 female servers worked an average of 36 hours per week, with a standard deviation of 2. A random sample of 62 male servers worked an average of 29.5 hours per week, with a standard deviation of 4. O Chi-Square Test for Independence O One Sample z Test OTwo Proportion z Test Z. OLinear Regression t Test O Chi-Square Goodness of Fit Test O Paired Samples t Test OTwo Independent Samples t Test O One Sample t Test O One Proportion z Test OANOVA
In order to compare the means of two normal populations that are known to have equal variances, independent random samples are taken of sizes n1 = 10 and n2 data yield: = 15. The results from the sample Sample 1 Sample 2 sample mean = 52 sample mean = 45 S1 5 S2 = 3 Sp 3.9065 %3D To test the null hypothesis Ho: H1 - H2=0 versus the alternative hypothesis H,: µ1 -µ2> 0 at the 0.01 level of significance, the most accurate statement is O The value of the test statistic is -8.48 and the critical value is -2.5 O The value of the test statistic is 4.39 and the critical value is 1.96 O The value of the test statistic is -4.39 and the critical value is -1.96 The value of the test statistic is 3.79 and the critical value is 2.326 The value of the test statistic is 8.48 and the critical value is +2,50

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Why is the null hypothesis for a dependent-samples t- test always μD=0 μD=0 ?