The null and alternate hypotheses are:He : Hd soна: на>0The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sampleof 4 days last month.Day2134Day shift.11 12 13 19Afternoon shift 9 10 14 15At the 0.03 significance level, can we conclude there are more defects produced on the day shift? Hint. For the calculations, assumethe day shift as the first sample.Required:a. What is the p-value? NOTE: P-value calculators are available through online search. (Round your answer to 4 decimal places.)p-value

The question is about hypothesis testing.Given :Null and alternative hypothesis :Data :To find :Is…

Answered: The null and alternate hypotheses are:…

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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Below are the SPSS results of a related (dependent) samples t-test. Use the SPSS results to answer Questions 5, 6, and 7. Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 Statistics Posttest 57.1250 40 16.73962 2.64677 Statistics Pretest 49.9250 40 15.09014 2.38596 Paired Samples Test Paired Differences t df Sig. (2-tailed) Mean Std. Deviation Std. Error Mean Pair 1 Statistics Posttest - Statistics Pretest 7.200 12.313 ( a ) ( b ) ( c ) .001 Find the mean of the differences ( and standard deviation of the differences (Sd). Replace “a,” “b,” and “c” in the table above with the correct values. Round to the third decimal place. Do you think there is a statistically significant difference between the pretest and posttest at the .05 level of significance?
Given below are the analysis of variance (ANOVA) results from a Minitab display. Assume that you want to use a 0.01 significance level in testing the null hypothesis that the samples come from populations with the same mean. Source DF SS MS F P-val Factor 2. 2280.0 1140.0 7.9 5.39E–04 Error 144 20691.7 143.7 Total 146 22971.8 What can you conclude about the equality of the population means? (circle one) Reject H0 since the p-value is greater than the significance level. Fail to reject H0 since the p-value is greater than the significance level. Reject H0 since the p-value is less than the significance level. Fail to reject H0 since the p-value is less than the significance level.
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A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is less than 0.5. Assume that sample data consists of 78 girls in 169 births, so the sample statistic of 613 results in a z score that is 1 standard deviation below 0. Complete parts (a) through (h) below.Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING...Question content area bottomPart 1a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.A.H0: p≠0.5H1: p<0.5B.H0: p=0.5H1: p≠0.5C.H0: p=0.5H1: p<0.5D.H0: p=0.5H1: p>0.5Part 2b. What is the value of α?α=enter your response here (Type an integer or a decimal.)Part 3c. What is the sampling distribution of the sample statistic? Normal distribution Student (t) distribution χ2Part 4d. Is the test two-tailed, left-tailed, or right-tailed? Right-tailed…
calculate ss variance and standard deviation for the following population of N=5 scores: 2,13,4,10,6
8.What is the sample variance and the estimated standard error for a sample of n = 9 scores with SS = 72? a. s2= 9 and sM = 3 c. s²= 3 and sM = 3 b. s? = 9 and SM = 1 d. s2 = 3 and sM= 1 %3D
Q.3 For the following data: a) Compute SST, SSTR, SSE b) Construct a one-way ANOVA table c) Decide at the 5% significance level, whether the data provide sufficient evidence to conclude that the means of the populations from which the samples are drawn are not the same. Sample 1 Sample 2 Sample 3 Sample 4 11 16 6 2 10 7 4 10 3. we have: ng = 3 n3= 3 n4 = 3 Ta 24 = 1s Ty= 36 THa9 n z Enj = 12 Exz= IT;= &4 ..... Ex - (Ex:*/n = SSTR = I (Tim;) - (Iri)/n = ()+( )+( )+( ) -- 2. SST = 12 SSE = 5sT - SSTR = Fostatistic df MS 2 df :عاطء Souree SS MSTR/MSE 138 Treatment where: MSE- SSE/(n-k) MSTR = SSTR/(K-1) Enron Tofal df z ( 3, 8) caitical Valme s ? Compare test- statistic to cv. Do you rejuctNull or Do not rejrit what dees it mean ? write here
Given below are the analysis of variance results from a stat software display comparing sample data for the meanmileage for 4 different types of cars. Assume that you want to use a 0.05 significance level in testing the null hypothesisthat the different samples come from cars with the same mileage. What can you conclude about the equality of the population means?Source DF SS MS F pFactor 3 13.500 4.500 5.17 0.011Error 16 13.925 0.870Total 19 27.425 12)What is the null and alternative hypothesis for this scenari?Based on this One Way Anova software analysis, would you reject of fail to reject the nul hypothesis?What does this result mean in context?
Question 16: Please use the approporate stats table attached. Do not use a software for this problem.a. What was the observed difference between the mean PCV levels for the two groups? b.Assuming that the two groups have similar variances, estimate the standard error of the difference between the sample means c. Conduct an appropriate test to assess the difference between the two means. What do you conclude?
Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level). One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p Between-group (treatment) 2 469.1827 2677.864 2.30 0.104 Within-group (error) 118 203.6918 11745.05 Total 120
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