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2. A sample of n = 16 scores has a mean of M = 83 and a standard deviation of s = 12. A. Explain what is measured by the sample standard deviation. b. Compute the estimated standard error for the sample mean and explain what is measured by the standard error. Sample standard deviation is an estimate of the population standard deviation. The observations in the sample is concentrated about a central value. This phenomenon is called central tendency. The central value is usually estimated by mean, median, mode etc.Another property of a sample of observations is the opposite of central tendency and is called dispersion. It is the tendency of the observations to spread away from the central value. The important measures of dispersion are the range, the standard deviation, the mean deviation, and the standard deviation. The most popular measure of dispersion is the standard deviation.Sample standard deviation measures the dispersion or spread of the observations from the central value of the sample. Population standard deviation measures the dispersion or spread of the observations from the central value of the population. The population standard deviation is usually estimated by the sample standard deviation.The standard error of the sample mean is defined as the standard deviation of the sampling distribution of the sample mean and is equal to σn, where σ is the population standard deviation and n is the sample size. The estimated standard error of the sample mean is sn, where s is the sample standard deviation.Estimated standard error of the sample mean = sn=1216=124=38.. The following sample was obtained from a population with unknown parameters. Scores: 6, 12, 0, 3, 4 a. Compute the sample mean and standard deviation. (Note that these are descriptive values that summarize the sample data). b. Compute the estimated standard error for M. (Note that this is an inferential value that describes how accurately the sample mean represents the unknown population mean). Sample mean = 6+12+0+3+45=255=5Sample variance =s2= 6-52+12-52+0-52+3-52+4-525=1+49+25+4+15=805=16Sample standard deviation = s=s2=16=4Estimated standard error of sample mean = sn=45=1.788910. To evaluate the effect of a treatment, a sample of n = 9 individuals is obtained from a population with a mean of u = 40, and a treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 33. a. If the sample standard deviation is s = 9, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with x = .05? b. If the sample standard deviation is s= 15, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with x = .05? Part an=9 μ=40 M=33 s=9 α=0.05 Compute student’s t statistic.t=nM-μs=933-409=-2.33 Critical value of t with n-1=9-1=8 degrees of freedom is 2.31t=-2.33=2.33>2.31=t(crtical) The test statistic is significant.The data is sufficient to conclude that the treatment has a significant effect. Part bn=9 μ=40 M=33 s=15 α=0.05 Compute student’s t statistic.t=nM-μs=933-4015=-1.4 Critical value of t with n-1=9-1=8 degrees of freedom is 2.31t=-1.4=1.4<2.31=t(crtical) The test statistic is not significant.The data is not sufficient to conclude that the treatment has a significant effect.Chapter 10: 2. Describe what is measured by the estimated standard error in the bottom of the independent-measure t statistic. The estimated standard error of the sample mean s/n is at the bottom of the t statistic given by t=M-μs/n.s/n is the standard deviation of the sampling distribution of the sample mean M.s is the sample standard deviation.6. One sample has SS = 48 and a second has SS = 32. a. Assuming that n= 5 for both samples, calculate each of the sample variances, then calculate the pooled variance. Because the samples are the same size, you should find that the pooled variance is exactly halfway between the two sample variances. b. Now assume that n = 5 for the first sample and n = 16 for the second. Again, calculate the two sample variances and the pooled variance. You should find that the pooled variance is closer to the variance for the larger sample. Part aSS1 = 48SS2 = 32n1 = 5n2 = 5s12=SS1n1-1=485-1=484=12 s22=SS2n2-1=325-1=324=8 s2=n1-1s12+n2-1s22n1+n2-2=4×12+4×88=808=10Verify thats2=12s12+12s22 10=1212+128 10 = 6 +4Part bSS1 = 48SS2 = 32n1 = 5n2 =16s12=SS1n1-1=485-1=484=12 s22=SS2n2-1=3216-1=3215=2.1333 s2=n1-1s12+n2-1s22n1+n2-2=4×12+15×2.133319=8019=4.2105Verify thats2=4.2105 is nearer to s22=2.1333 than to s12=12.10. Two separate samples receive two different treatments. One sample has n = 9 scores with SS = 710 and a second sample has n = 6 scores with SS = 460. a. Calculate the pooled variance for the two samples. b. Calculate the estimated standard error for the sample mean difference. c. If the sample mean difference is 10 points, is this enough to reject the null hypothesis for a two-tailed test with x = .05? d. If the sample mean difference is 13 points, is this enough to reject the null hypothesis for a two-tailed test with x = .05? Part an1 = 9n2 = 6SS1 = 710SS2 = 460Pooled variance = s2=SS1+SS2n1+n2-2=710+4609+6-2=117013=90Part bEstimated standard error for the sample mean difference = s21n1+1n29019+16=903+218=25=5Part c Student’s t =sample mean differencestandard error=105=25% critical value of t with n1+n2-2=9+6-2=13 df is 2.16Computed value of t is < 2.16T is not significant.The information is not sufficient to reject the null hypothesis.Par dt=sample mean differencestandard error=135=2.6>2.16The information is sufficient to reject the null hypothesis. ................
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