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Statistical Review Solutions: Normal Distribution, Confidence Intervals, Hypothesis Testin, Study notes of Statistics

Solutions to various statistical review problems, covering topics such as normal distribution, confidence intervals, hypothesis testing, and regression analysis. It includes calculations for finding sampling distributions, confidence intervals, and p-values for given data.

Typology: Study notes

Pre 2010

Uploaded on 08/19/2009

koofers-user-gv0
koofers-user-gv0 🇺🇸

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Download Statistical Review Solutions: Normal Distribution, Confidence Intervals, Hypothesis Testin and more Study notes Statistics in PDF only on Docsity! 1 Solutions to review problems #1 Here are solutions to the …rst set of review problems: 4.77: The sampling distribution of y is normal with mean y = 60 and standard deviation y = = p n = 5= p 16 = 1:25: Since the distribution of y is normal, approximately 95% of the values of y should fall within y  1:96 y which is 60  1.96(1.25), giving the interval (57.55, 62.45). 5.41 a) A 99% con…dence interval for  is given by y t:005;14 s= p n = 31.47  2:977 (5:04)= p 15 = 31.47  3:87, which gives the interval (27.6, 35.34). b) The question is a bit unclear on exactly what a null hypothesis might be, but if we choose H0 :  = 35, then at the = :01 signi…cance level we will fail to reject H0, since the 99% con…dence interval includes  = 35: 11.65 a) Yes, the plotted points seem to follow a line. b) From the printout, byi = 12:51 + 35:83xi . 11.66 a) b2" = 1n2P(yi byi)2 =MSE = 1:069: b) From the printout, s:e:(b 1) = 6:96: c) For this research hypothesis, H0 : 1 = 0 and Ha : 1 > 0 since they are interested in detecting a positive relationship. The p value in the printout for H0 : 1 = 0 is p = :0004, but the printout is for the two-sided alternative hypothesis Ha : 1 6= 0: Thus, to get the p value for our one-sided alternative hypothesis we divide the printed p value by 2, yielding p = :0004=2 = :0002: 11.30 a) The plot looks good, there could be one or more in‡uential points. b) The estimated regression equation is byi = 99:78 + 51:92xi, and the residual standard deviation isb" = pMSE = p148:999 = 12:21: c) A 95% con…dence interval for 1 is given by b 1  t:025;28 s:e:(b 1); yielding 51.92  2:048 (:586) or 51.9  1:2, giving an interval of (50.7, 53.1). 11.31 a and b) From the printout, t = 88:53; with a p value of p < :0001: 11.32 a) F = 7837:26; and p < :0001: b) They are equal, because both tests are testing the same null hypothesis when we do simple linear regression. The test statistics are related by t2 = F: 1
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