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Week 3 Electronic Text MicroSoft Word Solutions (10.32+10.40+10.42+10.58+11.22) - All work shown + megastat output

by: smartwriter Notetaker

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Week 3 Electronic Text MicroSoft Word Solutions (10.32+10.40+10.42+10.58+11.22) - All work shown + megastat output

Marketplace > Week 3 Electronic Text MicroSoft Word Solutions 10 32 10 40 10 42 10 58 11 22 All work shown megastat output
smartwriter Notetaker
CSU - Dominguez hills
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Week 3 Electronic Text MicroSoft Word Solutions (10.32+10.40+10.42+10.58+11.22) - All work shown + megastat output
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Date Created: 11/15/15
H0: The Fortune 100 companies do not have a greater proportion of women members  than the Fortune 500 companies   p1=p2∨p1−p2=0 Ha: The Fortune 100 companies have a greater proportion of women members than the  Fortune 500 companies  p1≠p2∨p1−p2≠0 This is the formula to use: So, let’s calculate the necessary things. 202 p = =0.169 1 1195 779 p2= 5727 =0.136 n =1195 1 x =202 1 n2=5727 x2=779 x1+x 2 202+779 981 pooledestimate∨p−dash= = = ≈0.142 n1+n 2 1195+5727 6922 The decision rules are as follows: ±z 0.05z 0.0251.96 2   Reject 0 if −1.96≥ z∨z≥1.96 RejectH if −1.96<z<1.96 a Calculation of the test stat: p1−p2 z= pdash(1−pdash) 1+ 1 √ (n1 n2 0.169−0.136 z= 0.142(1−0.142) 1 + 1 √ (1195 5727 ) 0.033 z=√0.142∗0.858∗0.0010114316 0.033 z= 0.0111008458 z=2.972746455 z≈2.973 Let’s apply the decision rule. Reject 0 if −1.96≥2.973∨2.973≥1.96 Reject a if −1.96<2.973<1.96 Looking at these, it can be said that we can clearly reject the null hypothesis. Conclusion: At a = 0.05, It appears that the Fortune 100 companies have a greater  proportion of women members than the Fortune 500 companies. Some factors are that women are getting more educated to a higher level, and as they see  women are working in fortune companies, they also want to work there. Here we can conduct a two sample hypothesis test of proportions. The hypotheses are: H 0 p1=p 2p1−p2=0 H a p1>p2∨p 1p >02 This is the formula to use: So, let’s calculate the necessary things. 18 p1= =0.3 60 32 p2= =0.4 80 n =60 1 x1=18 n2=80 x2=32 x +x pooledestimate∨p−dash= 1 2= 18+32 = 50 = 5 ≈0.357 n1+n2 60+80 140 14 The decision rules are as follows: z =z =1.645 0.1 0.05 2 Reject H0if z≥1.645 RejectH af z<1.645 Let’s calculate the test statistic. p1−p 2 z= 1 1 pdash (−pdash )(n)+ n √ 1 2 0.3−0.4 z= 1 1 0.35(1−0.357( + ) √ 60 80 z= −0.1 √0.357∗0.643∗0.013 −0.1 z= √0.000058 −0.1 z= 0.0076 z≈13.1578 Now let’s apply the decision rule. RejectH0if 13.1578≥1.645 RejectHaif 13.1578≮≮1.645 We can clearly reject the null hypothesis, and conclude that kids with PS3s play more  than kids with Xboxs.  Group 1 90% Confidence interval ­ proportion 90% confidence level 0.5 proportion 24 n 1.64 5 z 0.16 8 half­width 0.66 8 upper confidence limit 0.33 2 lower confidence limit 95% Confidence interval ­  proportion 95% confidence level 0.5 proportion 24 n 1.96 0 z 0.20 0 half­width 0.70 0 upper confidence limit 0.30 0 lower confidence limit 99% Confidence interval ­  proportion 99% confidence level 0.5 proportion 24 n 2.57 6 z 0.26 3 half­width 0.76 3 upper confidence limit 0.23 7 lower confidence limit Group 2 90% Confidence interval ­ proportion 90% confidence level 0.125 proportion 24 n 1.645 z 0.111 half­width 0.236 upper confidence limit 0.014 lower confidence limit 95% Confidence interval ­ proportion 95% confidence level 0.125 proportion 24 n 1.960 z 0.132 half­width 0.257 upper confidence limit ­0.007 lower confidence limit 99% Confidence interval ­ proportion 99% confidence level 0.125 proportion 24 n 2.576 z 0.174 half­width 0.299 upper confidence limit ­0.049 lower confidence limit Group 3 90% Confidence interval ­ proportion 90% confidence level 0.04166666 proportion 48 n 1.645 z 0.047 half­width 0.089 upper confidence limit ­0.006 lower confidence limit 95% Confidence interval ­ proportion 95% confidence level 0.04166666 proportion 48 n 1.960 z 0.057 half­width 0.098 upper confidence limit ­0.015 lower confidence limit 99% Confidence interval ­ proportion 99% confidence level 0.04166666 proportion 48 n 2.576 z 0.074 half­width 0.116 upper confidence limit lower confidence limit ­0.033 Some do overlap, and the meaning of this is that true population proportions may be the  same, and that only happens when the sample is greater. So, it shows the greater the  sample size, the more overlap there will be. The hypotheses for these tests are: H : p =p 0 1 2 H a p 1 p 2 The decision rules are: alpha=0.05 ±z 0.05±z 0.0251.96 2 We reject the null if: z≥1.96∨z≤−1.96 We reject the alternative if: −1.96<z<1.96 Hypothesis test for two independent proportions p1 p2 pc  p (as  0.5 0.125 0.4583 decimal)  p (as    12/24 0      12/27 fraction) 12. 0.375 12.375  X 24 3 27  n 0.375  difference 0.  hypothesized difference 0.3051  std. error 1.23  z .2191  p­value (two­tailed) ###### confidence interval 95.% lower 0.7993 confidence interval 95.% upper 0.4243    half­width From doing his test, we can say that we can’t reject the null (0.2191>alpha level) and  accept the null that the proportions are the same. The hypothesis test was better since it  gave the truth with sufficient evidence using the p value. Normality is ensured due to the  sample size and the rules for normalcy.  The test statistic qualifies for the second statement −1.96<1.23<1.96 Meaning we do not reject the null hypothesis, and this is now proven in two ways. H :μ =μ o 1 2 There is no significant difference between the hat sales in the two regions H aμ ≠1μ 2 There is a significant difference between the hat sales in the two regions We are doing a two tailed test with degrees of freedom: n1+n −2=5+5−2=10−2=8 Alpha=0.05 t0.025,82.3060 We reject the null if: t≤−2.3060∨t ≥2.3060 We reject the alternative if: −2.3060<t<2.3060 Let’s calculate the test statistic. Hypothesis Test: Independent Groups (t­test, pooled  variance) Group 1 Group 2 529.8 0  596.20  mean std.  68.75  102.77  dev. 5 5 n 8   df difference (Group 1 ­ Group  ­66.400  2) 7,644.20 0  pooled variance 87.431  pooled std. dev. 55.296  standard error of difference 0 hypothesized difference ­1.20  t  p­value (two­ .2642 tailed) confidence interval 95.%  ­193.913  lower confidence interval 95.%  61.113  upper 127.513     half­width F­test for equality of variance 10,561.7 0  variance: Group 2 4,726.70  variance: Group 1 2.23 F .4553 p­value Let’s apply the decision rule. We reject the null if: −1.2≤−2.3060∨−1.2≥2.3060 We reject the alternative if: −2.3060<−1.2<2.3060 Since the second statement is true, we do not reject the null hypothesis. There is no  sufficient statistical evidence of a significant difference between the hat sales in the two  regions. H 0μ 1μ 2μ 3 H :Atleast oneof these meansarenotequal¿eachother a Reject the null if the p­value is less than 0.05.  278 205 240 260 270 258 265 220 233 245 240 256 258 255 233 217 242 266 244 239 249 240 228 One factor ANOVA Std.   Mean n Dev   Group 245.2608696 261.2  5 11.95  1 Group 245.2608696 238.0  10 21.24  2 Group 245.2608696 244.4  8 9.46  3   245.3  23 17.91  Total   ANOVA table          p­ Source SS df MS F value 1,803.7 901.88 Treatment 6  2 0  3.43 .0523 5,256.6 262.83 Error 8  20 4  7,060.4 Total 3  22       Comparison of Groups 290.0  280.0  270.0  260.0  250.0  240.0  230.0  220.0  210.0  200.Group 1 Group 2 Group 3 At the 0.05 significance level, we can’t reject the null hypothesis since the p­value is  greater than the significance level, so we can say that the means are the same.

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