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Week 2 Electronic Text MicroSoft Word Solutions (9.52+9.58+9.60+9.66) - All work shown + megastat output

by: smartwriter Notetaker

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Week 2 Electronic Text MicroSoft Word Solutions (9.52+9.58+9.60+9.66) - All work shown + megastat output

Marketplace > Week 2 Electronic Text MicroSoft Word Solutions 9 52 9 58 9 60 9 66 All work shown megastat output
smartwriter Notetaker
CSU - Dominguez hills
GPA 3.0

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Week 2 Electronic Text MicroSoft Word Solutions (9.52+9.58+9.60+9.66) - All work shown + megastat output
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Date Created: 11/15/15
A.)We know the distribution is normally distributed. We have to look at the sample  size to decide to use the normal distribution, or the t distribution. If the sample  size is greater than or equal to 30, we use the normal. If it’s less than 30 and more  than 0, then we use the t distribution. Here are the rules in numerical format: n≥30−usethenormal distribution n<30−usethetdistribution As we see, the sample size is 10. If we apply the rule: ' 10≱30−can tusethenormaldistribution 10<30−usethet distribution Since the sample size is less than 30, we use the t distribution.  B.) Since the hypothesis has a two­tailed decision rule, we have a two­tailed  hypothesis. H 0μ=520(Thetruemeanis520ml) Thetruemeanisdifferent H aμ≠520(¿520ml) We are using the t distribution to figure out the decision rule, so we need to  calculate the degrees of freedom. D.F .=n−1=10−1=9 α=0.05 The critical value  ±t α ,9±t 0.025,92.262  (according to technology/t distribution  2 table) The rejection decisions are: Reject 0 if t≥2.262∨t≤−2.262 RejectH if −2.262<t<2.262 a C.) Let’s calculate the test statistic. μ=520 x=515 σ=4 n=16 x−μ t= σ √ n 515−520 t= 4 √16 −5 t= 4 4 t=−5 Let’s apply the decision rule: Reject H if −5≥2.262∨−5≤−2.262 0 RejectHaif −2.262≮−5≮2.262 Since the decision rule for reject the null hypothesis works (­5 is less than ­2.262), we  reject the null hypothesis. Since we reject the null hypothesis, this data does contradict  that the true mean is 520. Hypothesis Test: Mean vs. Hypothesized Value 520.0 hypothesized  0  value 515.0 0  mean Data std.  4.00  dev. std.  1.00  error Excel Output 16   n 15   df ­5.00  t  p­value (two­ .0002 tailed) 512.8 confidence interval 95.%  7  lower 517.1 confidence interval 95.%  3  upper 2.13     half­width A.)The hypotheses are: Thetrue population proportionisequal H 0π=0.25(¿0.25) H :a>0.25(Thetrue population proportionis morethan0.25) The decision rule is:  z =1.645 0.05 Reject 0 if z≥1.645 RejectHaif z<1.645 Let’s calculate the test statistic. 31 p−¿ =0.31 100 n=100 π=0.25 0.31−0.25 0.06 0.06 Test stat=¿ = = 0.0462493243≈1.2973 0.3( 1−0.3) 0.2139 √ 100 √ 100 We can apply the decision rule here. 1.2973≯1.645 1.2973<1.645 We don’t reject the null hypothesis since the test statistic is not greater than the critical  value and does not fall in the rejection region as shown above. We don't have sufficient  evidence to conclude that the percentage of single­earner or individual mortgages has  risen. B.) Calculate the p­value. P(z>1.2973)=0.0973 This is not a close decision since the p­value 0.0973 is almost twice as large as the  significance level 0.05. C.) Normalcy assumption required p−¿n=0.31∗100=31 31>5 ^ 1−p−¿ ¿ ¿ 69>5 Both values are greater than 5, so the normalcy assumption is content.  Ages of Boston Red Sox Pitchers, October 2005 Arroyo 28 Foulke 33 Mantei 32 Timlin 39 Clement 31 Gonzalez 30 Miller 29 Wakefield 39 Embree 35 Halama 33 Myers 36 Wells 42 The hypotheses are: Thetruemeanageislessthan∨equal H 0u≤30(¿30 years) H au>30(Thetruemeanageismorethan30years) Decision rule: If the sample size is greater than or equal to 30, we use the normal. If it’s less than 30 and more than 0, then we use the t distribution. Here are the rules in numerical  format: n≥30−usethenormaldistribution n<30−usethetdistribution As we see, the sample size is 10. If we apply the rule: 11≱30−can tusethenormaldistribution 11<30−usethetdistribution D.F .=n−1=12−1=11 t0.05,11796 Reject H 0f z≥1.796 RejectH if z<1.796 a Calculate the test statistic. μ=30 x≈33.917 σ≈4.379 n=12 x−μ t= σ √n 33.917−30 t= 4.379 √12 3.917 t= 4.379 3.464101615 3.917 t= 1.264108414 t=3.098626634 t≈3.0986 Let’s apply the rule: RejectH0if z≥1.812 RejectHaif z<1.812 3.0986≥1.796 3.0968≮≮1.796 We reject the null hypothesis, and it appears the true mean is greater than 30 years. The  difference is significant since the test statistic is greater than 1.796 by a lot.  The p­value is: P(z>3.0986)=0.0051 This shows the probability can be as extreme as the test statistic (0.0051 is very much  smaller than 0.05) Hypothesis Test: Mean vs. Hypothesized  Value 30.000  hypothesized value 33.917  mean Data 4.379  std. dev. Excel Output 1.264  std. error 12   n 11   df 3.10  t .0051  p­value (one­tailed, upper) 31.134  confidence interval 95.% lower 36.699  confidence interval 95.% upper 2.782     half­width The hypotheses are: Thetruemeanrepairtimeislessthan5∨equal  – Goal H 0u≤5(¿5days) H au>5(Thetruemeanrepairtimeisgreaterthan5days) Decision rule: If the sample size is greater than or equal to 30, we use the normal. If it’s less than 30 and more than 0, then we use the t distribution. Here are the rules in numerical  format: n≥30−usethenormaldistribution n<30−usethetdistribution As we see, the sample size is 10. If we apply the rule: ' 12≱30−can tusethenormaldistribution 12<30−usethet distribution D.F .=12−1=12−1=11 0.05,11796 RejectH 0f z≥1.796 RejectH af z<1.796 Calculate the test statistic: u=5 x≈5.083 σ≈2.999 n=12 x−μ t= σ √n 5.083−5 t= 2.999 √12 t≈ 0.083 2.999 3.464101615 t≈ 0.083 0.8657367286 t≈0.0958721021 t≈0.0959 Let’s apply the decision rules: RejectH 0f z≥1.796 RejectH if z<1.796 a 0.0959≱1.796 0.0959<1.796 As it seems, we do not reject the null hypothesis, but reject the alternative. Since we fail  to reject the null, the goal is being met.  We can also conclude this using the p­value. P (z>0.0979 =0.4619 Since the p­value is greater than the alpha level, we do not reject the null hypothesis. 0.4619>0.05 Hypothesis Test: Mean vs. Hypothesized Value hypothesized  5.000 value 5.083 mean Data std.  2.999 dev. Excel Output std.  0.866 error 12   n 11   df 0.10  t .4625  p­value (one­tailed, upper) confidence interval 95.%  3.178 lower confidence interval 95.%  6.989 upper 1.905    half­width

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