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- (^) Please use a table to provide a concept mapping that appropriately categorizes and summarizes all key concepts of Chapter 10.
Inference About Means and Proportions with Two Populations
Inferences About the Difference Between Two Population Means Based on Independent Samples: Sigma1 and Sigma2 Known
- μ1 = mean of population 1
- μ2 = mean of population 2
- Sigma1 and Sigma2: two population standard deviations
- n1 and n2: sample size of a simple random sample selected from population 1 and 2, respectively (independent simple random samples).
- Interval Estimation of μ1 - μ 2
- Point estimate of the difference between two population means: X1−X
- Standard error of X1−X2: Sigma(X1−X2)= pow((Sigma12)/n1+ ( Sigma22/n2),0.5)
- (^) Interval estimate: X1−X2 +- margin of error
- X1−X2 +- z(alpha/2)* pow((Sigma12)/n1+ ( Sigma22/n2),0.5)
- Z= (X1−X2-D0) / pow((Sigma12)/n1+ ( Sigma22/n2),0.5). In many cases, D0 =
Inferences About the Difference Between Two Population Means: Sigma and Sigma
- Interval Estimation of μ1 - μ 2
- Point estimate of the difference between two population means: X1−X
- df = pow((Sigma12)/n1+ ( Sigma22/n2),2)/ ((Sigma12)/n1)2(1/ n1-1) + (Sigma22)/n2)2(1/n2-1))
- (^) Standard error of X1−X2:
- Interval estimate: X1−X2 +- t(alpha/2)* pow((Sigma12)/n1+ ( Sigma22/
Unknown n2),0.5)
- t= (X1−X2-D0) / pow((Sigma12)/n1+ ( Sigma22/n2),0.5). In many cases, D0 =
Inferences About the Difference Between Two Population Proportions
- p1: proportion for population 1
- p2: proportion for population 2
- p1bar: sample proportion for a simple random sample from population 1
- p2bar: sample proportion for a simple random sample from population 2
- Interval Estimation of p1-p
- Point estimator of the difference between two population proportion :p1bar-p2bar
- (^) Standard error of p1bar-p2bar :pow((p1(1-p1))/n1 + p2(1-p2))/n2 ),0.5)
- Margin of error = z(alpha/2)* :pow((p1(1-p1))/n1 + p2(1-p2))/n2 ),0.5)
- Interval estimate: p1bar-p2bar+- z(alpha/2)* :pow((p1(1-p1))/n1 + p2(1-p2))/n2 ),0.5)
- Please use a table to provide a concept mapping that appropriately categorizes and summarizes all key concepts of Chapter 11.
Inference About Population Variances
Sampling Distribution of (n-1)s2/ Sigma
- The sample variance
- s2=(summation(X1-Xbar)2)/n-
- Whenever a simple random sample of size n is selected from a normal population, the sampling distribution of (n-1)s2/Sigma*2 has a chi-square distribution with n-1 degrees of freedom
- Interval Estimation:
- (n-1)s2/chi(alpha/2)2< =Sigma2<=(n-1)s2/chi(1-alpha/2)
90% confidence interval estimate of the population standard deviation: [CL_Lower0.5, CL_Upper0.5]=[0.47,0.84]
Hypothesis testing for H0: variance>=0.5, Ha: H0: variance<0. P_value=chi2.cdf((n-1)* sample_var/0.5,n-1) Print(P_value) P_value= 0. As P_value >0.05 so we accept null hypothesis and therefore we conclude that the population variance for the weight of this drug is greater than or equal to 0.
a. Investors commonly use the standard deviation of the monthly percentage return for a mutual fund as a measure of the risk for the fund; in such cases, a fund that has a larger standard deviation is considered more risky than a fund with a lower standard deviation. The standard deviation for the American Century Equity Growth fund and the standard deviation for the Fidelity Growth Discovery fund were recently reported to be 15.0% and 18.9%, respectively. Assume that each of these standard deviations is based on a sample of 60 months of returns. Do the sample results support the conclusion that the Fidelity fund has a larger population variance than the American Century fund? Which fund is more risky?
s1=0.15;s2=0.189;n1=60;n2=60;alpha=0.
H0: Sigma12>=Sigma2 Ha: Sigma12<Sigma2
F= s12/s2 print(F) F= 0.
P_value= f.cdf(F,n1-1,n2-1) print(P_value) P_value= 0.
For 95% confidence: As P_value<0.05 so we reject null hypothesis and we conclude that the standard deviation for the American Century Equity Growth fund is less than the standard deviation for the Fidelity Growth Discovery fund so American Century Equity Growth fund is less risky
For 99% confidence: As P_value>0.01 so we accept null hypothesis and we conclude that the standard deviation for the American Century Equity Growth fund is more than the standard deviation for the Fidelity Growth Discovery fund so American Century Equity Growth fund is more risky
- (^) Par, Inc., is a major manufacturer of golf equipment. Management believes that Par’s market share could be increased with the introduction of a cut-resistant, longer-lasting golf ball. There- fore, the research group at Par has been investigating a new golf ball coating designed to resist cuts and provide a more durable ball. The tests with the coating have been promising. One of the researchers voiced concern about the effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances comparable to those of the current-model golf ball. To compare the driving distances for the two balls, 40 balls of both the new and current models were subjected to distance tests. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models. The results of the tests, with distances measured to the nearest yard, follow. These data are below. Complete the following.2 02 8 1.a. Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls.
Variance= 97. Count=
a. What is the 95% confidence interval for the population mean driving distance of each model, and what is the 95% confidence interval for the difference between the means of the two populations?
For current:
Since there 20 data values in the sample so degree of freedom is df=20-1= mean=270. standard deviation=8. t_critical = t.ppf(1-0.05/2,19)= 2. 95% confidence interval for current model: [270.28-2.098.76/pow(40,0.5),270.28+2.098.76/pow(40,0.5)]= [267.38, 273.17]
For new: Since there 20 data values in the sample so degree of freedom is df=20-1= mean=267. Standard deviation=9. t_critical = t.ppf(1-0.05/2,19)= 2. 95% confidence interval for new model: [267.50-2.099.9/pow(40,0.5),267.50+2.099.9/pow(40,0.5)]= [264.22, 270.77]
def se_for_two_means(s1,s2,n1,n2): return pow(s1s1/n1 +s2s2/n2,0.5) x1=270.28;s1=8.76;n1=40;x2=267.50;s2=9.90;n2=
SE=se_for_two_means(s1,s2,n1,n2) df= t=t.ppf(1-alpha/2,df)
MOE=t*SE
So required confidence interval is:- [x1-x2-MOE,x1-x2+MOE]= [-1.38, 6.94]
a. Do you see a need for larger sample sizes and more testing with the golf balls? Discuss
Yes, we need for larger sample sizes and more testing with the golf balls
