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HYPOTHESIS TESTING AND CONFIDENCE INTERVALSStudentUniversityEXECUTIVE SUMMARYThe manager’s speculation that the average annual income of customers is less than $50,000 was true according to the hypothesis analysis. The p-value measures the strength of the evidence against the null hypothesis. The z test in this speculation has a p-value of 0.999 which is greater than the confidence interval of 0.05. This is an indication that we do not have enough evidence to refute the speculation based on the data provided. However, the hypothesis test on the data indicated that we reject the assumption that the average annual income is greater than $50,000 and concur with his speculations that it is rather less than $50,000. The p-value does not necessarily indicate that the null hypothesis is true hence we would rely more on the z statistics to make conclusions.The second speculation that the population proportion of credit customers who live in urban areas exceed 40% is false according to the hypothesis test. The p-value is 0.418 which I greater than the significance level of 0.05 which indicates that we should not reject the null hypothesis. The z statistics test and the p-value agree with the null hypothesis. Therefore, it will be wrong to say that the population proportion of credit customers who live in urban areas exceed 40%. The hypothesis test shows that it is less than 40%.The third speculation is that an average number of years lived in the current home is less than 13 years. We tested the null hypothesis that the years lived in the current home were more than 13 years against the manager's speculation. The analysis indicated that we reject the null hypothesis, therefore, agreeing with the manager's speculation. The p-value for the data is 0.0001 which is less than the 0.005 indicating that we should also reject the null hypothesis. Therefore, both the z statistics and the p-value indicate that the manager's speculations could be true.The final speculation was that the average credit balance for suburban customers is almost $4,300. We tested the null hypothesis that average credit balance for suburban customers is more than $4,300 against the manager’s speculation. The final results of the test indicated that we should not reject the null hypothesis hence refuting the manager’s speculation. The z statistics showed that the average credit balance for suburban customers is more than $4,300. The p-value for the data is 0.0001 which is smaller than the significance level of 0.05 hence we reject the null hypothesis. The two values provide conflicting evidence but a smaller p-value does not necessarily indicate that the null hypothesis is false.Confidence intervalsConfidence interval (CI) of the mean or proportions is a range with upper and lower numbers that describe possible values that could represent the mean or proportions. The confidence interval of population’s annual average income for the credit consumers is 0.1277 and the sample mean is 44.527. This means that we are 95 percent confident that the population’s annual average income for the credit customers is between 44.399 and 44.655 while we are 5 percent confident that it does not. The confidence interval for the proportion of the population for credit customers is -0.253 and the sample proportion in 40 percent. This means that we are 95 percent confident that the proportion of the population of credit customers is between 39747 and 40.253.The confidence interval for average number of years lived in current home is 0.042 and the sample mean is 12.109. Therefore, we are 95 percent confident that the population’s average number of years lived in current home is between the range of 12.067 and 12.151 and 5 percent confident that it is not between these intervals. The confidence interval for the credit balance for suburban customers is 1.95 and the sample mean is 4738.824. Therefore, we are 95 percent confident that the population’s average credit balance for suburban customers ranges between 4727.824 and 4749.774 and we are 5 percent that it does not range between the two points.APPENDIXThe average annual income of customers is less than $50,000Null hypothesis H0:?≥$50,000. This means that the average annual income of customers is greater than $50,000.Alternative hypothesis HA: ??$50,000. That is to say, the average annual income of customers is less than $50,000. Test statistics – z=X-$50σX=X-$50sn=44.527-$502.036=-2.688Rejection region – We choose α=0.05; this is one tailed test and critical value=1.960. We reject the null hypothesis if z ?1.645.Assumption: The population is not normal, and the variance is unknown. Experiment and calculation of test statistics: Conclusion: The z statistics fall in the rejection region z?1.645 hence at α=0.05; we reject the null hypothesis. X≥$50,000 and concludes that the X?$50,000. This means that average annual income of customers is less than $50,000.The true population proportion of credit customers who live in urban areas exceeds 40%.Null hypothesis H0:p=40%. This means that the population proportion of credit customers who live in urban areas does not exceed 40%Alternative hypothesis HA: p ? 40%. This means that the population proportion of credit customers who live in urban areas exceed 40%Test statistics z=p-40σp=43.64-40126.62=0.345Rejection region - We choose α=0.05; we carry out a test for one proportion and critical value=1.645. We reject the null hypothesis if z ?1.645. Assumption: The sample is selected from a binomial population.Experiment and calculation of the test statisticsConclusionThe z statistics does not fall in the rejection region z ?1.960 hence at α=0.05, we do not reject the null hypothesis. p=40% and concludes that the proportion of the population of credit customers who live in urban areas does not exceed 40%.The average number of years lived in the current home is less than 13 years.Null hypothesis H0:?≥13 years. This means that the average number of years lived in the current home is more than 13 yearsAlternative hypothesis HA: ??13 years. That is to say, the average number of years lived in the current home is less than 13 years. Test statistics – z=X-13σX=X-13sn=12.11-130.676=-1.317Rejection region - We choose α=0.05; this is one-tailed test and critical value=-1.645. We reject the null hypothesis if z ?-1.645.Assumption: The population is not normal, and the variance is unknown.Experiment and calculation of test statistics: Conclusion: The z statistics falls in the rejection region z ? -1.645 hence at α=0.05, we reject the null hypothesis. X≥13 years and concludes that the X? 13years. Therefore, the average number of years lived in the current home is less than 13 years. The average credit balance for suburban customers is almost $4,300.Null hypothesis H0:?≥$4,300. This means that the average credit balance for suburban customers is more than $4,300.Alternative hypothesis HA: ? ? $4,300. That is to say, the credit balance for suburban customers is less than $4,300.Test statistics – z=X-13σX=X-13sn=4738.834-4300174.55=2.514Rejection region - We choose α=0.05; this is one tailed test and critical value=2.210. We reject the null hypothesis if z ?2.210.Assumption: The population is not normal, and the variance is unknown.Experiment and calculation of test statistics: Conclusion: The z statistics does not fall in the rejection region z ? 2.210 hence at α=0.05, we do not reject the null hypothesis. X≥4,300 and concludes that the average credit balance for suburban customers is more than $4,300. ................
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