# Statistics, Course Project Part B

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Keller Graduate School of Management Project Part B: Answer a. In this part of the project I am going to test the manager’s claim that the average mean annual income was less than \$50,000. The hypothesis testing is done through the following steps: The null hypothesis H0 is that the mean annual income μ is equal to \$50,000. H0:μ = \$50,000. The alternative hypothesis H1 is that the mean annual income μ is less than \$50,000. H1:μ &lt; \$50,000 Significance level chosen is 5% or α = 0.05 Here, the population standard deviation is unknown. Hence, we use a t statistic Therefore the test statistic used is t = X-μS/n follows a t distribution with n-1 degrees of freedom From the t table corresponding to 0.05 probability, the critical value tα =1.6766. Hence the critical region is t &lt; -1.6766. Alternatively, we reject the null hypothesis, if the p value is less than the significance level Substituting the value we get t = 43.74-5014.6396/50 = -3.02 The p value corresponding to t = -3.02 and 49 d.f. is 0.002 which is smaller than the significance level. The value of the test statistic is in the critical region and hence it is significant. Therefore, we reject the null hypothesis at 5% level of significance. The p-value is 0.002 which is smaller than the significance level. Hence there are 0.002 chances that such a small value can observe when the null hypothesis is true. Since, this probability is smaller than the significance level we reject the null hypothesis at 5% level. Hence, we conclude that there is enough evidence to support the manager’s claim that the average annual income was less than \$50,000. b. The claim is that the true population proportion of customers who live in an urban area exceeds 40%. In the given sample of size 50, thirteen are from the ‘Urban’. The null hypothesis H0