Graded HW 4 S22 (1) (AutoRecovered)

d. (3pts) Using StatCrunch , conduct a significance test to see if there is evidence to support the research hypothesis that the proportion of the population who voted for Brown is greater than 50%.  Insert an image of the output. One sample proportion summary hypothesis test: p : Proportion of successes H 0 : p = 0.5 H A : p > 0.5 Hypothesis test results: ProportionCountTotalSample Prop. Std. Err. Z-Stat P-value p 2065 3889 0.530984830.00801772273.8645424<0.0001 e. (4pts) The media used the results of this exit poll to predict a win for Brown. Based on what you’ve seen in this exercise , answer the questions: - If only half the population voted for Brown; would it then be surprising that 53.1% of the sampled individuals voted for him? No, it would not be surprising that 53.1% of the sampled individuals voted for him because our hypothesis was that 50% voted for him, so 53.1 is close enough to 50%, meaning that the result from the sampled individuals matches. - Does it make sense that the race was called for Brown?  Explain your thoughts using at least 3 sentences. It does not make sense that the race was called for Brown. This is because this was selected at random so there is no definite number. 2. In HW3, you also investigated the 2014 gubernatorial election in Florida, where, in an exit poll of 2696 voters, 50.5% said they voted for Rick Scott and 49.5% said they voted for Charlie Crist. a. (6pts) By hand , find a 95% confidence interval for the proportion of the population who voted for Scott. Show all of your work, and clearly state:  Your value of ^ π = (1361/2696)= 50.5  Your value of z Z = 1.96  The standard error (rounded to 4 decimal places) .0096  The confidence interval and its interpretation . 505 +/-1.96(.0096) . 505+ 0.019= 0.523 . 505 – 0.019= 0.485 Interpretation: We can be 95% confident that the proportion who voted for Scott lies between .485 and .523

One sample T confidence interval: μ : Mean of variable 99% confidence interval results: Variable Sample Mean Std. Err. DF L. Limit U. Limit IDEAL NUMBER OF CHILDREN 2.49336870.02616128711302.42586772.5608697 Interpretation: Approximately, the ideal number of children is 2 so we can be 99% confident that the true proportion of the population wants a minimum of 2.4 or 2 children and maximum of 2.6 or 3 children. b. (6pts) In StatCrunch, run a significance test for the mean for this same variable, letting μ 0 = ¿ 2, 2.5, and then 3, in each case selecting “ ≠ ” in the alternative hypothesis.  Insert images of the three different output windows  Interpret the P -value (state your conclusion) in each case . One sample T hypothesis test: μ : Mean of variable H 0 : μ = 2 H A : μ ≠ 2 Hypothesis test results: Variable Sample Mean Std. Err. DF T-Stat P-value IDEAL NUMBER OF CHILDREN 2.49336870.026161287113018.858732<0.0001 - Interpretation: The p-value is much less than 1 so we can say that the mean is not accurate, and it can be rejected. Input 2: One sample T hypothesis test: μ : Mean of variable H 0 : μ = 2.5 H A : μ ≠ 2.5 Hypothesis test results: Variable Sample Mean Std. Err. DF T-Stat P-value IDEAL NUMBER OF CHILDREN 2.49336870.0261612871130-0.25347758 0.7999 - Interpretation: Even though the p-value is now closer to 1 than before, the mean is still inaccurate in comparison so it must be rejected. Input 3: One sample T hypothesis test: μ : Mean of variable H 0 : μ = 3 H A : μ ≠ 3 Hypothesis test results: