**Height Advantage in Presidential Candidates**A popular theory suggests that presidential candidates have an advantage if they are taller than their main opponents. Below are the heights (in centimeters) of randomly selected presidents along with the heights of their main opponents. Complete parts (a) and (b) below.| Height (cm) | President | Main Opponent ||------------------|-----------|---------------|| | 180 | 164 || | 170 | 182 || | 182 | 167 || | 175 | 174 || | 199 | 179 || | 177 | 176 |1. **Identify the test statistic.** - **t =** (Round to two decimal places as needed.)2. **Identify the P-value.** - **P-value =** (Round to three decimal places as needed.)3. **What is the conclusion based on the hypothesis test?** - Since the P-value is [dropdown] the significance level, [dropdown] the null hypothesis. There [dropdown] sufficient evidence to support the claim that presidents tend to be taller than their opponents.4. **Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion?** - [Additional details needed here, based on further calculations and input selections.]**Note:** This exercise involves calculating statistical values (test statistic, P-value, confidence interval) to assess whether there is an advantage for taller presidential candidates. **Exploring the Relationship Between Height and Presidential Success**A popular theory suggests that presidential candidates may have an advantage if they are taller than their main opponents. The data below provides heights (in centimeters) of randomly selected presidents alongside the heights of their main opponents. Analyze the parts (a) and (b) to understand better.**Height Data:**- **Height (cm) of President:** 180, 170, 182, 175, 199, 177- **Height (cm) of Main Opponent:** 164, 182, 167, 174, 179, 176---### Part (a)**Statistical Analysis:**Since the P-value is [dropdown: select an option], [dropdown: less than, greater than] the significance level, [dropdown: reject, do not reject] the null hypothesis. There [dropdown: is, is not] sufficient evidence to support the claim that presidents tend to be taller than their opponents.### Part (b)**Confidence Interval Construction:**- The confidence interval is [input box] cm < μd < [input box] cm (Round to one decimal place as needed.)**Conclusion from Confidence Interval:**What feature of the confidence interval leads to the same conclusion reached in part (a)?Since the confidence interval contains [dropdown: zero, only positive values, only negative values], [dropdown: reject, do not reject] the null hypothesis.This exercise helps in understanding the application of hypothesis testing and confidence intervals in real-world data analysis, exploring if height may influence presidential success.

Hypothesis test is the type of statistical test which are of two types. They are parametric and non parametric test. If the sample size is small, then the t test is used and the sample size is larger with known standard deviation, then z test is used. Confidence interval for the paired t test is defined as : d¯±tα,n-1×sdna) Given : President candidate and their main opponent heights are measured. Claim : President taller than their main opponent Let the null and alternate hypothesis : H0 : μd≤0H1 : μd>0 Level of significance is 0.01 Height of president, x Height of main opponent, y d=x-y d-d¯2 180 164 16 84.034 170 182 -12 354.682 182 167 15 66.700 175 174 1 34.024 199 179 20 173.370 177 176 1 34.024 Sum 41 746.833 Average 416=6.833 Standard deviation ∑d-d¯2n-1=746.8336-1=12.222 From the table, Test statistic : t=d¯-0sdn=6.83312.2226=1.37 Test statistic is 1.37 P value : Degree of freedom : df=n-1=6-1=5 P value with test statistic is 1.37 with 5 degree of freedom is 0.115 Excel code for p value is T.DIST.RT(1.37,5) P value is 0.115 Since, the p value is greater than significance level, the null hypothesis is failed to be rejected. There is no sufficient evidence to support the claim that president taller than their main opponent. b) Given : Critical value at 0.01 level of significance for 5 degree of freedom is 4.03 Critical value code for excel is T.INV(1-0.01/2,5) Confidence interval : d¯±t0.01×sdn=6.833±4.03×12.2226=6.833±20.108=6.833-20.108,6.833+20.108=-13.275,26.941 Confidence interval will be (-13.3,26.9) Since, the confidence interval contain 0 then the null hypothesis is failed to be rejected.