An experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. The mean tip percentages and standard deviations are provided in the accompanying table along with the sample sizes. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b).---Table:| | \( n \) | \( \bar{x} \) | \( s \) ||-----------------|------|-----------|-------|| No candy | 27 | 18.35 | 1.44 || Two candies | 27 | 21.08 | 2.55 |---**a.** Use a 0.05 significance level to test the claim that giving candy does result in greater tips.What are the null and alternative hypotheses?- **A.** - \( H_0: \mu_1 \neq \mu_2 \) - \( H_1: \mu_1 < \mu_2 \)- **B.** - \( H_0: \mu_1 = \mu_2 \) - \( H_1: \mu_1 > \mu_2 \)- **C.** - \( H_0: \mu_1 = \mu_2 \) - \( H_1: \mu_1 < \mu_2 \)- **D.** - \( H_0: \mu_1 = \mu_2 \) - \( H_1: \mu_1 \neq \mu_2 \) a. Use a 0.05 significance level to test the claim that giving candy does result in greater tips. What are the null and alternative hypotheses?- A. \( H_0: \mu_1 = \mu_2 \) \( H_1: \mu_1 < \mu_2 \)- B. \( H_0: \mu_1 = \mu_2 \) \( H_1: \mu_1 \neq \mu_2 \)- C. \( H_0: \mu_1 = \mu_2 \) \( H_1: \mu_1 > \mu_2 \)- D. \( H_0: \mu_1 \neq \mu_2 \) \( H_1: \mu_1 > \mu_2 \)The test statistic, \( t \), is \(\square\) (Round to two decimal places as needed.)The \( P \)-value is \(\square\) (Round to three decimal places as needed.)State the conclusion for the test.- A. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that giving candy does result in greater tips.- B. Reject the null hypothesis. There is sufficient evidence to support the claim that giving candy does result in greater tips.- C. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that giving candy does result in greater tips.- D. Reject the null hypothesis. There is not sufficient evidence to support the claim that giving candy does result in greater tips.b. Construct the confidence interval suitable for testing the claim in part (a).\(\square < \mu_1 - \mu_2 < \square\)(Round to two decimal places as needed.)

Two independent samples are compared with the population using the hypothesis testing. If the samples are taken randomly and the mean of two samples are compared using two sample independent t test. Test statistic for two sample independent t test is defined as : t=x¯1-x¯2n1-1s12+n2-1s22n1+n2-21n1+1n2 Confidence interval at various confidence level is defined as : x¯1-x¯2±tα,df=n1+n2-2×sp21n1+1n2a) Given : Experiment has been made to estimate whether giving candy result in greater tips. Let the null and alternate hypothesis : H0 : μ1=μ2H1 : μ1<μ2 Option C is correct Incorrect options : Option A - Null hypothesis does not include inequality Option B - Greater symbol is not used Option D - Two tailed is not used The information given as : Sample Sample size Sample mean Sample standard deviation No candy 27 18.35 1.44 Two candies 27 21.08 2.55 Now, t=x¯1-x¯2n1-1s12+n2-1s22n1+n2-21n1+1n2=18.35-21.0827-11.442+27-12.55227+27-2127+127=-4.84 Test statistic is -4.84 P value : Degree of freedom : df=n1+n2-2=27+27-2=52 Level of significance is 0.05 The test statistic with degree of freedom with 0.05 level of significance is 0.000006 i.e. P value is 0.000 Excel code for p value is 1-T.DIST.RT(-4.84,52), since left tail is used Conclusion : P value is less than level of significance, the null hypothesis is rejected. Hence, we conclude that giving candy result in greater tips. Option B is correct Incorrect options : Option A - Conclusion and decision are wrong Option C - Null hypothesis is rejected Option D - Decision is correct but the conclusions are wrongb) Given : Confidence interval : Critical value at 0.05 level of significance with 52 degree of freedom is 2.01 i.e. Excel code for critical value is T.INV.2T(0.05,52) Pooled variance : sp2=n1-1s12+n2-1s22n1+n2-2=27-11.442+27-12.55227+27-2=4.2881 Now, x¯1-x¯2±t0.05,df×sp21n1+1n2=18.35-21.08±2.01×4.2881×127+127=-2.73±2.01×0.5636=-2.73±1.13=-2.73-1.13,-2.73+1.13=-3.86,-1.60 Confidence interval at 95% is -3.86 and -1.60