**Hypothesis Testing: Comparing Mean Daily Sales**Based on these data, we aim to determine if, at the 0.10 level of significance, the mean daily sales of Store 1 exceed those of Store 2. This evaluation involves performing a hypothesis test regarding \( \mu_d \) (the population mean daily sales difference between the two stores). Assume this population of differences (Store 1 minus Store 2) follows a normal distribution.### Steps for Conducting the Hypothesis Test:#### (a) Formulate Hypotheses- **Null Hypothesis \( H_0 \):** \( \mu_d \leq 0 \)- **Alternative Hypothesis \( H_1 \):** \( \mu_d > 0 \)#### (b) Determine the Type of Test StatisticChoose an appropriate test statistic based on the sample data provided.#### (c) Calculate the Test StatisticFind the test statistic value and round it to three or more decimal places.#### (d) Determine the Critical ValueIdentify the critical value at the 0.10 level of significance, rounding to three or more decimal places.#### (e) ConclusionAt the 0.10 significance level, can the owner conclude that Store 1's mean daily sales exceed Store 2's?- **Options:** Yes or No.### Explanation of Diagram/Chart:**Data Table:**- **Differences (Store 1 - Store 2):** 113, 47, -68, 137, 232, 11, 223, 26, 318, 148**Calculator Tools:** On the right side, a calculator pad is available showing symbols for statistical calculations such as population mean (\( \mu \)), standard deviation (\( \sigma \)), and various data entry options for performing basic arithmetic and statistical computations.### Additional Resources:Consult the provided list of formulas if necessary to assist with this process. **Hypothesis Test for the Difference of Population Means: Paired Samples**The owner of a chain of mini-markets wants to compare the sales performance of two of her stores, Store 1 and Store 2. Though the two stores have been comparable in the past, the owner has made several improvements to Store 1 and wishes to see if the improvements have made Store 1 more popular than Store 2. Sales can vary considerably depending on the day of the week and the season of the year, so she decides to eliminate such effects by making sure to record each store's sales on the same 10 days, chosen at random. She records the sales (in dollars) for each store on these days, as shown in the table below.| Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 ||-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|| Store 1 | 618 | 784 | 516 | 617 | 996 | 618 | 337 | 887 | 815 | 803 || Store 2 | 505 | 671 | 583 | 685 | 685 | 307 | 311 | 569 | 667 | 655 || Difference (Store 1 - Store 2) | 113 | 47 | -67 | -68 | 311 | 311 | 26 | 318 | 148 | 148 |Based on these data, can the owner conclude, at the 0.10 level of significance, that the mean daily sales of Store 1 exceeds that of Store 2? Answer this question by performing a hypothesis test regarding \( \mu_d \) (which is \( \mu \) with a "d" subscript), the population mean daily sales difference between the two stores. Assume that this population of differences (Store 1 minus Store 2) is normally distributed.Perform a **one-tailed test**. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified. (If necessary, consult a list of formulas.)**(a)** State the **null hypothesis** \( H_0 \) and the **alternative hypothesis** \( H_1 \).- \(

The question is about paired t test Given : To find : To test the hypo. that mean daily sales for store 1 exceeds that of store 2Firstly, need to find the mean and sample std. deviation of the difference Total no. of days ( n ) = 10 The sample mean can be find as d¯ = ∑ dn = 118710 = 118.7 And, The sample std. deviation can be find as sd = ∑ ( d - d¯ )2n-1 = 124852.110 - 1sd = 117.7814 a ) Null and alt. hypo. H0 : μd ≤ 0H1 : μd > 0 b ) Type of test stat is t c ) The formula of test stat is given by t = d¯sdn = 118.7117.781410t = 3.186941 d ) df = 10 - 1 = 9 Since alt. hypo. has > sign, so it is right tailed test Given : Level of signif. ( α ) = 0.10 tcritical = t1-α,df = t1-0.1, 9 = t0.9,9 Critical value can be find by using excel command, =T.INV.RT ( Probability, df ) So, tcritical = 1.383029 e ) Decision rule : If test stat > tcritical then reject H0 otherwise fail to reject H0 Here, test stat > 1.383029 Hence, reject H0 Conclusion : There is suffi. evid. to conclude that mean daily sales for store 1 exceeds that of store 2 That is, Ans : Yes