Question

BUSI 1013 Assignment 004 Winter 2020 Part B Use this information to answer Question 13 to Questi lation to answer Question 13 to Question 16: Waiting time for checkout line at two stores of a supermarket chain were measu stomers at each store. The chain wants to use this data to test the research (alternative hypothesis that the mea duive hypothesis that the mean waiting time for checkout at Store 1 is lower than that of Store 2. arket chain were measured for a random Store I 461 384 167 293 187 115 195 280 228 315 205 197 Store II 264 308 266 224 244 178 279 289 253 223 13. (2 points) What are the mull and alternative hypothesis for this research? 14. (4 points) Compute the test-statistic. Da7f14

Answer 1

The data provided is:

	Store I	Store II
	461	264
	384	308
	167	266
	293	224
	187	244
	115	178
	195	279
	280	289
	228	253
	315	223
	205
	197
Count	12	10
Mean	252.25	252.8
St Dev	98.54222	37.61442

13.

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: Mi = 42

Ηα: μι < μ2

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

14.

Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t = X1 - X2 252.25 -252.8 (ni-1) +(n2-1) / 1 1 ) ( 1 + /(12-1)98.542215232+(10-1)37.614418042/ 1 ) 3042 (1 + ) ni+n2-2 nin2 V 12+10-2 I -0.017

15.

Decision about the null hypothesis

Since it is observed that t=−0.017≥tc​=−1.725, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.4935, and since p=0.4935≥0.05, it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is less than μ2​, at the 0.05 significance level.

Graphically

$normaldistributiongrapher.php?mean=0&sig$

16.

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: (nu – 1) si + (n2 – 1)s ni +n2 – 2 - 1) 98.5422 + (10 – 1) x 37.6142 12+10 – 2 = 77.314 Since we assume that the population variances are equal, the standard error is computed as follows: se = $11+1 PV ni ne = 1 1 77.314 x V 10 + 10 = 33.104 Now, we finally compute the confidence interval: CI = (#1 - 72 – tę X se, X1 - X2 + te X se) = (252.25 – 252.8 - 2.086 x 33.104, 252.25 - 252.8 +2.086 x 33.104) = (-69.604, 68.504)

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