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Assignment: Chapter 10 Humework Questions Exerclze 10.06 Agortthmic Assignment Score: 34.31% ave ubnsit Nssignment ter Gracing Question 3 of il Check y Work cBook Suppose that ycu are raspensible for making arrangemants for a businass convention and that vou hae bean charged with choasing a ciy for the convantion that has the least xpansva rooms. You hae narrawad your chaicos to Atlanta and Hauston. The tahla balow contains samples af prices or moms n Atlanta and Ho ston. Because considerable historical data on the pries of rooms n both dries are ava able he pop ist on standard dev at nns or the prices can be assumed to he 36 n Atlanta and in Houston. AtlantaHoucton Atlanta Houston Atlant Houston Atianta Houston 100 105 105 110 110 140 0. 100 1. 115 115 125 150 145 115 145 140 Rase n the Aan ple data can you conclude that ne mean nrice ถ a hotel ro m n Atlanta s lower than one in Houston, U A α 05 and null hypothesls ls g : μΑ ag > 0 Fnter neaative values as nega ve numbers. 1.27 to 2 decimals) p-vlue 102 tn 4 dacimals) that the mean price of a hotel room in Alanta is lower than the mean price of a hotel room in Houston Hide Feedhackk Partially Correct

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Answer #1

First of all we compute the mean prices for both Atlanta and Houston

We get :

For Atlanta

X1-91.43

For Houston :

103.38

Now, we conduct Z-test for two Means, with Known Population Standard Deviations :

The provided sample means are shown below: X1 91.43 X2 103.38 Also, the provided population standard deviations are σ2 27 and the sample sizes are n 35 and n 40 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: This corresponds to a left-tailed test, for which a z-test for two population means, with known population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the critical value for a left- tailed test is ze-1.64 The rejection region for this left-tailed test is R- {z: z<-1.64)

(3) Test Statistics The z-statistic is computed as follows 91.43 103.38 382/35 + 272/40 -1.549 (4)_Decision about the null hypothesis Since it is observed thatz1.549 1.64, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.0606, and since p- 0.06060.05, it is concluded that the null hypothesis is not rejected.

Z-value = - 1.55 ( 2 decimal places)

P-value = 0.0606

We cannot conclude that the mean price of a hotel room in Atlanta is lower than mean price of a hotel room in Houston.

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