Question

ASSIGNMENT - HYPOTHESIS TEST FOR TWO MEANS, INDEPENDENT POPULATIONS 50 Points Total of interest is whether the mean hotel price for a weekend two-night stay is the same in City A as it is in City B. Conduct a hypothesis test, at a significance level of 0.01, to test the claim that the City A hotel prices have a significantly higher mean than the City B hotel prices. (40 points) Also construct an appropriate C.L. (10 points) Summarized hotel price data is as follows (data is in US dollars, $USD): (City A) n =16:3, = 377.54: 5, -15.228796 (City B) n, = 21; , - 364.66 ; 8, -18.611564 Define The Parameter(8) Of Interest (3 points) State The Hypotheses (6 points) Sketch The Distribution of Interest (4 points) Confidence Level and Significance Level (2 points) Calculated Test Statistic (TS) (8 points) Critical Value (CV) (3 points) p-Value (3 points) Compare Critical Value To Test Statistic (3 points) Compare p-Value To Significance Level (3 points) Conclusion (Sentence(s)) (5 points) Confidence Interval (10 points) Use the back of this sheet.

Answer 1

The parameter of interest is mean hotel price in City A and B.

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ > µ₂

The distribution is:

Distribution Plot T, df=35 0.4 0.3 Density 0.2 0.1 0.05 0.0 1.690 0 X

The confidence level is 99% and the significance level is 0.01.

sp2 = (16 - 1)*15.228796^2 + (21 - 1)*18.611564^2/(16 + 21 - 2) = 297.32999

t = (377.54 - 364.66)/$\sqrt{}$297.32999*(1/16 + 1/21) = 2.251

The critical value is 1.690.

The p-value is 0.0154.

Since 2.251 > 1.690, we can reject the null hypothesis.

Since the p-value (0.0154) is less than the significance level (0.01), we can reject the null hypothesis.

Therefore, we can conclude that µ₁ > µ₂.

The 99% confidence interval is:

= (377.54 - 364.66) $\pm$ 2.72*($\sqrt{}$297.32999*(1/16 + 1/21))

= (-2.71, 28.47)