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"A study was conducted to compare the costs of supporting a family of four Canadians for a year in different foreign cities. The lifestyle of living in Canada on an annual income of 75,000 dollars was the standard against which living in foreign cities was compared. A comparable living standard in Perth, Australia, and Mexico City was attained for about $64,000. Suppose an executive wants to determine whether there is any difference in the average annual cost of supporting her family of four in the manner to which they are accustomed in Perth and Mexico City. She uses the following data, randomly gathered from 11 families in each city, and an alpha of 0.01 to test this difference. She assumes the annual cost is normally distributed and the population variances are equal. What does the executive find?"

Perth, Australia

                
68600,
64700,
67500,
64700,
66700,
68000,
65000,
68600,
71000,
68500,
67500
Mexico

64000,
64000,
66400,
64900,
62000,
60500,
63200,
63000,
64500,
63500,
61800

My attempt:

H0 u1-u2=0

Ha u1-u2=/=0

n=11

a=0.01

xbar1=67345.4545

sample sd1=1955.1796011434

xbar2=63436.3636

sample sd2=1621.8956361448

t=xbar

t=((x1-x2)/root((s1^2(n1-1))+(s2^2(n2-1))/n1+n2-2)*root((1/n1)+(1/n2))

-7280.7803/1486681.264/20*(root(22)/11=-0.35

The numerical answer is wrong so what mistake did I make? Whats the correct equation to solve for observed t.

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  • $\begingroup$ It is clear you have made (at least) one computational error, because the difference in means is nearly 4000 while the standard error of that difference must be a fraction of the typical SDs (which you report correctly as being about 2000 and 1600), whence the difference in means must be many standard errors in size rather than -0.35. Recommendation: double-check your arithmetic. $\endgroup$ – whuber Apr 3 at 12:59
  • $\begingroup$ @whuber thanks for id the error, is my formula correct then? I redid my calculations by hand,calculator and excel so I'm not sure where I errored, i found mean using sum of all/total # n $\endgroup$ – JanusP Apr 3 at 20:10
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It seems you are trying to compute a two-sample t test statistic. It is not clear whether you are doing a Welch two-sample t test (which would be best because you have no reason to believe the variance of living costs is the the same in the two cities), or a pooled t test. I will show results for both, using Minitab, which gives more intermediate results than some other statistical software. [Information about confidence intervals is not shown.]

I will leave it to you to find where your calculations differ from the Minitab output.

Descriptive Statistics: Perth, Mex.C 

Variable   N   Mean  StDev  Minimum     Q1  Median     Q3  Maximum
Perth     11  67345   1955    64700  65000   67500  68600    71000
Mex.C     11  63436   1622    60500  62000   63500  64500    66400

enter image description here

Pooled t test

Two-Sample T-Test and CI: Perth, Mex.C 

Two-sample T for Perth vs Mex.C

        N   Mean  StDev  SE Mean
Perth  11  67345   1955      590
Mex.C  11  63436   1622      489

Difference = μ (Perth) - μ (Mex.C)
Estimate for difference:  3909
T-Test of difference = 0 (vs ≠): 
  T-Value = 5.10  P-Value = 0.000  DF = 20
Uses Pooled StDev = 1796.2840

Welch two-sample t test.

Difference = μ (Perth) - μ (Mex.C)
Estimate for difference:  3909

T-Test of difference = 0 (vs ≠): 
   T-Value = 5.10  P-Value = 0.000  DF = 19

Because sample sizes are equal, the pooled and Welch two-sample t statistics (5.1037) are equal; the standard errors in their denominators can also be shown to be equal when sample sizes are equal.

Both methods use the difference [(67345.45 - 63436.36) = 3909.09] between the two sample means in the numerator.

As a 'correction' for slightly unequal sample variances, the Welch test uses degrees of freedom DF = 19, instead of DF = 20 for the pooled test.

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