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The Washington Metropolitan Area Transit Authority (WMATA) wants to promote an i

ID: 3128180 • Letter: T

Question

The Washington Metropolitan Area Transit Authority (WMATA) wants to promote an image of reliability by encouraging its drivers to maintain consistent schedules. As a standard policy the company would like arrival times at bus stops to have low variability. In terms of the variance of arrival times, the company standard specifies an arrival time variance of 4 or less when arrival times are measured in minutes. WMATA wants to determine whether the arrival time population variance is excessive, so it takes a random sample of 24 GCRTA bus arrivals taken at a downtown intersection. The sample variance turns out to be s2 = 4.9.

I am given a partial excel output and need to figure out how to get the numbers

_____hypothesized variance

4.900 observed variance of the Arrival Times

N 24

______DF(need to find)

_____Test Statistic(need to find)

_____P-Value(need to find)

____confidence interval 95% lower(need to find)

____confidence interval 95% upper(need to find)

Explanation / Answer

The Washington Metropolitan Area Transit Authority (WMATA) wants to promote an image of reliability by encouraging its drivers to maintain consistent schedules. As a standard policy the company would like arrival times at bus stops to have low variability. In terms of the variance of arrival times, the company standard specifies an arrival time variance of 4 or less when arrival times are measured in minutes. WMATA wants to determine whether the arrival time population variance is excessive, so it takes a random sample of 24 GCRTA bus arrivals taken at a downtown intersection. The sample variance turns out to be s2 = 4.9.

Chi-square Variance Test

4.000

hypothesized variance

4.900

observed variance

24

n

23

df

28.18

chi-square

.2092

p-value (one-tailed, upper)

2.960

confidence interval 95.% lower

9.642

confidence interval 95.% upper

Calculated chi square =28.18

Table value of chi square =35.17

Calculated chi square =28.18 < 35.17, null hypothesis is not rejected.

We conclude that an arrival time variance is 4 or less.

95% CI for variance = (2.960, 9.642).

Chi-square Variance Test

4.000

hypothesized variance

4.900

observed variance

24

n

23

df

28.18

chi-square

.2092

p-value (one-tailed, upper)

2.960

confidence interval 95.% lower

9.642

confidence interval 95.% upper

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