Use StatCrunch to conduct a 2 sample t-test to see if there is a statistically s

Question

Use StatCrunch to conduct a 2 sample t-test to see if there is a statistically significant difference between mean LFPR in 1972 and 1968, using =0.05. Use 1972 as sample 1 and 1968 as sample 2, and do not pool the variances. Answer the following questions, and round to 4 decimal places.

City   1972   1968   diff (1972 - 1968)
N.Y.   0.45   0.42   0.030000000000000027
L.A.   0.5   0.5   0
Chicago   0.52   0.52   0
Philadelphia   0.45   0.45   0
Detroit   0.46   0.43   0.030000000000000027
San Francisco   0.55   0.55   0
Boston   0.6   0.45   0.14999999999999997
Pitt.   0.49   0.34   0.14999999999999997
St. Louis   0.35   0.45   -0.10000000000000003
Connecticut   0.55   0.54   0.010000000000000009
Wash., D.C.   0.52   0.42   0.10000000000000003
Cinn.   0.53   0.51   0.020000000000000018
Baltimore   0.57   0.49   0.07999999999999996
Newark   0.53   0.54   -0.010000000000000009
Minn/St. Paul   0.59   0.5   0.08999999999999997
Buffalo   0.64   0.58   0.06000000000000005
Houston   0.5   0.49   0.010000000000000009
Patterson   0.57   0.56   0.009999999999999898
Dallas   0.64   0.63   0.010000000000000009

a. The difference in sample means is:?

b. The standard error of the difference in sample means is:?

c. The test statistic is t = ?

d. The p-value is:?

e. The statistical decision is:

*FTR Ho; there is not enough evidence to conclude that the means differ significantly

*FTR Ho; there is enough evidence to conclude that the means differ significantly    

*Reject Ho; there is not enough evidence to conclude that the means differ significantly

*Reject Ho; there is enough evidence to conclude that the means differ significantly

Explanation / Answer

The Statcrunch output for this problem is:

Two sample T hypothesis test:
1 : Mean of 1972
2 : Mean of 1968
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 0
(without pooled variances)

Hypothesis test results:

0.0533

From this, we can see:

a) difference = 0.0445 (Rounded off to 4 decimal places)

b) Standard error = 0.0222 (Rounded off to 4 decimal places)

c) Test statistic (t) = 2.0028 (Rounded off to 4 decimal places)

d) p - value = 0.0533 (Rounded off to 4 decimal places)

e) Level of significance = 0.05

Since p - value > 0.05, we fail to reject Ho.

Hence,

Option A is correct.

0.0533

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