Question

In: Statistics and Probability

It is commonly believed that cities with wind speeds of 10 or more have different average...

It is commonly believed that cities with wind speeds of 10 or more have different average temperature from the cities with winds of less than 10 (mp/h). Use Pollutiondata and your statistical expertise to answer the questions: Is this a reasonable belief?

4. What test/procedure did you perform?

a. One-sided t-test
b. Two-sided t-test
c. Regression
d. Confidence interval

5. Statistical Interpretation

a. Since P-value is small we are confident that the slope is not zero.
b. Since P-value is small we are confident that the averages are different.
c. Since P-value is too large the test is inconclusive.
d. None of these.

6. Conclusion

a. Yes, I am confident that the above belief is correct.
b. No, I cannot claim that the above belief is correct.

CITY    SO2     MANUF   POP     TEMP    WIND    PRECIP-INCHES   PRECIP-#DAYS
Phoenix 11      213     582     70.3    6       7.05    36
Little Rock     15      91      132     61      8.2     48.52   100
San Francisco   16      453     716     56.7    8.7     20.66   67
Denver  24      454     515     51.9    9       12.95   86
Hartford        82      412     158     49.1    9       43.37   127
Wilmington      43      80      80      54      9       40.25   114
Washington      30      434     757     57.3    9.3     38.89   111
Jacksonville    18      136     529     68.4    8.8     54.47   116
Miami   14      207     335     75.5    9       59.8    128
Atlanta 32      368     497     61.5    9.1     48.34   115
Chicago 131     3344    3369    50.6    10.4    34.44   122
Indianapolis    40      361     746     52.3    9.7     38.74   121
Des Moines      20      104     201     49      11.2    30.85   103
Wichita 10      125     277     56.6    12.7    30.58   82
Louisville      35      291     593     55.6    8.3     43.11   123
New Orleans     9       204     361     68.3    8.4     56.77   113
Baltimore       47      625     905     55      9.6     41.31   111
Detroit 46      1064    1513    49.9    10.1    30.96   129
Minneapolis-St. Paul    42      699     744     43.5    10.6    25.94   137
Kansas City     18      381     507     54.5    10      37      99
St. Louis       61      775     622     55.9    9.5     35.89   105
Omaha   17      181     347     51.5    10.9    30.18   98
Albuquerque     15      46      244     56.8    8.9     7.77    58
Albany  56      44      116     47.6    8.8     33.36   135
Buffalo 11      391     463     47.1    12.4    36.11   166
Cincinnati      27      462     453     54      7.1     39.04   132
Cleveland       80      1007    751     49.7    10.9    34.99   155
Columbus        27      266     540     51.5    8.6     37.01   134
Philadelphia    79      1692    1950    54.6    9.6     39.93   115
Pittsburgh      63      347     520     50.4    9.4     36.22   147
Providence      136     343     179     50      10.6    42.75   125
Memphis 10      337     624     61.6    9.2     49.1    105
Nashville       23      275     448     59.4    7.9     46      119
Dallas  11      641     844     66.2    10.9    35.94   78
Houston 10      721     1233    68.9    10.8    48.19   103
Salt Lake City  28      137     176     51      8.7     15.17   89
Norfolk 38      96      308     59.3    10.6    44.68   116
Richmond        38      197     299     57.8    7.6     42.59   115
Seattle 40      379     531     51.1    9.4     38.79   164
Charleston      40      35      71      55.2    6.5     40.75   148
Milwaukee       20      569     717     45.7    11.8    29.07   123

Expert Solution

The data can be divided into two groups as follows:

CITY	SO2	MANUF	POP	TEMP	WIND	PRECIP-INCHES	PRECIP-#DAYS	Group
Phoenix	11	213	582	70.3	6	7.05	36	0
Little	15	91	132	61	8.2	48.52	100	0
San	16	453	716	56.7	8.7	20.66	67	0
Denver	24	454	515	51.9	9	12.95	86	0
Hartford	82	412	158	49.1	9	43.37	127	0
Wilmington	43	80	80	54	9	40.25	114	0
Washington	30	434	757	57.3	9.3	38.89	111	0
Jacksonville	18	136	529	68.4	8.8	54.47	116	0
Miami	14	207	335	75.5	9	59.8	128	0
Atlanta	32	368	497	61.5	9.1	48.34	115	0
Indianapolis	40	361	746	52.3	9.7	38.74	121	0
Louisville	35	291	593	55.6	8.3	43.11	123	0
New	9	204	361	68.3	8.4	56.77	113	0
Baltimore	47	625	905	55	9.6	41.31	111	0
St.	61	775	622	55.9	9.5	35.89	105	0
Albuquerque	15	46	244	56.8	8.9	7.77	58	0
Albany	56	44	116	47.6	8.8	33.36	135	0
Cincinnati	27	462	453	54	7.1	39.04	132	0
Columbus	27	266	540	51.5	8.6	37.01	134	0
Philadelphia	79	1692	1950	54.6	9.6	39.93	115	0
Pittsburgh	63	347	520	50.4	9.4	36.22	147	0
Memphis	10	337	624	61.6	9.2	49.1	105	0
Nashville	23	275	448	59.4	7.9	46	119	0
Salt	28	137	176	51	8.7	15.17	89	0
Richmond	38	197	299	57.8	7.6	42.59	115	0
Seattle	40	379	531	51.1	9.4	38.79	164	0
Charleston	40	35	71	55.2	6.5	40.75	148	0
Chicago	131	3344	3369	50.6	10.4	34.44	122	1
Des	20	104	201	49	11.2	30.85	103	1
Wichita	10	125	277	56.6	12.7	30.58	82	1
Detroit	46	1064	1513	49.9	10.1	30.96	129	1
Minneapolis-St.	42	699	744	43.5	10.6	25.94	137	1
Kansas	18	381	507	54.5	10	37	99	1
Omaha	17	181	347	51.5	10.9	30.18	98	1
Buffalo	11	391	463	47.1	12.4	36.11	166	1
Cleveland	80	1007	751	49.7	10.9	34.99	155	1
Providence	136	343	179	50	10.6	42.75	125	1
Dallas	11	641	844	66.2	10.9	35.94	78	1
Houston	10	721	1233	68.9	10.8	48.19	103	1
Norfolk	38	96	308	59.3	10.6	44.68	116	1
Milwaukee	20	569	717	45.7	11.8	29.07	123	1

Group 0 has wind speed less than 10. Group 1 has wind speeds more than equal to 10

So we have the following data of temperature:

Group 0	Group 1
70.3	50.6
61	49
56.7	56.6
51.9	49.9
49.1	43.5
54	54.5
57.3	51.5
68.4	47.1
75.5	49.7
61.5	50
52.3	66.2
55.6	68.9
68.3	59.3
55	45.7
55.9
56.8
47.6
54
51.5
54.6
50.4
61.6
59.4
51
57.8
51.1
55.2

We need to test the hypothesis:

Ho: u1 = u2

Ha: u1 =/ u2

4. This is a two-sided t-test. B) is correct.

5. Let's compute the t-statistic and calculate the p-value.

Difference Scores Calculations

Treatment 1

N₁: 27
df₁ = N - 1 = 27 - 1 = 26
M₁: 57.18
SS₁: 1217.63
s²₁ = SS₁/(N - 1) = 1217.63/(27-1) = 46.83

Treatment 2

N₂: 14
df₂ = N - 1 = 14 - 1 = 13
M₂: 53.04
SS₂: 713.79
s²₂ = SS₂/(N - 1) = 713.79/(14-1) = 54.91

T-value Calculation

s²_p = ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂) = ((26/39) * 46.83) + ((13/39) * 54.91) = 49.52

s²_M₁ = s²_p/N₁ = 49.52/27 = 1.83
s²_M₂ = s²_p/N₂ = 49.52/14 = 3.54

t = (M₁ - M₂)/√(s²_M₁ + s²_M₂) = 4.14/√5.37 = 1.79

p = 0.08. We cannot reject the null hypothesis. We do not have enough evidence.

5. C) is correct as p-value is too large for the test to be conclusive

6. We cannot claim that the belief is correct

orchestra answered 1 year ago

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