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

Solutions

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

s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((26/39) * 46.83) + ((13/39) * 54.91) = 49.52

s2M1 = s2p/N1 = 49.52/27 = 1.83
s2M2 = s2p/N2 = 49.52/14 = 3.54

t = (M1 - M2)/√(s2M1 + s2M2) = 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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