In the natural gas consumption case, the estimated consumption is 12.003 in a week when the average hourly temperature is 30 degrees. If the distance value is 0.2642 when xo 30, compute a 95% conficdence interval for the mean consumption for all weeks having average hourly temperature of 30 degrees. Recall that n= 8 and s=0.6542 O m35, 12.656) Reason: Use 9 toass ) distance = 12.003 (2.447),6542)5140) 12.003 823 = [1.18O, 12.826] Om 580, 12.426) Reason: Use 9= tozss) distance 12.003 (2.447K.6542 5140) 12.003 : 823 (1180, 12.826) O118012 826) (11739, 12.267) Reason: Use 9s toass)distance #12.003 (2.447)6542) 5140) = 12.003823 [11180. 12.826] Answer

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In the natural gas consumption case, the estimated consumption is 12.003 in a week when the average hourly temperature is 30 degrees. If the
distance value is 0.2642 when xo 30, compute a 95% confidence interval for the mean consumption for all weeks having average hourly temperature
of 30 degrees. Recall that n= 8 and s=0.6542
O 135, 12.656)
Reason: Use tozss ) distance = 12.003 : (2.447),6542)5140) = 12.003 823 = (1.180, 12.826]
O m 580, 12.426)
Reason: Use 9 = tozss) distance = 12.003 (2.447K.6542 5140) = 12.003 1823 = [11180, 12.826)
O 18012 826)
(11.739, 12.267)
Reason: Use tozss) distance #12.003 1(2.4476542 5140) = 12.003823 [11180. 12.826)
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Transcribed Image Text:In the natural gas consumption case, the estimated consumption is 12.003 in a week when the average hourly temperature is 30 degrees. If the distance value is 0.2642 when xo 30, compute a 95% confidence interval for the mean consumption for all weeks having average hourly temperature of 30 degrees. Recall that n= 8 and s=0.6542 O 135, 12.656) Reason: Use tozss ) distance = 12.003 : (2.447),6542)5140) = 12.003 823 = (1.180, 12.826] O m 580, 12.426) Reason: Use 9 = tozss) distance = 12.003 (2.447K.6542 5140) = 12.003 1823 = [11180, 12.826) O 18012 826) (11.739, 12.267) Reason: Use tozss) distance #12.003 1(2.4476542 5140) = 12.003823 [11180. 12.826) Correct Answer Read
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