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%...


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

Extracted 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

Jun 01, 2022
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