The data in below, provides a sample of 48 data points showing the (centigrade)

Question

The data in below, provides a sample of 48 data points showing the (centigrade) temperature in each of two parallel reactor vessels, measured
at identical relative locations (the top of each vessel) every hour, on the hour, for 48 hours.
1. Estimate the difference between the two means to a 95% confidence level.
2. Conduct a hypothesis test to see if the first mean is greater than the second, to
a 95% confidence level.
3. If you want to estimate the difference to within 0.2 degrees, what sample size
should you use (95% confidence level)?
4. What is the observed significance level (p-value)?

91.43633779 91.48868405 90.85576476 91.04109869 90.15669379 90.94344212 91.45552625 91.33170686 91.88287238 90.0670231 91.09038167 90.52179888 90.73831978 91.41108115 91.39719055 90.86051058 90.02400295 90.09622094 90.86213103 91.47928982 90.89022294 90.88460223 91.2711666 89.69815986 91.52917961 90.67852957 90.98126679 90.89173407 91.00719722 91.16529831 90.30356823 90.73851505 90.6792784 90.29943028 91.45455432 90.11974603 90.90128277 90.39083511 90.83172729 91.05104127 90.144227 91.50408243 91.09616033 90.82399882 91.13096679 89.27720407 90.32283368 90.64861804 89.99545446 91.26932079 90.77313254 90.49140672 91.0868432 90.42552398 90.55683739 91.24498551 90.14144413 91.3604756 91.82463499 91.51793556 90.76620055 91.19984319 90.79111174 89.98381703 90.93122012 91.23487075 91.60096586 90.27246616 90.46496491 92.42142432 91.52103862 90.10637649 91.36386141 90.81024485 90.66887068 90.59942908 91.07122817 91.12455052 91.66311618 90.70718927 91.24848607 90.76597182 91.24701963 91.19262967 91.08701172 90.48562109 89.91305629 90.66469129 91.79894644 91.51030342 91.11763398 91.12099438 91.25428588 91.25921143 90.70009701 91.09922481

Explanation / Answer

1)

2)

p-value = 0.1384

As p-value is greater than the significance level of 0.05, we failed to reject the null hypothesis.

3)

n = 64

4)

p-value = 0.1384

x1(bar) 90.9590 x2(bar) 90.8392 s1 0.5044 s2 0.5667 n1 48 n2 48 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 0.0053 (s22/n2) 0.0067 SE 0.1095 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 0.000 [ (s22 / n2)2 / (n2 - 1) ] 0.00 (s12/n1 + s22/n2)2 0.00 DF = 93

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.