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RES 342 Week 1 E-Text Assignments

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8.48
Condence level = 95%
Mean = 346.5
Standard deviation = 170.38
T (95%) = 2.093
Degree of freedom= 20-1=19
The condence interval:
Lower Limit 346.5 – 2.093 * 170.378/ sqrt (20) = 266.76
Upper Limit 346.5 + 2.093 * 170.378/ sqrt (20) = 426.24
Normality might denitely be an issue since it is not known if the pages
were taken randomly and the sample size is relatively small.
Z for 99% condence interval is 2.5758
E=zs/sqrt(n)
N= (zs/e) ^2
N=(2.5758*170.38/10)^2
N=1926.023
The sample size should be at 1927
This is certainly not a reasonable requirement with the phone directory
being 1591 pages. The same size requirement would become 1116
and reasonable if the condence level is taken slightly down to 95%
8.64
The sample proportion, p=x/n = 86/773 = 0.1113
The z-value =0.10 is 1.645
The 90% condence interval for the proportion of all kernels that would
not pop is: (p-z* sqrt [p (1-p), p+z* sqrt [p(1-p)/n])
= (0.1113 -1.645*Sqrt [(0.1113*0.8887)/ 773], 0.1113+1.645*Sqrt
[(0.1113*0.8887)/ 773])
= (0.1113 - 0.0186, 0.1113 + 0.0186)
= (0.0927, 0.1299)
np= 773*0.1113 = 86 and n(1-p)= 773*0.8887 = 687.
Since both np and n(1-p) are greater than 5, the normality assumption
is this equation is satised.
The Very Quick Rule does not work well here. It requires for p be close
to 0.5.

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8.48 Confidence level = 95% Mean = 346.5 Standard deviation = 170.38 T (95%) = 2.093 Degree of freedom= 20-1=19 The confidence interval: Lower Limit 346.5 - 2.093 * 170.378/ sqrt (20) = 266.76 Upper Limit 346.5 + 2.093 * 170.378/ sqrt (20) = 426.24 Normality might definitely be an issue since it is not known if the pages were taken randomly and the sample size is relatively small. Z for 99% confidence interval is 2.5758 E=zs/sqrt(n) N= (zs/e) ^2 N=(2.5758*170.38/10)^2 N=1926.023 The sample size should be at 1927 This is certainly not a reasonable requirement with the phone directory being ...
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