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# RES 342 Week 1 E-Text Assignments

Homework

### Rating

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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.

The sample might not be typical. The sampling is not to be random.
The self confessed connoisseur of cheap popcorn might be better at
making popcorn than other people and might have a lower number of
unpopped kernels.