A) A random sample of 100 observations produced a sample proportion of .25. An approximate 90% confidence interval for the population proportion p is
|
a. |
.248 and .252 |
|
b. |
.179 and .321 |
|
c. |
.423 and .567 |
|
d. |
.246 and .254 |
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e. |
None of the above answers are correct. |
B)
Suppose that an investigator believes that virtually all values in the population are between 38 and 70. The appropriate sample size for estimating the true population mean μ within 2 units with 95% confidence level is approximately
|
a. |
61 |
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b. |
62 |
|
c. |
15 |
|
d. |
16 |
|
e. |
None of the above answers are correct. |
Solution:Given that,{:[n=100],[ hat(p)=0.25]:}confidence level =90%{:[ hat(p)=0.25],[ hat(q)=1- hat(p)],[=1-0.25],[=0.75]:}At 90% confidence level{:[1-alpha=0.90],[alpha=1-0.90],[alpha=0.10],[(alpha)/(2)=0.05]:}z(alpha)/(2)=z 0.05=(1.645)/(" (vsing standard normal tabie) ")" Margin of error "{:[E=2(alpha)/(2)xxsqrt((( hat(p))-( hat(q)))/(n))],[=1.645 xxsqrt(((0.25)*(0.75))/(100))],[E=0.071]:}confidence intervel for ' p '{:[ hat(p)-E <= P <= hat(P)+E],[0.25-0.071 <= P <= 0.25+0.071],[0.179 <= P <= 0.32 ... See the full answer