Let X_1, X_2 and X_3 represent a random sample of size three selected from some
ID: 3218816 • Letter: L
Question
Let X_1, X_2 and X_3 represent a random sample of size three selected from some population having mean mu and standard deviation sigma. You want to estimate mu and you must choose between two estimators M_1 = (X_1 + X_2 + 2X_3)/4 and M_2 = (xX_1 + X_2 + 2X_3)/7 to calculate your point estimate for mu. Given that both estimators are unbiased for u, which to estimator would you prefer? Justify your answer. A random sample of 16 lightning flashes recorded in a certain region yielded a sample average radar echo duration of 0 71 seconds with a sample standard deviation of .22 seconds. Give an expression for a 90% two sided confidence interval for the true average echo duration assuming the distribution of radar echo durations is normal. A machine in a certain factory must be repaired if it produces more that 10% defective among a certain large lot of it produces in a day. A random sample of 100 items from the day's production items 15 defectives, and the foreman says that the machine must be repaired. Does the sample evidence support his decision at the 0.05 significance level? a. State the null hypothesis. b. State the alternative hypothesis. c. Find the P-value of your test and state your conclusion.Explanation / Answer
Q5.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=0.71
Standard deviation( sd )=0.22
Sample Size(n)=16
Confidence Interval = [ 0.71 ± t a/2 ( 0.22/ Sqrt ( 16) ) ]
= [ 0.71 - 1.753 * (0.055) , 0.71 + 1.753 * (0.055) ]
= [ 0.614,0.806 ]
Interpretations:
1) We are 90% sure that the interval [0.614 , 0.806 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
Q6.
Given that,
possibile chances (x)=15
sample size(n)=100
success rate ( p )= x/n = 0.15
success probability,( po )=0.1
failure probability,( qo) = 0.9
null, Ho:p<0.1
alternate, H1: p>0.1
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.15-0.1/(sqrt(0.09)/100)
zo =1.6667
| zo | =1.6667
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.667 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 1.66667 ) = 0.04779
hence value of p0.05 > 0.04779,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.1
alternate, produces more that 10% defectives among a large H1: p>0.1
test statistic: 1.6667
critical value: 1.64
decision: reject Ho
p-value: 0.04779
we support thet it produces more that 10% defectives among a large lot of items
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