Question

In the EAl sampling problem, the population mean is $51,900 and the population standard deviation is $4,000. When the sample size is n 30, there is a 0.5878 probability of obtaining a sample mean within+$600 of the population mean. Use z-table. 

a. What is the probability that the sample mean is within $600 of the population mean if a sample of size 60 is used (to 4 decimals)?  

b. What is the probability that the sample mean is within $600 of the population mean if a sample of size 120 is used (to 4 decimals)?

According to Reader's Digest, 42% of primary care doctors think their patients receive unnecessary medical care. 

a. Suppose a sample of 350 primary care doctors was taken, Show the sampling distribution of the proportion of the doctors who think their patients receive u medical care. Use z-table.

b. What is the probability that the sample proportion will be within +0.03 of the population proportion. Round your answer to four decimals. 

c. Whet is the probability that the sample proportion will be within +0.05 of the population proportion. Round your answer to four decimals 

d. What would be the effect of taking a larger sample on the probabilities in parts (b) and (c)? Why? 

The population proportion is 0.40. What is the probability that a sample proportion will be within ±0.02 of the population proportion for each of the following samps Round your answers to 4 decimal places. Use z-table 

Answer 1

1)
a)

std error=σx̅=σ/√n=

516.3978

probability =

P(51300

=

P(-1.16

0.8770-0.1230=

0.7540

b)

probability =

P(51300

=

P(-1.64

0.9495-0.0505=

0.8990

2)

a)E(p)=0.42

std error of proportion=σp=√(p*(1-p)/n)=

0.0264

b)

probability =

P(0.39

=

P(-1.14

0.8729-0.1271=

0.7458

c)

probability =

P(0.37

=

P(-1.9

0.9713-0.0287=

0.9426

d)the proportion would increase,,,,the std error decreases

3)

a) 0.3182

b)0.4380

c)0.6372

d)0.8030