A sample of 100unit is found to have 5 Ibs as mean.could it be regarded as a simple random sample froma large population whose mean is 5.64 Ibs and standard deviation 1.5 Ibs.
Answers
Answer:
Do it yourself
Explanation:
OK u understand
Answer:
The deviation is highly significant, therefore the given sample can not be regarded as a simple sample from a large population with a mean =5.6 and s.d. = 1.5.
Explanation:
Method 1:
The standard normal statistics
z = (x’ - μ)/(σ/√n) = (x’ - μ)/(σ/10)
= {(x’ - μ)×10}/σ = {(5 - 5.6)x10}/1.5 ==>
|z| = 4 > 3 . Hence the deviation is highly significant, therefore the given sample can not be regarded as a simple sample from a large population with a mean =5.6 and s.d. = 1.5.
Method 2:
The sample size is 100 should allow us to invoke the central limit theorem and consider the population of sample means to be normal. Since we don’t have a standard deviation of the population, we can use the t-test statistic in a test of mean, using the sample size of 100, the difference between the sample mean (5) and an assumed population mean (5.6), and the sample standard deviation (1.5).
Then use a claim that the population means is equal to 5.6.
The test statistic is -4 and the p-value is 0.0001.
Using a significant value as small as 0.001, we would still reject the assumption of 5.6 as the mean. There is sufficient sample evidence to reject the claim of 5.6
Method 3:
It is a normal sample of n=100 with those parameters
Using R:
library(tigerstats)
qnormGC(0.99,region="between",mean=5.6,sd=1.5,graph=TRUE) #### 99% of the data are between the values:
[1] 1.736256 9.463744
