A standardized test is used to assess the students knowledge

A standardized test is used to assess the students knowledge (national reported mean=65, sd=5). A sample of 30 students are tested (sample mean=58, standard error=3.2). Compute a 99 percent confidence interval based on this sample\'s data. How do these students compare to the national sample?

So what is a good sample size to use for national reported mean=65, sd=5? or how do you calculate a for a sample size or do I even need to?

THe second question I have is there a way to change the standard error to a standard deviation? then compute the confidence interval?

Solution

A standardized test is used to assess the students knowledge (national reported mean=65, sd=5). A sample of 30 students are tested (sample mean=58, standard error=3.2). Compute a 99 percent confidence interval based on this sample\'s data. How do these students compare to the national sample?

Confidence Interval Estimate for the Mean

Data

Sample Standard Deviation

3.2

Sample Mean

58

Sample Size

30

Confidence Level

99%

Intermediate Calculations

Standard Error of the Mean = sd/sqrt(n)=

0.584237395

Degrees of Freedom

29

t Value

2.7564

Interval Half Width= t*se =

1.6104

Confidence Interval

Interval Lower Limit

56.39

Interval Upper Limit

59.61

99% CI =(56.39,59.61) national mean of 65 not in the interval. There is significant different with the national mean.

So what is a good sample size to use for national reported mean=65, sd=5? or how do you calculate a for a sample size or do I even need to?

For calculating sample size we need the margin of error

If we have margin of error =d

Z= 2.576 at 99% level

Then Sample size =z²*sd²/d² round up to whole number

THe second question I have is there a way to change the standard error to a standard deviation? then compute the confidence interval?

You can use this relation Standard deviation =stand error *sqrt(n)