Question

A parent population has a true mean equal to 25. A random sample of n=16 is taken and the estimated standard deviation is 6.0

What is the value of the standard error of the mean in this instance?

What percentage of sample means would be expected to lie within the interval 23.5 to 26.5?

What is the t-value (in absolute value) associated with a 95% symmetric confidence interval?

What is the lower bound of the 95% confidence interval?

T/F A value of the sample mean of 29 would be highly unlikely in this instance.

The probability that a sample mean exceeds 29 when the true value is 25 with a standard deviation of 6 and a sample size of 16 is?

T/F The null hypothesis for a one-sided test that the mean is less than 25 is Ho: mu <25; and the alternative is H1: mu > 25.

T/F The null hypothesis for a one-sided test, Ho: mu <=25 against the alternative is H1: mu > 25 would be rejected at reasonable alpha levels when the sample mean is 29?

Answer 1

1) the standard error of the mean = 6/sqrt(16)= 6/4= 1.5

2) percentage of sample means would be expected to lie within the interval 23.5 to 26.5

μ=25 and σ=1.5 we have:
P ( 23.5<X<26.5 )=P ( 23.5−25< X−μ<26.5−25 )=P ((23.5−25)/1.5<(X−μ)/σ<(26.5−25)/1.5)
Since Z=(x−μ)/σ , (23.5−25)/1.5=−1 and (26.5−25)/1.5=1 we have:
P ( 23.5<X<26.5 )=P ( −1<Z<1 )
Use the standard normal table to conclude that:
P ( −1<Z<1 )=0.6826

3) df=n-1=16-1=15 and 0.95 confidence level

t critical= 2.13

4) the lower bound of the 95% confidence interval= ( , infinity +)

= (29-1.75*1.5, infinity +)

= (26.375, infinity +)

NOTE: I HAVE DONE THE FIRST FOUR QUESTIONS PLEASE REPOST THE REST. THANK YOU.