Can u help me with this
2 answers:
You might be interested in
Answer:
95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].
Step-by-step explanation:
We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.
Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;
P.Q. = ~
where, = sample mean procrastination score = 41
s = sample standard deviation = 6.89
n = sample of students = 69
= population mean estimate
<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>
So, 95% confidence interval for the true mean, is ;
P(-1.9973 < < 1.9973) = 0.95 {As the critical value of t at 68 degree
of freedom are -1.9973 & 1.9973 with P = 2.5%}
P(-1.9973 < < 1.9973) = 0.95
P( <
<
) = 0.95
P( <
<
) = 0.95
<u>95% confidence interval for </u> =[
,
]
= [ ,
]
= [39.34 , 42.66]
Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].