You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients.

Isa Trevino

Isa Trevino

Answered question

2021-08-14

You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 14 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.
9610151962326191611251611
a) To compute the confidence interval use a t or z distribution.
b) With 95% confidence the population mean number of visits per physical therapy patient is between "?" and "?" visits.
c) If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About "?" percent of these confidence intervals will contain the true population mean number of visits per patient and about "?" percent will not contain the true population mean number of visits per patient.

Answer & Explanation

smallq9

smallq9

Skilled2021-08-15Added 106 answers

Step 1
a) Here, the population standard deviation is unknown, and the sample standard deviation must be used for the calculation of the confidence interval.
Thus, to compute the confidence interval, the t-distribution must be used.
Step 2
b) The 1001-α% confidence interval for the population mean, μ when the population standard deviation, σ is unknown and the sample standard deviation, s is used, is the t-confidence interval as follows:
x¯-t/2;n-1  αsn, x¯+t/2;n-1  αsn
such that:
Px¯-t/2;n-1  αsnμx¯+t/2;n-1  αsn=1-α
Here,
E=t/2;n-1  αsn is the margin of error.
Here, n is the sample size, x¯ is the sample mean, and t/2;n-1  α is the critical value of the t-distribution with n-1, above which, 100α2% or α2 proportion of the observations lie, and below which, 1001-α+α2%=1001-α2% or 1-α2 proportion of the observations lie.
The sample size is, n=14, so that the degrees of freedom is, n-1=13.
The means and standard deviations are obtained as follows:
xx29816361010015225193616362352926676193611625611121256251625611121 x=212 x2=3.784
x¯=1ni=1nxi
=21214
15.143
s=1n-1i=1nxi2-nx¯2
=114-13,784-14·15.1432
6.643.
The desired confidence level is 95%. Thus,
1001-α=95
1-α=0.95
α=0.05
α2=0.025
1-α2=0.975.
This implies that,
t {α}{2.n1}=t0.025.13,
such that:
P(tn1 t {α}{2.n1})=P(t13 t0.05.11)
=0.975
=Pt132.16
Using the t-distribution table, or the Excel formula:=T.INV0.975,13,Pt132.16=0.975
 t0.025.13=2.16
First, the margin of error is calculated below:
E=t {α}{2.n1}sn
=t0.025.13sn
=2.16·6.64314
3.835
The confidence interval can be obtained as follows:
(x¯t {α}{2.n1}sn,x¯+t {α}{2.n1}sn)=(x¯E,x¯+E)
15.143-3.835,15.143+3.835
=11.308,18.978.
Hence, with “95% confidence, the population mean number of visits per physical therapy patient is between 11.308 (or approximately 11) and 18.978 (or approximately 19) visits.”
Step 3
c) As it can be said that 95% of the confidence intervals calculated on from several samples of the same size will contain the true population mean, therefore, the remaining 5%=100%-95% of the confidence intervals will not contain the true population mean.
Hence, if “many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population mean number of visits per patient and about 5 percent will not contain the true population mean number of visits per patient.”

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