Question

We wish to gather some information relating to the average time a hospitalized Eagle Flu patient remains in the hospital. We randomly sample 16 cured formerly hospitalized Eagle Flu patients and these were the number of days spent in the hospital:

6.09, 8.04, 6.43, 8.58, 10.34, 5.40, 9.87, 9.24, 9.18, 5.06, 5.39, 5.89, 11.91, 8.94, 5.68, 9.85


Many surveys of flu hospital stays show that hospitalization length is normally distributed. So, we assume that our sample comes from a normal population with unknown mean of μ days and an unknown standard deviation of σ days. We would like to test whether the average hospital stay length is less than 8.1 days..
The null hypothesis is thus

H₀:μ=8.1

. We will test this against the alternative

H_a

.

We want to test at the 8% level.

Let x = the sample mean and s = the sample standard deviation.

a) What should the alternative hypothesis,

H_a

, be?

H_a:μ<8.1

H_a:μ=8%

    

H_a:μ=8.1

H_a:μ≠8.1

H_a:μ>8.1

b) What is the formula for your test statistic?

T =

x -8%

s

T =

x-8.1

s 15

     T =

x-8.1

s 15

T =

x-8.1

s 16

T =

x-8.1

s 16

c) What value does your test statistic,T, take on with the sample data?
d) What type of probability distribution does your test statistic,T, have?

Cauchy binomial     normal t Chi-Squared

e) How many degrees of freedom does T have?

Answer 1

(a)

Since ; So, it's a One Tailed ( Left Tailed ) Hypothesis Test.

(b) To test the we use the t - Statistic

NOTE; Since it's the LEFT TAILED HYPOTHESIS TEST; So, we shouldn't take MODULOUS in the Numrator of the formula.

Where = Sample Mean

Where = Sample Standared Deviation

n = Sample Size

(c)

	x
	6.09
	8.04
	6.43
	8.58
	10.34
	5.4
	9.87
	9.24
	9.18
	5.06
	5.39
	5.89
	11.91
	8.94
	5.68
	9.85
MEAN	7.8681
S.D	2.0875

(e) Degrees of freedom (d.f)

We have n = 16

Degress of freedom = n-1

Therefore d.f = 16-1 = 15

Therefore d.f = 15.

Given =  8% = 0.08

we reject Ho

(d) we follow t distribution since n is a small sample,