Question

8. (12 pointsThe recovery time for a flue is known to follow normal distribution with mean of 7 days. A medicine was developed. The medicine developing company stated that the recovery time decreases with the medicine. 25 patients diagnosed as the flue were randomly selected. They were given the medicine. After the experiment, we get the sample mean of 6.5 days, and sample standard deviation of 1.5 days. (1) (4pts) Let significance level a = 0.1, implement hypothesis test if the medicine de- creases the recovery time, apply the reject region method to solve the question. (2) 4pts) Apply the sample information to compute the 99% two-sided confidence inter- val for the true recovery time with the medicine treatment. (3)(4pts) For the hypothesis test in (1), compute the type II error (6).

Answer 1

1)

Given:

Mu = 7 days ( Population mean )

X-bar = 6.5 days ( Sample mean )

s = 1.5 days ( Sample standard deviation )

n = 25 ( Sample size )

Alpha = 0.1 ( Significance level 10% )

Hypothesis:

H0 : Mu = 7

H1 : Mu < 7 ( One-tailed test: Left tailed test )

Test statistic:

Since Population standard deviation is not given and n < 30,

We use one-sample t-test to test whether a population mean is significantly less from the assumed value.

Test statistic formula:

X-μ

Now, substitute given values in the above formula and calculate t-test statistic.

t = (6.5 - 7)/( 1.5/sqrt(25))

t = - 1.6667...........((1)

Critical value for rejection region method:

Degrees of freedom = n-1 = 24

Alpha = 0.10 ( One tailed test )

t-critical value = - 1.318............(2) [ Refer below t-Distribution Table ]

From (1) and (2);

t-statistic < t-critical value implies Reject H0.

Conclusion : Medicine decreases the recovery time.

t-Distribution Table The shaded area is equal to a for t = ta. t 100 t.050 t.025 0.010 t.005 2.518 2.508 1.323 1.321 1.319 1.318 1.316 1.721 1.717 1.714 1.711 1.708 2.080 2.074 2.069 2.064 2.060 2.500 2.831 2.819 2.807 2.797 2.787 2.492 2.485

2)

99% Confidence Interval for true recovery time :

Given:

Alpha = 0.01 ( 1% )

But we take Alpha/2 for two sided Confidence Interval;

ie : Alpha = 0.005

Confidence Interval Formula:

Calculation:

From the above t-Distribution Table;

t ( 0.005 ) = 2.797

Now, substitute all values in the CI formula;

CI - Lower limit = 6.5-(2.797*(1.5/sqrt(25)))

CI - Lower limit = 5.6609

CI - Upper limit = 6.5+(2.797*(1.5/sqrt(25)))

CI - Upper limit = 7.3391

3)

Type-II error ( Beta ) for the following Hypothesis:

Hypothesis:

H0 : Mu = 7

H1 : Mu < 7 ( One-tailed test: Left tailed test )

P ( x-bar > 6.5 / Mu=7)

= P (t>(6.5 - 7)/( 1.5/sqrt(25))

= P( t > -1.67 )

= 1 - 0.05 ( Refer above t-Distribution Table , where 1.67 is approximately near to 1.711 at df=24)

= 0.95

Beta = 0.95 ( Approximately )