Question

Help with Q9 pls

Q8

Q9

A pharmaceutical company has created a new drug to decrease how long patients suffer from a runny nose while having a cold. It is known that without treatment, on average, people report having a runny nose for 60 hours while having a cold. A test is organised for this new drug, hoping to prove the drug indeed decreases the duration of symptoms: 154 patients use it at the first sign of a runny nose and record how long they suffer from this runny nose; on average the recorded duration is 58 hours. We suppose the observations are independent and identically distributed, and the duration of symptoms during a cold follows an exponential distribution. Thus X, ~Exp(), where X, is the duration of runny nose recorded by patient i, measured in hours. We want to test the efficiency of this drug to decrease the duration of a runny nose. Give appropriate null and alternative hypotheses for this problem. In hours, the null hypothesis is Ho 60 60 while the alternative hypothesis is HI : A pharmaceutical company has created a new drug to decrease how long patients suffer from a runny nose while having a cold. It is known that without treatment, on average, people report having a runny nose for 60 hours while having a cold. A test is organised for this new drug, hoping to prove the drug indeed decreases the duration of symptoms: 154 patients use it at the first sign of a runny nose and record how long they suffer from this runny nose; on average the recorded duration is 58 hours. We suppose the observations are independent and identically distributed, and the duration of symptoms during a cold follows an exponential distribution. Thus Xi Exp(X), where Xi is the duration of runny nose recorded by patient i, measured in hours Assuming the null hypothesis from the last exercise holds, specify an expression for the standardised version Z of the estimator X of u such that ZN(0, 1). Then calculate the observed value z of the test statistic, and specify the critical region at level of significancea 0.04 Use the logic section of the virtual keyboard to enter intervals. In terms of Y- Xn a suitable test statistic is Z | The observed statistic z is (to 2 decimal places) The critical region is (write an interval or union of intervals with 2 decimal places)

Answer 1

Given,

X, ~ Exp(A)

Then,

$\mu = 1/\lambda$

$\sigma = 1/\lambda$

Using Central limit theorem,

$Y = \bar{X_n} \sim N(\mu = 1/\lambda , \sigma = 1/n \lambda)$

Test Statistic,  $Z = (\bar{X_n} - \mu) / \sigma = (\bar{X_n} -1 /\lambda) / (1 /n\lambda)$

For null hypothesis, rate = $1 /\lambda = 60$

Test Statistic,  

(x,ー60)/60/154) z

Z (58-60)// (60/154)--5.13

For
α 0.04
, and left tailed test, the critical region is z < -1.75 or ($\infty$, -1.75)