Question

1- 2. A certain STAT 410 instructor claims that the time to complete a particular STAT 410 homework assignment is normally distributed with (unknown) mean u and standard deviation o = 7.2 minutes. A random sample of n= 81 students is obtained.

Answer 1

a) The test statistic here is computed as:

$z^* = \frac{\bar X - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{76.2 - 75}{\frac{7.2}{\sqrt{81}}} = 1.5$

As this is a two tailed test, the p-value here is obtained from the standard normal tables as:
p = 2P(Z > 1.5) = 2*0.0668 = 0.1336

therefore 0.1336 is the required p-value here.

b) For 0.05 level of significance, we have from the standard normal tables here :
P(-1.96 < Z < 1.96) = 0.95

$\frac{\bar X_{crit} - 75}{\frac{7.2}{\sqrt{81}}} = 1.96$

$\bar X_{crit} = 76.568$

Therefore, $\bar X_{crit} = 75 - 1.96*\frac{7.2}{\sqrt{81}} = 73.432$

therefore the rejection region here is given as:

Reject H0 if:

$\bar X < 73.432 \ or \ \bar X > 76.568$

c) (i) given that the true mean is 74.6 minutes, the power of the test is defined as the probability of rejecting a false null hypothesis.

= 1 - Probability of not rejecting the null hypothesis

$= 1 - P( 73.432 < \bar X < 76.568)$

Converting it to a standard normal variable, we have here:

$= 1 - P( \frac{73.432 - 74.6}{\frac{7.2}{\sqrt{81}}} < Z < \frac{76.568 - 74.6}{\frac{7.2}{\sqrt{81}}})$

Getting it from the standard normal tables, we have here:

Therefore 0.079 is the required power of the test here. that is 7.9% is the power of the test here.

ii) Similar to the above part, the power is computed here as:

$= 1 - P( \frac{73.432 - 76}{\frac{7.2}{\sqrt{81}}} < Z < \frac{76.568 - 76}{\frac{7.2}{\sqrt{81}}})$

Getting it from the standard normal tables, we have here:

Therefore 0.2396 is the required power of the test here that is 23.96% here.