Question

2c

Answer 1

2c

i.

Standard error of mean = 10 /$\sqrt{9}$ = 3.33

critical value c = 7.55

Then, c = t * SE

where z is the t statistic at level of significance

=> 7.55 = z * 3.33

=> z = 7.55 / 3.33 = 2.27

From z distribution table, confidence level = 0.988

For two level test, level of significance = 2 * (1 - confidence level ) = 2 * (1 - 0.988) = 0.024

ii.

Power = P(Reject H0 | $\mu$ = 151)

= P( T2 > 7.55 | $\mu$ = 151)

= P( |x - 152| > 7.55 |  $\mu$ = 151)

= P(x < 152 - 7.55 |  $\mu$ = 151) + P(x > 152 + 7.55 |  $\mu$ = 151)

= P(x < 144.45 |  $\mu$ = 151) + P(x > 159.55 |  $\mu$ = 151)

= P[z < (144.45 - 151) / 3.33] + P[z > (159.55 - 151) / 3.33]

= P[z < -1.96] + P[z > 2.57]

= 0.025 + 0.005

= 0.03

iii.

Power = P(Reject H0 | $\mu$ = 148)

= P( T2 > 7.55 | $\mu$ = 148)

= P( |x - 152| > 7.55 |  $\mu$ = 148)

= P(x < 152 - 7.55 |  $\mu$ = 148) + P(x > 152 + 7.55 |  $\mu$ = 148)

= P(x < 144.45 |  $\mu$ = 148) + P(x > 159.55 |  $\mu$ = 148)

= P[z < (144.45 - 148) / 3.33] + P[z > (159.55 - 148) / 3.33]

= P[z < -1.07] + P[z > 3.47]

= 0.1423 + 0.0003

= 0.1426