Question

9-60. Cloud seeding has been studied for weather modification procedure (for an intere subject, see the article in Technometrics, "A sis of a Multiplicative Treatment Effect in Weather Mod tion," Vol. 17, pp. 161-166). The rainfall in cubic 0 clouds that were selected at random and seeded with sily nitrate follows: 22.2 x 10, 37.9 x 103, 24.4 x 103, 33.4 x 7.5 x 10, 23.2 x 103, 39.2 x 103, 28.9 x 103, 26.1 x 103, 34.4 x 10, 39.3 × 10, 33.4 × 10, 30.8 x 10, 30.4 x 10, 33.2x 103 26.9 х 10, 36.2x 10, 42.98 10, 32.9 10, and 38.9 x 103 (a) Can you support a claim that the mean rainfall from seeded many decades as a Ba c-meters from er 33.4 x 10 clouds exceeds 30 x 103 m3? Úse α = 0.01 . Find the P-value. (b) Check that rainfall is normally distributed (c) Compute the power of the test if the true mean rainfall is 33 x 103 m3 (d) What sample size would be required to detect a true mean rainfall of 33.5 x 103 m if we wanted the power o to be at least 0.9?

Answer 1

a) we will set up the null hypothesis

Ho : μ = 30 × 103111

$H_{a}:\mu\neq30\times 10^{3}m^{3}$

Under the null hypothesis the test statistics

$t=\frac{\bar{x}-\mu_{0}}{\frac{\hat{\sigma}}{\sqrt{n}}}$ where $\bar{x}=\frac{\sum_{i=1}^{n} x_{i}}{n}=32.10\times 10^{3}$ , and n=20

= 5.90 × 103 rl

$\therefore t=\frac{32.10\times 10^{3}-32.10\times 10^{3}}{\frac{5.90\times 10^{3}}{\sqrt{20}}} = 1.60$

The corresponding p.value = 0.127

Since calculated p.value = 0.127 which is greater then 0.01 therefore we will accept our null hypothesis.

b) The normality test is obtained with the qq-plot below

Probability Plot of x Normal Mean 32.10 StDev 5.901 20 0.216 P-Value 0.820 AD 90 C 60 U 50 + 40 45 20 25 30 35

the data is normally distributed as p.value = 0.82 for anderson darling.

c) We have

$Z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

The null hypothesis will be rejected if

$|\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}|\geq 2.58$ at 0.01 level of significance.

$|\frac{\bar{x} -30 \times 10^3 }{\frac{5.90 }{\sqrt{20}}}|\geq 2.58$

The null hypothesis will be rejected if  $\bar{x} \leq 26.6 \times 10^3$    or   $\bar{x} \geq 33.4 \times 10^3$

Power of the test is calculated

$P(\bar{x} \leq 26.6\times 10^3/_{\mu =33.5\times 10^3})=P(Z\leq\frac{26.6\times 10^3-33.5\times 10^3}{\frac{5.90}{\sqrt{20}}} )$

$P(\bar{x} \leq 26.6\times 10^3/_{\mu =33.5\times 10^3})=P(Z\leq -5.23)$

Also,

$P(\bar{x} \geq 33.4\times 10^3/_{\mu =33.5\times 10^3})=P(Z\geq\frac{33.4\times 10^3-33.5\times 10^3}{\frac{5.90}{\sqrt{20}}} )$

$P(\bar{x} \geq 33.4 \times 10^3/_{\mu =33.5\times 10^3})=P(Z\geq -0.075)$$=0.4721$

power of the test is = 0.000+0.4721=0.4721

d) Since power of the test = $=1-\beta=0.9$

which means $\beta=0.1$

= E

-P(Accept Ho/Hi) > 0.1

1.645 *5.9 2 rl = 97 0.1

The sample size must be 97