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) to support the claim, we need to do 1 sample t testing (one tail test)

Null hypothesis (Ho) : There is no significant difference between the mean rainfall and the average of 30*10^3 m^3

Alternate hypothesis (Ho) : The mean rainfall is greater than the average of 30*10^3 m^3

Running the test, we get the values

One-Sample T: C13

Descriptive Statistics

N	Mean	StDev	SE Mean	99% Lower Bound for μ
20	32105	5901	1320	28754

μ: mean of C13

Test

Null hypothesis	H₀: μ = 30000
Alternative hypothesis	H₁: μ > 30000

T-Value	P-Value
1.60	0.064

Since p-value is greater than the level of significance (0.01), we fail to reject the null hypothesis and conclude that the mean rainfall is not greater than 30 * 10^3 m^3

b) Checking for normality

Summary Report for Mean rainfall (*10л3) Anderson-Darling Normality Test A-Squared p-Value 0.22 0.820 32.105 5.901 34.823 0.013605 0.893643 20 22.200 1st Quartile 27.050 33.050 3rd Quartile 37.475 42.900 Mean StDev Variance Skewness Kurtosis Minimum Median Maximum 95% Confidence Interval for Mean 29.343 34.867 25 30 35 40 95% Confidence Interval for Median 27.829 35.777 95% Confidence Interval for StDev 8.619 4.488 95% Confidence Intervals Mean Median 28 30 32 34 36

The result from Anderson darling normality test shows that the p value is more than the level of significance of 0.05 (for this test only), we conclude that the data is normal. also the histogram shows data follows normality

c) If the true mean rainfall is 33 * 10^3 and the average we need to check for is 30*10^3 m^3

we find the difference of 3000 m^3 to be meaningful

Sample size = 20

Power and Sample Size

1-Sample t Test

Testing mean = null (versus > null)

Calculating power for mean = null + difference

α = 0.01 Assumed standard deviation = 5901

Results

Difference	Sample Size	Power
3000	20	0.415298

The power of the test is 0.41

d) To get the power of 0.9

sample size is

Power and Sample Size

1-Sample t Test

Testing mean = null (versus > null)

Calculating power for mean = null + difference

α = 0.01 Assumed standard deviation = 5901

Results

Difference	Sample Size	Target Power	Actual Power
3000	54	0.9	0.905446

Sample size should be 54