Question

Decade	Sample	Cat 3-5 Hurricanes
1851-1860	1	6
1861-1870	2	1
1871-1880	3	7
1881-1890	4	5
1891-1900	5	8
1901-1910	6	4
1911-1920	7	7
1921-1930	8	5
1931-1940	9	8
1941-1950	10	10
1951-1960	11	8
1961-1970	12	6
1971-1980	13	4
1981-1990	14	5
1991-2000	15	5
2001-2010	16	7

What is the mean, mode, median, range?

Statistical analysis of the above data reveals a confidence interval of 1.10652102+/- the mean.
What are the upper and lower confidence limits?

Suppose 10 hurricanes were to occur this decade.  Based on the 95%CI, what is the probability
of such an occurrence?

Based on statistical theory of the normal distribution, what conclusion can you draw?

Answer 1

First of all we need to obtain sample mean and sample variance from the given sample of 16 observations:

sample mean = (6+1+7+5+8+4+5+7+8+10+8+6+4+5+5+7)/16 = 96/16 = 6

sample variance = (36+1+49+25+64+25+49+64+100+64+36+16+25+25+49)/16 - 6*6 = 644/16 - 36 = 40.25-36 = 4.25

Now,

The upper and lower confidence limit is : 1.10652102 +/- mean = (6-1.10652102, 6+1.10652102) = (4.8935, 7.1065)

If 10 hurricane were to occur, mu = 10

Null Hypothesis : 10 hurricanes to occur this year

Alternate Hypothesis : hurricanes not equal to 10, mu = 10

Level of significance = 5% ( CI 95% )

Z = |x_bar- mu|/sqrt(sample variance/n) , sqrt : square root

Z = (6-10)/sqrt(4.25/16) = 4/0.515 = 7.77

Probability of such occurrence = 0.000 , using normal table

also 95% CI are , x_bar +/- (Zcritical)*sqrt(sample variance/n) , sqrt : square root

n = 16, sample variance = 4.25 , Zcritical at 95% CI is 1.96

95% CI : 6 +/- (1.96*sqrt(4.25/16)) = (6 - 0.515*1.96, 6 + 0.515*1.96) = (6 - 1.01, 6 + 1.01) = (4.99, 7.01)

now, using criteria, reject H0 if Z > Zcritical

or Z (7.77) > Zcritical (1.96)

Hence we reject H0

this decade hurricanes will not be 10