Statistics Confidence level and hypothesis testing

Statistics: Confidence level and hypothesis testing

Question 32

We will test if mean of rainfall=11.45

H0: u=11.45


Significance level =0.05

Z-value= α/2=0.05/2=0.025

The significance value of 0.1 a two tail is -1.96 and 1.96




-1.96 Ho 1.96

Z score=x-uσ=11.45-7.51.4=2.8214(4d.p)

Since the z-score is greater than 1.96 we reject null hypothesis and conclude that the average amount of rainfall during summer months for the northeast part is not 11.45 inches

Question 33


Sample Mean (μ)= 282518=156.9444(4d.p)

Sample Standard deviation=σ=62.2080(4d.p)

We will test if variance σ2=652=4225

H0: σ2=4225


Degree of freedom =n-1=18-1=17

Significance level=0.01

x21-α2=x21-0.12=x20.95(df=17)=8.672(from the x2 with 17 degree of freedom


From the x2 table with 17 degree of freedom we get the area on the left x20.95=8.672 and on right x20.05=27.587


Since the test statistic is inside the critical values we reject null hypothesis and conclude variance is not significantly different from the claimed value 4225

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