Statistics: Confidence level and hypothesis testing

Question 32

We will test if mean of rainfall=11.45

H0: u=11.45

Ha: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

104775027559000447675275567

77152518542000140970027114500

37147510414000

-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

55,200,105,210,95,225,215,110,235,195,110,220,115,215,105,235,115,65

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

Ha: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

x20.05(df=17)=27.587

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

X2=n-1S2σ2=17*62.208024225=15.5701(4.d.p)

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