Question

Past data indicate that the amount of money contributed by the working residents of large two cities to a volunteer rescue squad is normally distributed. Based on 9 observations, city 1 has a sample standard deciation of $1.10 and based on 6 observation from city 2, sample standard deviation is equal to $2.90. State and conclude your hypothesis at the 0.05 level of significance if the variance in city 1 is less than the variance in city 2.

Answer 1

`F=(s^2_1)/(s^2_2)=(1.10^2)/(2.9^2)=0.144`

`p_v =P(F_(8,5)<0.144)=0.0087`

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude tha the variance in city 1 is less than the variance in city 2

Explanation:

Data given

`n_1 = 9` represent the sampe size for group 1

`n_2 =6` represent the sample size for group 2

`s_1 = 1.10` represent the sample deviation for group 1

`s_2 = 2.90` represent the sample deviation for group 2

`\alpha=0.05` represent the significance level provided

Confidence =0.95 or 95%

Solution to the problem

System of hypothesis

We want to test if the variance in city 1 is less than the variance in city 2., so the system of hypothesis are:

H0:
`\sigma^2_1 \geq \sigma^2_2`

H1:
`\sigma^2_1 <\sigma^2_2`

Calculate the statistic

The statistic is given by:

`F=(s^2_1)/(s^2_2)`

Replacing we got:

`F=(s^2_1)/(s^2_2)=(1.10^2)/(2.9^2)=0.144`

We need to find the degrees of freedom. For the numerator we have
`n_1 -1 =9-1=8` and for the denominator we have
`n_2 -1 =6-1=5` and the F statistic present 8 degrees of freedom for the numerator and 5 for the denominator.

P value

`p_v =P(F_(8,5)<0.144)=0.0087`

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude tha the variance in city 1 is less than the variance in city 2