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.

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

Step-by-step explanation:

Data given

[tex]n_1 = 9 [/tex] represent the sampe size for group 1

[tex]n_2 =6[/tex] represent the sample size for group 2

[tex]s_1 = 1.10[/tex] represent the sample deviation for group 1

[tex]s_2 = 2.90[/tex] represent the sample deviation for group 2

[tex]\alpha=0.05[/tex] 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: [tex] \sigma^2_1 \geq \sigma^2_2[/tex]

H1: [tex] \sigma^2_1 <\sigma^2_2[/tex]

Calculate the statistic

The statistic is given by:

[tex]F=\frac{s^2_1}{s^2_2}[/tex]

Replacing we got:

[tex]F=\frac{s^2_1}{s^2_2}=\frac{1.10^2}{2.9^2}=0.144[/tex]

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

P value

[tex]p_v =P(F_{8,5}<0.144)=0.0087[/tex]

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