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

Answer:

$F=\frac{s^2_1}{s^2_2}=\frac{1.10^2}{2.9^2}=0.144$

$p_v =P(F_{8,5}$

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

$n_1 = 9$ 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: $\sigma^2_1$

Calculate the statistic

The statistic is given by:

$F=\frac{s^2_1}{s^2_2}$

Replacing we got:

$F=\frac{s^2_1}{s^2_2}=\frac{1.10^2}{2.9^2}=0.144$

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

$p_v =P(F_{8,5}$

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