Question

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research firm, using a standardized one-week shopping plan (grocery list), makes identical purchases at 10 of each chain’s stores. The stores for each chain are randomly selected, and all purchases are made during a single week. It is found that the mean and the standard deviation of the shopping expenses at the 10 Miller’s stores are $121.92 and $1.40, respectively. It is also found that the mean and the standard deviation of the shopping expenses at the 10 Albert’s stores are $114.81 and $1.84, respectively. Assuming normality, test to see if the corresponding population standard deviations differ by setting α equal to .05. Is it reasonable to use the equal variances procedure to compare population means?(a) Calculate the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.)
Test statistic ___
(b) Calculate the critical value. (Round your answer to 2 decimal places.)

Critical value ____

Answer 1

Answer:

a) $F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$, the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation  

$n_1 = 10$ represent the sampe size for the Miller's Stores

$n_2 =10$ represent the sample size for the Albert's stores

$\bar X_1 =121.92$ represent the sample mean for Miller's store

$\bar X_2 =114.81$ represent the sample mean for Albert's store

$s_1 = 1.4$ represent the sample deviation for the Miller's store

$s_2 = 1.84$ represent the sample deviation for the Albert's stores

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

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

Solution to the problem  

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \new \sigma^2_2$

a) Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =10-1=9$ and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

$p_v =2*P(F_{9,9}>1.727)=0.428$

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$, the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.