Question

The Russet Potato Company has been working on the development of a new potato seed that is hoped to be an improvement over the existing seed that is being used. Specifically, the company hopes that the new seed will result in less variability in individual potato length than the existing seed without reducing the mean length. To test whether this is the case, a sample of each seed is used to grow potatoes to maturity. The following information is given:Old SeedNumber of Seeds = 11Average Length = 6.25 inchesStandard Deviation = 1.0 inchesNew SeedNumber of Seeds = 16Average Length = 5.95 inchesStandard Deviation = 0.80 inchesBased on these data, if the hypothesis test is conducted using a 0.05 level of significance, the calculated test statistic is:______

Answer 1

Answer:

$F=\frac{s^2_2}{s^2_1}=\frac{0.8^2}{1.0^2}=0.64$

$p_v =P(F_{15,10}<0.64)=0.2105$

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 variation for the New sample it's significantly less than the variation for the Old sample at 5% of significance.  

Step-by-step explanation:

Data given and notation  

$n_1 = 11$ represent the sampe size for the Old

$n_2 =16$ represent the sample size for the New

$\bar X_1 =6.25$ represent the sample mean for Old

$\bar X_2 =5.95$ represent the sample mean for the New

$s_1 = 1.0$ represent the sample deviation for Old

$s_2 = 0.8$ represent the sample deviation for New

$\alpha=0.05$ 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 New sample it's lower than the variation for the Old sample, so the system of hypothesis are:

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

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

Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{0.8^2}{1.0^2}=0.64$

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 $n_2 -1 =16-1=15$ and for the denominator we have $n_1 -1 =11-1=10$ and the F statistic have 15 degrees of freedom for the numerator and 10 for the denominator. And the P value is given by:

P value

Since we have a left tailed test the p value is given by:

$p_v =P(F_{15,10}<0.64)=0.2105$

And we can use the following excel code to find the p value:"=F.DIST(0.64,15,10,TRUE)"

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 variation for the New sample it's significantly less than the variation for the Old sample at 5% of significance.