Question

Random samples of two species of iris gave the following petal lengths (in cm). x1, Iris virginica 5.5 5.9 4.5 4.9 5.7 4.8 5.8 6.4 5.3 5.9 x2, Iris versicolor 4.5 4.6 4.7 5.0 3.8 5.1 4.4 4.2 (a) Use a 5% level of significance to test the claim that the population standard deviation of x1 is larger than 0.55. What is the level of significance? State the null and alternate hypotheses. H0: σ = 0.55; H1: σ > 0.55 H0: σ = 0.55; H1: σ < 0.55 H0: σ > 0.55; H1: σ = 0.55 H0: σ = 0.55; H1: σ ≠ 0.55 Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom? What assumptions are you making about the original distribution? We assume a binomial population distribution. We assume a normal population distribution. We assume a uniform population distribution. We assume a exponential population distribution. Find or estimate the P-value of the sample test statistic. P-value > 0.100 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 0.005 < P-value < 0.010 P-value < 0.005 Will you reject or fail to reject the null hypothesis? Since the P-value > α, we fail to reject the null hypothesis. Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis. Since the P-value ≤ α, we fail to reject the null hypothesis. Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient evidence to conclude conclude that the standard deviation is greater than 0.55. At the 5% level of significance, there is sufficient evidence to conclude conclude that the standard deviation is greater than 0.55. (b) Find a 90% confidence interval for the population standard deviation of x1. (Round your answers to two decimal places.) lower limit upper limit (c) Use a 1% level of significance to test the claim that the population variance of x1 is larger than that of x2. Interpret the results. What is the level of significance? State the null and alternate hypotheses. H0: σ12 = σ22; H1: σ12 > σ22 H0: σ12 > σ22; H1: σ12 = σ22 H0: σ22 = σ12; H1: σ22 > σ12 H0: σ12 = σ22; H1: σ12 ≠ σ22 Find the value of the sample F statistic. (Round your answer to two decimal places.) What are the degrees of freedom? dfN = dfD = What assumptions are you making about the original distribution? The populations follow dependent normal distributions. We have random samples from each population. The populations follow independent normal distributions. We have random samples from each population. The populations follow independent chi-square distributions. We have random samples from each population. The populations follow independent normal distributions. Find or estimate the P-value of the sample test statistic. P-value > 0.100 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 0.001 < P-value < 0.010 P-value < 0.001 Will you reject or fail to reject the null hypothesis? At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence that the variance in iris petal length is greater in the first plot. Reject the null hypothesis, there is insufficient evidence that the variance in iris petal length is greater in the first plot. Reject the null hypothesis, there is sufficient evidence that the variance in iris petal length is greater in the first plot. Fail to reject the null hypothesis, there is insufficient evidence that the variance in iris petal length is greater in the first plot.

Answer 1

a)

H0: σ = 0.55; H1: σ > 0.55

b)

test statistic X2 =(n-1)s2/ σ2=

(10-1)*0.349/0.303=

10.38

degree of freedom=n-1=9

We assume a normal population distribution.

P-value > 0.100
Since the P-value > α, we fail to reject the null hypothesis.

At the 5% level of significance, there is insufficient evidence to conclude conclude that the standard deviation is greater than 0.55.

b)

Critical value of chi square distribution for n-1=9 df and 90 % CI
Lower critical value χ2L=		3.325	from excel: chiinv(0.95,9)
Upper critical valueχ2U=		16.919	from excel: chiinv(0.05,9)

for Confidence interval of standard deviation:
Lower bound =√((n-1)s2/χ2U)=	0.43
Upper bound =√((n-1)s2/χ2L)=	0.97

c)

H0: σ12 = σ22; H1: σ12 > σ22

value of the sample F statistic =(s1/s2)^2 =1.97

dfN = 9

dfD = 7

The populations follow independent normal distributions.

P-value > 0.100

At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

Fail to reject the null hypothesis, there is insufficient evidence that the variance in iris petal length is greater in the first plot.