Question

Camellia plants were grown in two different settings - low light and high light - and the leaf widths of the plants were measured (in millimeters). For the plants growing in low light areas, 25 leaves sampled returned a mean of 47.7 mm and a standard deviation of 3.96 mm. The sample of 25 plants in high-light areas had a mean leaf width of 55.8 mm with standard deviation of 7.26 mm. At the 5% significance level, do camellia plants in high light areas have significantly wider leaves?

(a) State the null and alternate hypothesis (symbolic or sentence form).

(b) State the p-value.

(c) Draw a conclusion about the claim.

Answer 1

n1 = 25

n 2= 25

s1 = 3.96

s2 = 7.26

claim : camellia plants in high light areas have significantly wider leaves

a)

Null and alternative hypothesis is

Vs

Level of significance = 0.05

Before doing this test we have to check population variances are equal or not.

Null and alternative hypothesis is

Vs

Test statistic is

F = Larger variance / Smaller variance = 52.7076 / 15.6816 = 3.361

Degrees of freedoms

Degrees of freedom for numerator = n1 - 1 = 25 - 1 = 24

Degrees of freedom for denominator = n2 - 1 = 25 - 1 = 24

Critical value = 1.984                ( using f-table )

F test statistic > critical value    we reject null hypothesis.

Conclusion: Population variances are unequal.

So we have to use unpooled variance.

b)

Test statistic is

               (Assume H0 is True )

p-value =0.00001   

p-value , Reject H0

c)

conclusion : There is a sufficient evidence to conclude that the camellia plants in high light areas have significantly wider leaves