Question

(Lengths of leaves) The length (cm) of leaves of a particular plant species follows the normal distribution
expectation µ and standard deviation σ = 4.0. Botanist thinks that the unknown expectation value would be
16.0. Test the validity of this hypothesis when the botanist measured five species
plant leaves and got an average of 19.0. Use the Normalized average as the Test Size
z(x)=m(x)-µ₀/(σ/$\sqrt{n}$) ,where m ( x) is the mean of the observed data points and µ0 is the null hypothesis of the plant population
expected value.
(a) Determine the p-value of the test for the observation above.
(b) Is the null hypothesis rejected at significance level α = 5%? What about significance level α =10%?

Answer 1

Sample size = n = 5

Sample mean = $\bar{x}$ = 19

Population standard deviation = $\sigma$ = 4

The null and alternative hypothesis is

$H0:\mu=16$

$H1:\mu\neq 16$

Level of significance = 0.05

Here population standard deviation is known so we have to use z-test statistic.

Test statistic is

$z =\frac{\bar{x}-\mu }{\sigma /\sqrt{n}}$

$z =\frac{19-16 }{4 /\sqrt{5}}=\frac{3}{1.788854}=1.68$

a) P-value = 2*P(Z > 1.68) = 2*0.0468 = 0.0935

b)

P-value > 0.05 we fail to reject null hypothesis.

Conclusion: Botanist thinks that the unknown expectation value would be 16.0