Question

A gardener plants 300 sunflower seeds (of a brand called KwikGrow) and, after 2 weeks, measures the seedlings’ heights (in mm). These height are: 38.6 39.1 50.4 49.2 46.2 0 43.1 49.6 16.1 17.9 11.6 50.2 36.5 50.6 40.3 36.3 14.3 40.8 12.1 43.7 47.2 49.9 37.9 49.1 53 47.7 13.8 38.3 49.2 50.6 49.6 52.3 19.8 12.8 46.9 35.3 38.7 39.3 12.4 51.9 36.9 20.8 51.7 44.3 49.1 10.7 14.9 48.2 14.8 38 41.4 39.4 11.9 13.8 0 35.1 37.3 47.5 12.5 11.8 16 15.7 38.1 58.6 51.2 37.4 36.4 40.8 35.2 15.5 55.9 42.1 47.9 41.1 38.5 51.1 41 53.8 41.5 38.6 48 50.4 17.9 50.8 39.2 13.5 35.9 12.4 16.5 47.9 38.4 11.7 49.4 44.7 45.8 14 40.4 48.9 44.6 17.6 12.4 43.1 18.3 20 17.1 50.1 57.6 50.9 50.2 18.3 0 14.5 40 49.3 51.9 16.1 47 14.6 48.7 38.1 39.7 39.3 0 37.1 13 17.4 37.7 41.3 39.6 18 17.4 38.3 48.9 54.9 41.5 13.8 36.6 56.7 35.6 42.4 0 49.7 11 17.1 47.4 17.1 14.2 43.3 52.4 14.4 18.7 16.7 50.9 15.2 12.4 14.6 43.5 46.8 45.6 42 49.8 50.4 51.5 14.7 50.9 15 34.8 52.3 35.5 13.6 44.6 14.1 47.5 16.3 40.9 52.6 44 49.8 0 40.3 50.9 18.6 56.9 40.1 49.3 45 0 38.7 14.4 15.2 48.9 53.6 42.6 14.6 39 49.5 42.3 54.5 12.5 14.2 51.5 41 12.3 51.2 43.2 17.8 34.8 50.1 53 53.1 13.4 16.5 17.7 45 39.2 47.2 37.9 45.6 7.6 49.4 48.7 40.2 15.6 50.5 48.2 0 41.8 45.6 40.9 38.2 0 52.1 9.3 17.9 36.8 39.6 11.3 48.5 55.6 38.1 37.8 52.4 40.5 46.5 38.7 15.4 53.1 20.8 49.2 49.5 42.6 10.1 45.8 42.6 42.2 36.5 45.2 41.5 43.3 39.9 17.1 15.7 32.9 40.8 38.5 39.8 52.2 15.9 13.6 20.7 44 13.7 0 50 13.6 38.7 51.5 16.4 37.2 15.1.

He is interested in testing whether the mean height of sunflowers grown from KwikGrow seeds is greater than 33 mm two weeks after planting.

a. What are the null and alternative hypotheses?

b. (1 mark) What is the value of the test statistic?

c. (1 mark) What is the approximate p-value? d. Which assumption is required for the p-value in c) to be valid? e. Using a significance level of alpha = 0.05, state your conclusions in the language of the problem.  

Answer 1

Her gardener plants is not 300 it is a 280 sunflower seeds

First enter Data into EXCEL

We have to find the sample mean.

Excel command is =AVERAGE(Select data)

sample mean = 34.4107

Now we have to find sample standard deviation.

Excel command is =STDEV(Select data)

standard deviation = 15.1938

n = 280

s = 15.1938

a)

Null and alternative hypothesis is

H0 : u = 33

H1 : u > 33

Level of significance   = 0.05

b)

Here population standard deviation is not known so we use t-test statistic.

Test statistic is

c)

Degrees of freedom = n - 1 = 280 - 1 = 279

P-value = 0.0607 ( using t table)

P-value > ? ,Failed to Reject H0

d)

at a significance level of alpha = 0.05,data are not significant