Question

In: Statistics and Probability

A gardener plants 300 sunflower seeds (of a brand called KwikGrow) and, after 2 weeks, measures...

A gardener plants 300 sunflower seeds (of a brand called KwikGrow) and, after 2 weeks, measures the seedlings’ heights (in mm). These heights are recorded below: height 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 38.8 41.9 18.4 41.4 48.7 16.3 50 13.8 50.3 47.6 42 14.9 41.1 43.7 10.1 36.4 40.1 14.9 50.3 12.3 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. He decides to conduct a hypothesis test by assuming that the sampling distribution of the sample mean has a normal distribution. For the purposes of this question, you may assume that the standard deviation of the sunflower heights is 13 mm.

a.Name one assumption that is required for the gardener to be able to use his sample to draw conclusions about KwikGrow seeds in general.

b.Name one key feature of the data that will assure the approximate validity of the gardener’s assumption that the sampling distribution of the sample mean has a normal distribution. Do you think his decision was wise considering the distribution of individual sunflower heights?

c.What are the null and alternative hypotheses?

d.What is the value of the test statistic?

e.What is the p-value (assuming the sample mean does indeed have a normal distribution)?

f.Using a significance level of alpha=0.05, state your conclusions in the language of the problem.

Solutions

Expert Solution

One-sample Kolmogorov-Smirnov test

D = 0.20244, p-value = 4.188e-11
alternative hypothesis: two-sided

So the observations do not follow normal distribution. However this is a continuous variable i.e.  seedlings’ heights and sample size is very large then from central limit theorem, we can approximate the sampling distribution of the sample mean by normal distribution.

and conclude that the mean height of sunflowers grown from KwikGrow seeds is greater than 33 mm two weeks after planting.


Related Solutions

A gardener plants 300 sunflower seeds (of a brand called KwikGrow) and, after 2 weeks, measures...
A gardener plants 300 sunflower seeds (of a brand called KwikGrow) and, after 2 weeks, measures the seedlings’ heights (in mm). These heights are recorded in the file Sunflower.csv. 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?...
An avid gardener plants a kiwi vine in the backyard. After many years, in spite of...
An avid gardener plants a kiwi vine in the backyard. After many years, in spite of luxuriant growth and many pollinating bees in the backyard, the tree does not bear fruit. All the blooms display lovely white flowers and a large carpel with long sticky stigmas in the center. What is a possible explanation for the problem?
1. Improvements in strength after a short period (such as 2 weeks) of resistance training are...
1. Improvements in strength after a short period (such as 2 weeks) of resistance training are mainly due to: Neural adaptations Both neural adaptations and increases in muscle size equally Increases in muscle size 2. What is "cross education" with respect to resistance training? The improvement in strength is greater in those with a history of strength training. Resistance training also results in improved endurance performance. Resistance training in one limb also results in muscular strength increase in the untrained...
2. (25pts) The decrease in cholesterol level after eating a certain brand of oatmeal for breakfast...
2. (25pts) The decrease in cholesterol level after eating a certain brand of oatmeal for breakfast for one month in people with cholesterol levels is Normally distributed. A random sample of 25 adults was selected and mean 8.5 and standard deviation 3. The brand advertises that eating its oatmeal for breakfast daily for one month will produce a mean decrease in cholesterol of more than 10 units for people with cholesterol levels. (α= 0.05) c. Report a 95% one-sided confidence...
A 2-year-old male was brought to the hospital after his behavior during the preceding several weeks...
A 2-year-old male was brought to the hospital after his behavior during the preceding several weeks indicated to his mother that he was experiencing otalgia of the right ear. The general physician examined his ears using an otoscope and observed myringitis. The physician referred the patient to an Otologist. Unfortunately 2 weeks later the infection had spread into the tympanic cavity causing otitis media. The tympanic membrane was surgically incised in a myringotomy procedure. 1. What was the patient experiencing?...
Your firm has a free cash flow of $300 at year 1, $360 at year 2, and $864 at year 3. After three years, the firm will cease to exist.
Your firm has a free cash flow of $300 at year 1, $360 at year 2, and $864 at year 3. After three years, the firm will cease to exist. As of today (i.e. at year 0), the firm is partially financed with a 1-year maturity debt, whose face value is $660 and interest rate is 10%. After the debt matures at year 1, the firm will not issue any more debt and will remain unlevered. Assume that the firm’s...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT