Question

A crop scientist evaluating lettuce yields plants 20 plots, treats them with a new fertilizer, lets the lettuce grow, and then measures yield in numbers of heads per plot, with these results: 145, 142, 144, 141, 142, 155, 143, 157, 152, 143, 103, 151, 150, 148, 150, 162, 149, 158, 144, 151

The scientist is interested in testing whether the lettuce data are compatible with a population median of 150, or rather are strong evidence of a median less than 150.

Define if the following statement is true or false:

Using α =5%, the scientist can conclude that the median number of heads per plot is less than 150.

Answer 1

A crop scientist evaluating lettuce yields plants 20 plots, treats them with a new fertilizer, lets the lettuce grow, and then measures yield in numbers of heads per plot, with these results: 145, 142, 144, 141, 142, 155, 143, 157, 152, 143, 103, 151, 150, 148, 150, 162, 149, 158, 144, 151

Our hypothesis is:

H0: Median=150

H1:Median<150

Assign a value of 0 to those with lettuce yields below your hypothesized median, and 1, to those above.

103      0

141      0

142      0

142      0

143      0

143      0

144      0

144      0

145      0

148      0

149      0

150      0

150      0

151      1

151      1

152      1

155      1

157      1

158      1

162      1

Find the obtained frequency (of). There are 13 plants with lettuce yields below the median (the median is 150), and 7 above.

of = 13 and 7

Find the expected frequency (ef). The total number of data points plants—is 20. If 150 was a true median, we’d expect to have 10 plants over median and 10 below median

ef = 10

Calculate the chi-square:

Χ2 = Σ [(of – ef)^2/ef].

This is:

(13 – 10)^2/10 + (7 – 10)^2/10

or 0.9 + 0.9 = 0.18.

Find the degrees of freedom. There are two observed frequencies (equal to two cells in a contingency table), so there is just one degree of freedom.

Use a Chi-squared table to find the critical chi-square value

for 1 degree of freedom and an alpha level of 0.05 (α = 0.05). This equals 3.84.

Since our Χ2 value of 0.18 is less than the critical Χ2 value of 3.84, we fail to reject the null hypothesis (i.e. we can keep it) —that the median number of heads per plot is 150.

The given statement is false that is Using α =5%, the scientist can conclude that the median number of heads per plot is less than 150.