Question

1. A flower shop wishes to add the valuable Waimea orchid to its product list. They purchase a large shipment of bulbs from a supplier in Kauai. It is established by Mendelian theory that the predominant colors in the Waimea orchid (blue, red, violet, orange) will occur with the ratio of 7:5:4:4. When the first 60 Waimea orchid bulbs bloom, the predominant colors are 29 blue, 18 red, 22 violet, 6 orange. The florists are concerned that these bulbs are not Waimea orchids, but a similar appearing (and more common, less valuable) hybrid. Use Minitab to conduct an appropriate hypothesis test at α = 0.05 to test this.

Answer 1

Answer:

The concern of the florist that these bulbs are not Waimea orchids, but a similar appearing hybrid maybe true.

Step-by-step explanation:

A Chi-square goodness of fit test can be used to perform the hypothesis test.

The hypothesis is defined as:

H₀: There is no difference between the observed and expected value.

Hₐ: There is a difference between the observed and expected value.

The test statistic is:

$\chi^{2}=\sum \frac{(O-E)^{2}}{E}$

The total number of Waimea orchid bulbs is, 60.

7x + 5x + 4x + 4x = 60

20x = 60

x = 3

The expected number of Waimea orchid bulbs are:

Blue = 7 × 3 = 21

Red = 5 × 3 = 15

Violet = 4 × 3 = 12

Orange = 4 × 3 = 12

Consider the table provided.

The test statistic value is:

$\chi^{2}=\sum \frac{(O-E)^{2}}{E}=14.981$

The decision rule:

The null hypothesis will be rejected if $\chi^{2}\geq \chi^{2}_{\alpha, (k-1)}$.

Compute the critical value as follows:

$\chi^{2}_{0.05, (4-1)}= \chi^{2}_{0.05, (3)}=7.815$

The test statistic value is 14.981.

$\chi^{2}=14.981> \chi^{2}_{\alpha, (k-1)}=7.815$

The null hypothesis will be rejected at 5% level of significance.

Conclusion:

As the null hypothesis was rejected it implies that there is a significant difference between the observed and expected value.

Thus, the concern of the florist that these bulbs are not Waimea orchids, but a similar appearing hybrid maybe true.