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Solnce55 [7]
3 years ago
6

Please help me........

Mathematics
1 answer:
NARA [144]3 years ago
3 0

Answer:

idk

Step-by-step explanation:

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1. A flower shop wishes to add the valuable Waimea orchid to its product list. They purchase a large shipment of bulbs from a su
AysviL [449]

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:

<em>H</em>₀: There is no difference between the observed and expected value.

<em>Hₐ</em>: 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.

4 0
3 years ago
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