Can someone please help me?!? 25 points
1 answer:
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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:
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:
The decision rule:
The null hypothesis will be rejected if .
Compute the critical value as follows:
The test statistic value is 14.981.
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.
To subtract fractions, we need a common denominator, which would be 10. To do this, multiply 3/5 by 2
3 * 2 = 6
5 * 2 = 10
Now subtract
3 * 2 = 6
5 * 2 = 10
Now subtract