Question

Wild fruit flies have red eyes. A recessive mutation produces white-eyed individuals. A researcher wants to assess the proportion of heterozygous individuals. A heterozygous red-eyed fly can be identified through its off-spring. When crossed with a white-eyed fly it will have a mixed progeny.
A random sample of 100 red-eyed fruit flies was taken. Each was crossed with a whiteeyed fly. Of the sample flies, 11 were shown to be heterozygous because they produced mixed progeny.

a) Check this data for the conditions necessary for the calculation of a large-sample confidence interval. Does it comply OR should you use the plus-four interval only?

b) Determine a 95% confidence interval for the proportion

c) Also use a test of significance at 5% to test the hypothesis that the proportion of heterozygous red-eyed flies is different from 10 %?

d) Compare the answer from this test at 5% in c) to that from the 95% confidence interval in b). Would you necessarily expect the same answer?

Answer 1

Answer:

a. The conditions are met to use a large-sample confidence interval.

b. [0.048; 0.171]

c. Decision: Not reject the null hypothesis.

d. Check explanation.

Step-by-step explanation:

Hello!

You have the following study variable:

X: red-eyed fruit fly that shows to be heterozygous due to producing mixed progeny after being crossed with s white-eyed fruit fly.

n= 100

x= 11

sample proportion 'p= 0.11

a)

Binomial criteria:

1. The number of observation of the trial is fixed (In this case n = 100)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial

3. The probability of success in the same from one trial to another (In this case our "success" is that the fruit fly show heterozygous trough its progeny)

So X≈ Bi (n;ρ)

Considering that the sample is big enough (n≥30), you can apply the Central Limit Theorem and approximate the distribution of the sample proportion to normal. So you can use the large-sample confidence interval.

b. The formula for the confidence interval is:

'p ± $Z_{1-\alpha /2}$*$\sqrt{\frac{'p(1-'p)}{n} }$

$Z_{1-\alpha /2} = Z_{0.975} = 1.96$

0.11 ± 1.96*$\sqrt{\frac{0.11*0.89}{100} }$

[0.048; 0.171]

At a confidence level of 95%, you'd expect that the interval [0.048; 0.171] will contain the value of the population proportion of heterozygous red-eyed flies.

c.

H₀: ρ = 0.10

H₁: ρ ≠ 0.10

α: 0.05

Z= $\frac{'p - p}{\sqrt{\frac{p(1-p)}{n} } }$

Z= $\frac{0.11 - 0.10}{\sqrt{\frac{0.10*0.90}{100} } }$

Z= 0.33

The two tailed p-value for this test is 0.7414. The p-value is greater than the level of significance so the decision is to not reject the null hypothesis.

d.

To be able to compare a Confidence Interval several conditions shoud be met.

1) The interval and the hypothesis test should be made for the sale population parameter.

2) The hypothesis has to be two-tailed

3) Both levels (confidence and significance) should be complementary.

4) Both have to be made with the information of the same sample. (Remember that the variable is random and the values will change from sample to sample therefore it makes no sense to compare an interval and hypothesis of different samples. You could reach a wrong conclusion)

If the conditions are met, you check the calculated interval to see if it contains the value of the parameter under the null hypothesis. If the interval contains the value (In this case 0.10) then the hypothesis is to be supported. If the interval doesn't include the value, then the hypothesis is to be rejected.

I hope this helps!