Question

2) we want to develop a 90% confidence interval for the fraction of fruit flies which possess a gene for red eyes. To this end we obtain a simple random sample of 200 fruit flies, we find that 35% of the flies in the sample possess the gene. Give a 90% confidence interval for the fraction of all fruit flies with the gene a) 0.29 to 0.41 b) 0.28 to 0.42 c) 0.26 to 0.44 d) 0.31 to 0.39 e) 0.25 to 0.45

Answer 1

Q.1) Given that, n = 100 and x = 70

sample proportion $(\hat p)$ = 70/100 = 0.7

The null and alternative hypotheses are,

$H_0: p = 0.65 \text { vs }H_1 : p > 0.65$

Test statistic is,

$Z = \frac {0.7 - 0.65}{\sqrt {\frac {0.65*(1-0.65)}{100}}} = 1.05$

p-value = P(Z > 1.05) = 1 - P(Z < 1.05)= 1 - 0.8531 = 0.1469

=> p-value = 0.1469

Answer : d) 0.1469

Q.2) Given that, n = 200 and

sample proportion $(\hat p)$ = 0.35

A 90% confidence level has significance level of 0.10 and critical value is, $Z_{\alpha/2} = 1.645$

The 90% confidence interval for the fraction of all fruit flies with the game is,

$0.35 - 1.645 * \sqrt {\frac {0.35*(1-0.35)}{200}} < p < 0.35 &plus; 1.645 * \sqrt {\frac {0.35*(1-0.35)}{200}}$

$0.35 - 0.06 < p < 0.35 &plus; 0.06$

$0.29 < p < 0.41$

Answer: a) 0.29 to 0.41