(ii) The head of the statistics department in a certain university believes that

Question

(ii) The head of the statistics department in a certain university believes that 70% of the department’s graduate assistantships are given to international students. A random sample of 50 graduate assistants is taken. a. Assume that the chairman is correct and p = 0.70. What is the sampling distribution of the sample proportion pˆ ? b. Find the expected value and the standard error of the sampling distribution of pˆ . c. What is the probability that the sample proportion pˆ will be between 0.65 and 0.73? d. What is the probability that the sample proportion pˆ will be within ±0.05 of the population proportion p?

Explanation / Answer

Answer to part a)

Since the probability is 0.70 and n = 50 , the sampling distribution is considered to be Binomial distribution

.
Answer to part b)

The expected value E = n*p

E = 50 * 0.70 = 35

Standard deviation = Square root [n*p*(1-p)]

SD = square root [0.7 *0.3/50]

SD = 0.0648

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Answer to part c)

Using central limit theorem , we can approximate this distribution to normal

We use the continuity correction 0.5/50 = 0.01

Thus to find P(0.65 < P < 0.73) We find P(0.65-0.01 < P < 0.73+0.01) = P(0.64 < P < 0.74)

P(0.64 < P < 0.74) = P(P < 0.74) - P(P < 0.64)

P(P<0.74) = (0.74 - 0.70) / 0.0648

P(P<0.74) = 0.62

P(P<0.74) = 0.7324

.

P(P<0.64) = (0.64-0.70)/0.0648

P(P<0.64) = -0.93

P(P<0.64) = 0.1762

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Thus P(0.64 < P < 0.73) = 0.7324 - 0.1762 = 0.5562

.

Answer to part d)

E = 0.05

n = 50

Z = 1.96

n = p (1-p) (Z/E)^2

50= p(1-p) (1.96/0.05)^2

0.0325 = p(1-p)

p^2-p+0.0325 =0

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