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According to a survey in a country, 35% of adults do not own a credit card suppose a simple random sample of 200 adus is obta
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Answer #1

Given: p = 0.35

n = 200

(a) Approximately normal because n ≤ 0.05N and np(1 - p) ≥ 10

Mean of the sampling distribution of \hat{p} , \mu_\hat{p} = 0.35

Standard deviation of the sampling distribution of \hat{p} ,

\sigma_\hat{p} = p(1-p) V n = 0.034

(b) The required probability = P(\hat{p} > 0.37)

= P{Z > (0.37 - 0.35)/0.034} = P(Z > 0.588)

= 0.2883​​​​​​

If 100 different random samples of 200 adults were obtained, one would expect 29 to result in more than 37% not owning a credit card

(c) The required probability = P(0.32 < \hat{p} < 0.37)

= P(-0.882 < Z < 0.588)

= 0.5328

If 100 different random samples of 200 adults were obtained, one would expect 53 to result in between 32% and 37% not owning a credit card

(d) 64/200 = 0.32

Now, P(\hat{p} < 0.32) = 0.1789

The result is not unusual because the probability that \hat{p} is less than or equal to the sample proportion is 0.1789 which is greater than 5%

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