Question

Suppose that a poll taken among 100 students at a university turns out that 57 of them favor a diet containing meats, while 39 of them favor a vegetarian diet and 4 of them did not answer the poll with either of the two options (meat or vegetarian).

(i) Compute the probability that the observed sample average differs from the true proportion of students at the university that favor a meat diet in less than 3%.

(ii) Based on this poll, can we say with 90% confidence that THE MAJORITY of students at the university favor a meat diet?

Answer 1

proportion of students favurs a meat diet = p = 57/100 = 0.57

standard error of proportion = se = sqrt [p * (1-p)/n] = sqrt [0.57 * 0.43/100] = 0.0495

Here let say true proportion is $\mu$

then we have to find P($\mu$ - 0.03 < $\bar x$ < $\mu$ + 0.03)

z₂ = 0.03/0.0495 = 0.606

z₁ = -0.606

P($\mu$ - 0.03 < $\bar x$ < $\mu$ + 0.03) = P(Z < 0.606) - P(Z < -0.606)

= 0.7277 - 0.2723 = 0.4555

(ii)

Here 90% confidence interval for favor a meat diet = p +- z_critical se

= 0.57 +- 1.645 * 0.0495

= (0.4886, 0.6514)

so here the interval contains the value of 50% that means we can't say that THE MAJORITY of students at the university favor a meat diet.