Answer to Question #258856 in Statistics and Probability for boo

Question

Suppose a study conducted on where people turn for news was based on 400 respondents. Of the

400 respondents, 199 got their news primarily from newspapers.

(a) Calculate the point estimate and standard error of the proportion of people who get their news

primarily from newspapers.

(b) Is there evidence that the proportion of people who get their news primarily from newspapers

is 0.5 or more? (use 𝛼 = 0.01)

(c) Determine the 𝑝-value in (b).

(d) Construct a 99% confidence interval estimate of the true population proportion of respondents

who get their news primarily from newspapers.

Answer 1

a)

point estimate is proportion of respondents who got their news primarily from newspapers:

$p=199/400=0.4975$

standard error:

$SE=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4975(1-0.4975)}{400}}=0.025$

b)

the null hypothesis: $H_0:p\ge0.5$

the alternative hypothesis: $H_a:p<0.5$

$z=\frac{0.5-p}{SE}=\frac{0.5-0.4975}{0.025}=0.10$

$P(z>0.10)=1-P(z<0.10)=1-0.5398=0.4602$

Since $P(z>0.10)>\alpha=0.01$ , we accept the null hypothesis. So, there is evidence that the proportion of people who get their news primarily from newspapers is 0.5 or more.

c)

p-value:

$1-P(z<0.10)=1-0.5398=0.4602>\alpha=0.01$

d)

$-2.576<z<2.576$

$-2.576<\frac{p-0.4975}{0.025}<2.576$

$-2.576\cdot0.025+0.4975<p<2.576\cdot0.025+0.4975$

$0.4331<p<0.5619$