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In: Statistics and Probability

A researcher thinks that more than 10% of of U.S. adult Twitter users get at least...

A researcher thinks that more than 10% of of U.S. adult Twitter users get at least some news on Twitter. In order to support the claim, the researcher conducted a poll survey of 251 random U.S adult Twitter users and asked if they get at least some news on Twitter. 38 out of 251 users answered that they got some new on Twitter. Use alpha = 0.05. (a) Designate the null and alternative hypotheses. (b) Verify the condition. (c) What is the test statistic? (d) What is the rejection region (or p−value)? (e) State the conclusion in the context of the original problem. (f) Construct a 95% confidence interval for the true mean foam height.

Solutions

Expert Solution

(a)

Ho: p = 0.10

Ha: p > 0.10

Null hypothesis states that proportion of U.S. adult Twitter users who get at least some news on Twitter is 0.10

Alternative hypothesis states that proportion of U.S. adult Twitter users who get at least some news on Twitter is greater than 0.10

(b) Conditions:

np ≥ 10

np = 251 * 0.10 = 25 ≥10

n(1-p) ≥ 10

n(1-p) = 251 (1-0.90) = 23

As both the conditions are met, we can conduct One sample proportion test

(c)

z = 2.714

(d) z critical for alpha = 0.05

Critical Value of Z (Right Tailed): 1.65

Rejection region : {z,  z stat is greater than 1.65} i.e. z stat falls in the rejection area.

Or if p value is less than 0.05

(e) As z stat (2.714) falls in the rejection area, we reject the Null hypothesis.

Hence we have sufficient evidence to belive that the proportion of U.S. adult Twitter users who get at least some news on Twitter is greater than 0.10

(f) CI = +/- E

= x/ n = 38/ 51 = 0.1514

E = 0.0443

CI = 0.1514 + /- 0.0443

CI = 0.1071 , 0.1957

orchestra answered 2 years ago
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