Question

A researcher thinks that more than 10% of the U.S. adult Twitter users get at least some news from Twitter. In order to support this claim, the researcher conducted a poll survey of 251 random U.S. adult Twitter users and asked if they get at least some news from Twitter. 38 out of 251 users answered that they got some news 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.

Answer 1

a)

H0 : proportion,p =0.10 ; 10% of the U.S. adult Twitter users get at least some news from Twitter

Ha: p > 0.10 ; more than 10% of the U.S. adult Twitter users get at least some news from Twitter

b)

n=251 which is  less than 10% of the population because here population is very big.

n*p = 251 * .1 = 25.1 > 10

n*q=251 * .9 = 225.9 >10

n*p*q =22.59 >10

So, follows approximately normal

c)

38 out of 251 users answered that they got some news on Twitter. i.e .

=38 / 251 = 0.1514

SD of proportion = = = 0.0189

test statistic, Z= ( - p) / SD of proportion = (0.1514 -0.10 ) / 0.0189 = 2.72

d)

Z=0.05 =1.645 , so, test statistic falls in  rejection region if test statistic > 1.645

p-value correspondinf to our test statistic = 1- P( Z> 2.72) = 1- .99674 = 0.00326

e)

Since our test statistic falls in the rejection region, or our p-value is very low (< 0.05) , So, there is enough evidence to reject the null hypothesis and say more than 10% of the U.S. adult Twitter users get at least some news from Twitter.

f)

So, 95 % confidence interval

= [ - Z/2=0.025 * SE (p) , + Z/2=0.025 * SE (p) ]

= [0.1514 - 1.96 * 0.0189 , 0.1514 + 1.96 * 0.0189 ]

=[0.1144 , 0.1884 ]

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