A Survey of 1, 695 randomly selected adults showed that 595 of the have heard of

Question

A Survey of 1, 695 randomly selected adults showed that 595 of the have heard of a new electronic reader The accompanying technology display results from a test of the claim that 38% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 level to complete parts (a) through (d. a. Is the test two-tailed, left-tailed, or right-tailed? Right tailed test Left-tailed test Two-tailed test b. What is the P-value? P-value = ___ (Round to four decimal places as needed.) c. What is the null hypothesis and what do you conclude about it? identify the null hypothesis A. H_0: p 0.38 C. H_0: p = 0.38 D. H_0: p notequalto 0.38

Explanation / Answer

Below are the null and alternate hypothesis for the given problem,
H0: p = 0.38
H1: p not equals to 0.38

This can also be written in the form of mean
H0: mu = 644.1
H1: mu not equals to 644.1

From the hypothesis, we can conclude that this is two tailed test.

Using normal approximation to binomial,
mean, mu = 0.38*1695 = 644.1
std. dev. = sqrt(1695*0.38*0.62) = 19.984

z = (595 - 644.1)/19.984 = -2.457
P(z<-2.457) = 0.007

Hence p-value = 0.007*2 = 0.014

As p-value is less than significance level, we reject the null hypothesis.

Lets solve the same in proportion form,
z = (0.3510 - 0.38)/sqrt(0.38*0.62/1695) = -2.46
P(z<-2.46) = 0.007

Hence p-value = 2*0.007 = 0.014

Below are the correct answers for each question

(a) Two tailed Test
(b) p-value = 0.014
(c) H0: p = 0.38

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