Question

In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike...

In a fishing lodge brochure, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. Suppose that last summer 59 out of a random sample of 81 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from 75% (either higher or lower)? Use α = 0.05.

What is the value of the sample test statistic? (Round your answer to two decimal places.)

(c) Find the P-value of the test statistic. (Round your answer to four decimal places.)

What is your favorite color? A large survey of countries, including the United States, China, Russia, France, Turkey, Kenya, and others, indicated that most people prefer the color blue. In fact, about 24% of the population claim blue as their favorite color.† Suppose a random sample of n = 55 college students were surveyed and r = 11 of them said that blue is their favorite color. Does this information imply that the color preference of all college students is different (either way) from that of the general population? Use α = 0.05.


What is the value of the sample test statistic? (Round your answer to two decimal places.)

(c) Find the P-value of the test statistic. (Round your answer to four decimal places.)

Homework Answers

Answer #1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.75
Alternative Hypothesis, Ha: p ≠ 0.75


Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.7284 - 0.75)/sqrt(0.75*(1-0.75)/81)
z = -0.45

P-value Approach
P-value = 0.6527
As P-value >= 0.05, fail to reject null hypothesis.

2)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.24
Alternative Hypothesis, Ha: p ≠ 0.24

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.2 - 0.24)/sqrt(0.24*(1-0.24)/55)
z = -0.69

P-value Approach
P-value = 0.4902
As P-value >= 0.05, fail to reject null hypothesis.


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