Eggs that are contaminated with salmonella can cause food poisoning among consumers. A large egg producer takes an SRS of 200 eggs from all the eggs shipped in one day. The laboratory reports that 11 of these eggs had salmonella contamination. Unknown to the producer, 0.2% (two-tenths of 1%) of all eggs shipped had salmonella. In this situation

0.2% is a parameter and 11 is a statistic.

11 is a parameter and 0.2% is a statistic.

both 0.2% and 11 are statistics.

both 0.2% and 11 are parameters.

Flag this QuestionQuestion 20.5pts Which of the following statements are true?

A statistic characterizes a population, whereas a parameter describes a sample.

A parameter characterizes a population, whereas a sttistic describes a sample.

You can have statistics and parameters from both samples and populations.

None of the above.

Flag this QuestionQuestion 30.5pts In a survey of sleeping habits, 8400 adults were selected randomly and contacted by telephone. Based on a medical associations recommended amount of sleep for adults, respondents were asked, Do you typically sleep more than six hours during the night? Of those surveyed, only 46% reported that they did. Which of the following is true with respect to this scenario?

8400 is the size of the population being studied.

46% is a statistic and represents an estimate of the unknown value of a parameter of interest.

46% is a parameter and represents an estimate of the unknown value of a statistic of interest.

None of the above.

Flag this QuestionQuestion 41pts As the number of estimates that you have for p increase, what happens to the distribution of all p?

It becomes more Normal.

Its center more closely approximates the true population proportion p.

Its standard deviation is approximately equal to the square root of p(1 p)/n.

All of the above.

Flag this QuestionQuestion 51pts Students conducted a survey and found out that 36% of their peers on campus had tattoos, but only 4% of their peers were smokers. If 100 students were surveyed, can these students use the Normal approximation to study the proportion of students in the population who have tattoos?

Yes, because both np and n(1 p) are less than 10.

Yes, because both np and n(1 p) are greater than 10.

No, because either np or n(1 p) are less than 10.

No, because either np or n(1 p) are greater than 10.

Flag this QuestionQuestion 61pts Students conducted a survey and found out that 36% of their peers on campus had tattoos, but only 4% of their peers were smokers. If 100 students were surveyed, can these students use the Normal approximation to study the proportion of students in the population who are smokers?

Yes, because both np and n(1 p) are less than 10.

Yes, because both np and n(1 p) are greater than 10.

No, because either np or n(1 p) are less than 10.

No, because either np or n(1 p) are greater than 10.

Flag this QuestionQuestion 71pts Students conducted a survey and found out that 36% of their peers on campus had tattoos, but only 4% of their peers were smokers. The standard error of p of tattooed students in the 100-student sample is

4%

5%

6%

It is not possible to tell from the information provided.

Flag this QuestionQuestion 81pts A news organization previously stated that three of four people believed that the state of the economy was the countrys most significant concern. They would like to test the new data against this prior belief. The most appropriate hypotheses are

H₀: p = 0.65, H_a: p > 0.65.

H₀: p = 0.65, H_a: p < 0.65.

H₀: p = 0.75, H_a: p > 0.75.

H₀: p = 0.75, H_a: p 0.75.

Flag this QuestionQuestion 91.5pts

Given the following information calculate the z test statistic, p-value and state whether or not to reject the null hypothesisbased on the signicance level.

H0:p=0.70,Ha:p<0.70

X=68,n=112,=0.05

Z = [ Select ] ["6.78", "2.14", "1.47", "-1.47", "-2.14"]

p-value = [ Select ] ["0.000", "0.016", "0.984", "0.032", "0.089"]

We [ Select ] ["reject", "fail to reject"] the null hypothesis.

Flag this QuestionQuestion 101.5pts

Given the following information calculate the z test statistic, p-value and state whether or not to reject the null hypothesisbased on the signicance level.

H0:p=0.39,Ha:p>0.39

X=44,n=89,=0.01

Z = [ Select ] ["-3.21", "-2.02", "3.21", "2.02", "1.65"]

p-value = [ Select ] ["0.008", "0.044", "0.022", "0.918", "0.003"]

We [ Select ] ["reject", "fail to reject"] the null hypothesis.