A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States. Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football.
(a) Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.
(b) Based on your answer to part (a), is it reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football? Justify your answer.

95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football

Given that,

A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States.

Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football.

We have to determine,

Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.

According to the question,

Sample proportion p = 37% = 0.37

Sample space n = 1000

To compute the critical value of z for 95% confidence interval as follows:

By using a z-table for the value.

Compute the 95% confidence interval for the population proportion p as follows:

Hence,  95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.

Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33

The hypothesis can be tested based on a confidence interval.

The (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected.

And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.

The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

The confidence interval does includes the null value of p, i.e. 0.33.

So, the null hypothesis will be rejected.

Hence, Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football

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The 95% Confidence Interval we will found for given case is:

(0.34, 0.40)

It is Not reasonable to believe that  33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.

When there is n sample size and the probability of success is p with level of significance calculated as:

Where is the z score   .

1 -   = 0.95 ( since it is given to find 95% CI)

Thus,    = 0.05

The z score at p value = 1 -  /2 = 1 -0.025 = 0.975 is 1.96

Thus we have:

Lower limit of CI = 0.34 and Upper limit of CI = 0.40.

The case to decide to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football can be solved as:

33% = 0.33 probability of actually liking watching american probability.

Since 0.33 is lower than the lower limit of CI for 0.37 probability, thus we reject the null hypothesis that both 0.37 and 0.33 are approximately same, and thus decide that it is false that 0.33 percent believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football.

Thus it is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football.

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a)

b) No, it is not reasonable

Step-by-step explanation:

According to the survey, 37 percent indicated that their favorite sport to watch on television was American football.

This means that:

and

Since , we have:

Substitute in the formula:

This gives us:

We evaluate to get:

We simplify to get:

This gives:

Or

Part b)

From a) we obtain the confidence interval as .

This means we are 95% confident that, the actual percent of people in the United States whose favorite sport to watch on television is American football is between 34% and 40%.

Since 33% is less than 34%, it is not reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

the answer is below

Explanation: The independent variable is the experimental factor which is manipulated and the variable whose effects are being studied. The dependent variable is a variable that could change due to the independent variable. The independent variable is the different ways that Jennifer treats her puppies in order to bring them closer to what she wants in the agility competition. The dependent variable is how they puppies perform each day.

Behavioral shaping is a conditioning procedure where the reinforcers guide behavior toward closer and closer approximations of a desired behavior. Jennifer guided Sasha closer and closer to completing a behavior correctly, thus utilizing shaping in order to bring the puppy closer to what she wanted.

Answer Part A:

1. There is a significant negative relationship between anxiety and academic performance, described as:

a. Subjects with lower anxiety scores have higher current GPAs, or;

b. Subjects who have lower current GPAs have higher anxiety scores.

Explanation:

We can tell that the two variables (anxiety and academic performance) have a negative relationship from its correlation coefficient (-.64). The negative sign shows the relationship, while the statistical significance (p < .01 or p < .05) shows whether there is a relationship or not.

Remember, the power of the correlation (as represented by the number 0.64) does not tell you whether a relationship actually happens or not. It can, however tell the strength of a relationship. A .64 is usually described as moderate to strong, depending on which reference you are using.

Also remember that this is not causation - we do not know which variable causes changes at the other. This is why it is possible to describe the relationship with either option (A) or (B). It is preferable to use a neutral sounding one such as the one stated above: By only saying that there is a negative relationship between the two variables.  

2. There is no relationship between hours sleeping per night and academic performance.

Explanation:

A correlation coefficient of 0.00 means that there is no relationship between two variables. The closer a correlation coefficient to 0, the less likely do the variables have a relationship with one another.

Answer Part B:

1. It contributes to the interpretation of the findings by discussing whether:

(1) The measurement used is appropriate for the research conducted,

(2) There are any limitations from the current method of measurement.

Explanation:

The operational definitions in the study are not explicitly stated, but we can come to our own conclusions from the description of the study. They are:

Academic performance is the current GPA as reported by the research subjects.

Anxiety is the anxiety score summed from the research subjects' responses to the anxiety scale.

Interpretation is usually done during the discussion part of a research journal, after the statistical results have been reported. Discussing operational definitions would help determine whether the right measurement is used. Is the method free of biases? For example, Dr. Aguilera used a 5-point scale to measure the student's anxiety levels; there is a possibility of central tendency bias occurring.

2. Cohort effects may contribute to a possible bias over the result of the research. The variations in both anxiety levels and GPA may not be due to each other's influences, but perhaps instead caused by other age-related experience variables such as social media use.

Explanation:

Cohort effects are defined as a condition shared with same-aged groups of people which affected the result of the research. It is a form of bias, when it is not controlled. Dr. Aguilera's study is at risk of being affected with this since her research subjects are all university students.

In the study, we saw that Dr. Aguilera measured social media use, however, she did not seem to control this variable. It is possible that the variation in anxiety is moderated by social media, which then affects the subjects' GPA.

Something that is persistent remains over time; this is a key characteristic of DDT, which also happens to be a pollutant/pesticide. DDT is able to persist in an environment (as a pollutant) in part due to a phenomenon known as biological magnification. In simpler terms. once DDT enters an ecosystem/trophic structure, it gets worse as you go up the trophic levels; in other words, this means that the worst effects/concentrations of the pesticide will be felt in the uppermost trophic levels (your consumers rather than producers).

The basis for this, however, lies in the fact that organisms of a trophic structure/ecosystem consume each other; they feed on each other to survive (e.g. consumers feeding on producers, higher-level consumers feeding on lower-level consumers, etc.). Thus, once DDT gets into an ecosystem, it can only persist and spread in that ecosystem. Hope this helps :)

The proportion of the available motorcycles that Molly can afford is 60%.

The 30th percentile for the motorcycle is $11920

It should be noted that $25,000 is considered an outlier because it is greater than $24792

Since, mean, μ =  $14,000 and standard deviation, σ = $4,000, the standard variable Z or x = $15,000 will be:

Z = (x - μ)/σ

Z = (15000 - 14000)/4000

Z = 0.25

We then check the table and see that it implies that Molly can buy 60% of the available motorcycles.

The 30th percentile for the prices of motorcycles of this type corresponds to a value of Z = -0.52. This will then be:

x = μ + Z*σ

x = 14000 + (-0.52) × 4000

x = $11920

Lastly, the third quartile value when added to 1.5 x IQR will be equal to:

= 16698 + (1.5 × 5396) = $24792

Therefore, $25,000 is considered an outlier because it is greater than $24792

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Yes, the business journal is interpreting the probability correctly, because 3 out of 5 randomly selected Internet users that will use the mobile payment application, means that, 3/5, or 0.6 of probability. That is exactly the probability reported by the business journal.

Explanation:

Probability that Internet users in the United States will use a mobile payment app is 0.60.

To generate objective knowledge about any area it is important to use the scientific method. The scientific method uses measurement, analysis, observation, and experimentation to obtain objective data to validate the hypotheses and generate knowledge.

In this case, City University carried out a scientific investigation to verify that its students were happy throughout the time in the university. For the investigation, they used an instrument that evaluates satisfaction. The results indicate that the senior-students are happier than the first-year students, for that the university officials conclude that the students are happier when they attend university for a longer time. This is convenient to attract new students and not change to another study center.

I hope this information can help you.