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Oduvanchick [21]
2 years ago
15

Do oddsmakers believe that teams who play at home will have home field advantage? Specifically, do oddsmakers give higher point

spreads when the favored team plays home games as compared to when the favored team plays away games? Two samples were randomly selected from three complete National Football League seasons (1989, 1990, and 1991). The first sample consisted of 50 games, where the favored team played in a home game, while the second sample consisted of 50 games, where the favored team played in an away game. The oddsmakers’ point spreads (which are the number of points by which the favored team is predicted to beat the weaker team) were then collected. The following hypotheses were tested: H0: µ1 = µ2 Ha: μ1 > μ2 Analyses were run. The following is the (edited) output for the test: Which of the following is an appropriate conclusion based on the output?1. The data provide sufficient evidence to reject H0; thus, we can conclude that the mean point spread for home games is higher than that of away games.2. The data provide sufficient evidence reject the H0; thus, we cannot conclude that the mean point spread of home games is higher than that of away games.3. The data do not provide sufficient evidence to reject H0; thus, we cannot conclude that the mean point spread of home games is higher than that of away games.4. The data do not provide sufficient evidence reject the H0; thus, we can conclude that the mean point spread for home games is higher than that of away games.

Mathematics
1 answer:
vladimir2022 [97]2 years ago
7 0

Answer:

Option 3 is correct.

The data do not provide sufficient evidence to reject H0; thus, we cannot conclude that the mean point spread of home games is higher than that of away games.

Step-by-step explanation:

Before anything else, we first give the null and alternative hypothesis for this question.

Null hypothesis would be that there isn't significant evidence to conclude that the mean point spread of home games is higher than that of away games.

H0: µ1 = µ2

And the alternative hypothesis would be that there is significant evidence to conclude that the mean point spread of home games is higher than that of away games.

Ha: μ1 > μ2

The data from the output of the analysis of the hypothesis test is missing from the question. It was obtained online and is attached to this solution of the question.

The table consists of the difference, the sample mean, the standard error of the mean, degree of freedom, the test statistic and most importantly, the p-value. It is the p-value that absolutely gives us the concluding statement of the hypothesis testing.

When the significance level for a test isn't provided, the convention is usually to use 5% significance.

Interpretation of p-value

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.4351

0.4351 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that 'The data do not provide sufficient evidence to reject H0; thus, we cannot conclude that the mean point spread of home games is higher than that of away games'.

Hope this Helps!!!

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Scorpion4ik [409]
<span><span>1.       </span>We have 2 boxes weighs:
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kirza4 [7]

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

An insurance company reported that, on average claims for a certain medical procedure are $942. an independent organization constructed a 95% confidence interval of ($930, $950) for the average amount claimed for the particular medical procedure. what conclusion best evaluates the truthfulness of the number reported by the insurance company?

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Confidence interval is used to express how confident we are that the population parameter that we are looking for is contained in a range of given values. Looking at the given confident interval, the lower limit is $930 and the upper limit is $950. We can see that the population mean, $942 lies within these values. The correct option would be

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