2024-11-01
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Steps for Success:
Generate a hypothesis (null hypothesis) about a population
Determine the alternative of that hypothesis
Set the confidence interval for the testing
Determine critical z-values based on the confidence interval
Calculate the z test score
Decide whether to accept or reject the null hypothesis
The hypothesis asks a question about a population.
Hypothesis for examination: Is it reasonable to claim that the average budget of a movie in IMDB is $12.5 million?
Null Hypothesis: It is reasonable to claim that the average budget of a movie in IMDB is $12.5 million.
Alternate Hypothesis: The average budget of a movie in IMDB cannot be reasonably claimed to be $12.5 million.
The confidence interval ensures that a result is statistically significant
With a proper confidence interval results are repeatable
The most common confidence interval is \(\alpha = .05\)
For this analysis the confidence interval will be \(\alpha = .05\)
Before beginning the analysis, here is a preliminary look at the distribution of budgets. The vast majority of movies have a low budget, but the average budget is unclear from this plot.
This plot makes it more clear that the mean is between $10-15 million
The critical Z values are determined based on the confidence interval selected and type of hypothesis tested.
Since the confidence interval is \(\alpha = .05\) and this is a two-tailed test, the critical Z values will be at \(\alpha / 2\) on both sides of the curve.
Use a table of values to determine the Z values at these points. For \(\alpha = .05\), \(Z < -1.96 \quad \text{and} \quad 1.96 < Z\).
The Z score is calculated according to the formula \(Z = (x - \mu) / (\sigma / \sqrt{n})\) x is the mean of the sample being evaluated, \(\mu\) is the mean of the population, \(\sigma\) is the standard of deviation of the population, n is the sample size R has a library named BSDA that has statistical tests, including Z tests, than can be installed and imported Run the test with the mean of $12.5 million that is being analyzed
z_test_result <- z.test(movies_filtered$budget,
mu = 12500000,
alternative = "two.sided",
sigma.x = sd(movies_filtered$budget),
conf.level = 0.95)
print(z_test_result)
## ## One-sample z-Test ## ## data: movies_filtered$budget ## z = 3.0625, p-value = 0.002195 ## alternative hypothesis: true mean is not equal to 12500000 ## 95 percent confidence interval: ## 12858322 14132323 ## sample estimates: ## mean of x ## 13495323
The Z score of 3.0625 is greater than the critical z value of 1.96, thus the null hypothesis would be rejected.
This is reflected in the p-value of 0.002195. Since \(.002195 < .05\), the result is not within the confidence interval.
Thus, it is not reasonable to state that the mean budget of movies on IMDB is $12.5 million.
This plot highlights how the confidence interval influences what estimates would be statistically valid.
