91Ó°ÊÓ

The report "Robot, You Can Drive My Car: Majority Prefer Driverless Technology" (Transportation Research Institute University of Michigan, www.umtri.umich.edu/what -were-doing/news/robot-you-can-drive-my-car-majority -prefer-driverless-technology, July \(22,2015,\) retrieved May 8 , 2017 ) describes a survey of 505 licensed drivers. Each driver in the sample was asked if they would prefer to keep complete control of the car while driving, to use a partially self-driving car that allowed partial driver control, or to turn full control over to a driverless car. Suppose that it is reasonable to regard this sample as a random sample of licensed drivers in the United States, and that you want to use the data from this survey to decide if there is evidence that fewer than half of all licensed drivers in the United States prefer to keep complete control of the car while driving. a. Describe the shape, center, and variability of the sampling distribution of \(\hat{p}\) for random samples of size 505 if the null hypothesis \(H_{0}: p=0.50\) is true. b. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.48\) for a sample of size 505 if the null hypothesis \(H_{0}: p=0.50\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.46\) for a sample of size 505 if the null hypothesis \(H_{0}: p=0.50\) were true? Explain why or why not, d. The actual sample proportion observed in the study was \(\hat{p}=0.44\). Based on this sample proportion, is there convincing evidence that fewer than \(50 \%\) of licensed drivers prefer to keep complete control of the car when driving, or is the sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.

Step by step solution

01

Shape

Since the sample size (505) is quite large, according to the Central Limit Theorem, the sampling distribution of the sample proportion \(\hat{p}\) will be approximately normally distributed.

02

Center

The center of the sampling distribution is given by the population proportion under the null hypothesis: \(\mu_{\hat{p}} = p = 0.50\).

03

Variability

The variability of the sampling distribution is given by the standard error, which is computed as: \[\text{SE}(\hat{p}) = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.50(1-0.50)}{505}} = \approx 0.022.\] #b. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.48\) for a sample of size 505 if the null hypothesis \(H_{0}: p=0.50\) were true? Explain why or why not. #

04

Z-score Calculation

To determine if a sample proportion of \(\hat{p} = 0.48\) is surprising, we can calculate the Z-score by using the formula \[Z = \frac{\hat{p} - p}{\text{SE}(\hat{p})} = \frac{0.48 - 0.50}{0.022} \approx -0.91.\]

05

Conclusion

The Z-score of -0.91 indicates that the sample proportion of 0.48 is approximately 0.91 standard errors below the population proportion, which is not too far from the expected value under the null hypothesis. Based on this Z-score, we would not be surprised to observe a sample proportion of 0.48 if the null hypothesis were true. #c. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.46\) for a sample of size 505 if the null hypothesis \(H_{0}: p=0.50\) were true? Explain why or why not. #

06

Z-score Calculation

We can follow the same procedure as before to determine if a sample proportion of \(\hat{p} = 0.46\) is surprising. The Z-score for this sample proportion is: \[Z = \frac{0.46 - 0.50}{0.022} \approx -1.82.\]

07

Conclusion

The Z-score of -1.82 indicates that the sample proportion of 0.46 is approximately 1.82 standard errors below the population proportion. This is quite far from the expected value under the null hypothesis and could be considered somewhat surprising, but not highly surprising. #d. The actual sample proportion observed in the study was \(\hat{p}=0.44\). Based on this sample proportion, is there convincing evidence that fewer than 50% of licensed drivers prefer to keep complete control of the car when driving, or is the sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation. #

08

Z-score Calculation

We can calculate the Z-score for the observed sample proportion \(\hat{p} = 0.44\): \[Z = \frac{0.44 - 0.50}{0.022} \approx -2.73.\]

09

P-value Calculation

Now we calculate the P-value, which is the probability of obtaining a sample proportion as extreme as or more extreme than the observed sample proportion if the null hypothesis is true. In this case, we want to find the probability of obtaining a sample proportion less than or equal to 0.44 if \(H_0: p = 0.50\) is true: \[P(\hat{p} \leq 0.44 | H_0) = P(Z \leq -2.73) \approx 0.0032.\]

10

Conclusion

Since the P-value of approximately 0.0032 is quite small (below any commonly used significance levels such as 0.05 or 0.01), we can conclude that there is convincing evidence that fewer than 50% of licensed drivers prefer to keep complete control of the car when driving. The observed sample proportion of 0.44 is not consistent with what we would expect to see if the null hypothesis were true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental principle in statistics that describes how the distribution of sample means will tend to follow a normal distribution, especially as sample sizes increase. This theorem is incredibly useful because it allows us to make inferences about population parameters, even when the underlying distribution is not normal.

For instance, in the problem where licensed drivers were surveyed on their preference for car control, despite not knowing the exact distribution of preferences in the entire population, we can still conclude that the sampling distribution of the sample proportion \(\hat{p}\) is approximately normal, thanks to the large sample size of 505. The CLT is crucial for this inference because it justifies the use of normal probability methods to calculate probabilities and z-scores associated with sample proportions.

Null Hypothesis

The null hypothesis (\(H_0\)) is a statement made for the purpose of statistical testing. It posits there is no effect or no difference, and it's what we assume to be true before collecting any data. In the context of our survey, the null hypothesis states that exactly half of all licensed drivers (\(p=0.50\)) prefer to keep complete control of their cars while driving. When we analyze sample data, we are essentially testing whether this hypothesis is likely to be true or if the evidence suggests it should be rejected.

The null hypothesis serves as a starting benchmark; it asserts a specific value of the population proportion, which can then be tested against the sample data we have collected to draw conclusions about the population based on sample evidence.

Standard Error

Standard error (\(SE\)) measures the variability or precision of a sample statistic as an estimate of the corresponding population parameter. For proportions, the formula for standard error is \(SE(\hat{p}) = \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the population proportion and \(n\) is the sample size. The standard error tells us how spread out the sampling distribution of the sample proportion would be if we took many random samples of the same size from the population.

In our problem, the standard error is used to assess the variability of sampling distribution of \(\hat{p}\) around the population proportion when the null hypothesis is true. It plays a crucial role in calculating the Z-score to determine how unusual or extreme the observed sample proportion is.

Z-score Calculation

The Z-score is a statistical measure that indicates how many standard deviations a data point (in this case, the sample proportion) is from the mean. It's calculated by subtracting the assumed population proportion under the null hypothesis from the sample proportion and then dividing by the standard error. The formula for this is \(Z = \frac{\hat{p} - p}{SE(\hat{p})}\).

A Z-score helps us understand where a sample proportion falls within the sampling distribution under the null hypothesis. It provides a way to quantify how extreme the sample proportion is. In the example of licensed drivers, we calculate Z-scores to assess how surprising the sample proportions (0.48, 0.46, and 0.44) are, given the null hypothesis that half prefer to keep complete control.

P-value Calculation

The P-value is the probability of obtaining test results at least as extreme as the results actually observed during the test, under the assumption that the null hypothesis is correct. It's a crucial concept in hypothesis testing because it helps determine whether to reject the null hypothesis. The lower the P-value, the stronger the evidence against the null hypothesis.

To find the P-value, we could look up the Z-score in standard normal distribution tables, or use software to find it for us. A small P-value (typically ≤ 0.05) indicates that the observed sample proportion is unlikely to have occurred by random chance if the null hypothesis were true, enabling us to conclude there's statistically significant evidence against the null hypothesis, as seen when we calculated a P-value of approximately 0.0032 for the sample proportion of 0.44 in our exercise.