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Chapter 9: Problem 58

The police that patrol a heavily traveled highway claim that the average driver exceeds the 65 miles per hour speed limit by more than 10 miles per hour. Seventy-two randomly selected cars were clocked by airplane radar. The average speed was \(77.40\) miles per hour, and the standard deviation of the speeds was \(5.90\) miles per hour. Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.02\) ?

Short Answer

Expert verified
After getting the z-value, one should find corresponding p-value from the Z-table. If that p-value is less than 0.02, the null hypothesis is rejected, indicating the police's claim stands. If it isn't, then there isn't enough evidence to support the police's claim. Refer to step 5 for the final conclusion based on the calculated p-value.

Step by step solution

01

State Hypotheses

The null hypothesis is \(H_0: \mu = 75\) miles per hour, meaning that the average speed does not exceed the speed limit by more than 10 mph. The alternative hypothesis is \(H_1: \mu > 75\) miles per hour, representing the police claim that the average speed is indeed more than 75 mph.
02

Calculate the Test Statistic

Let's calculate the test statistic z. The formula for finding z in this case is \[ z = \frac{{\bar{x} - \mu_0}}{{s / \sqrt{n}}}\], where \(\bar{x}\) is the sample mean (77.40 mph), \(\mu_0\) is the assumed population mean (75 mph), s is the sample standard deviation (5.9 mph), and n is the sample size (72). Plugging our numbers into this equation gives \[ z = \frac{{77.40 - 75}}{{5.9 / \sqrt{72}}}\]. Calculate this to get the z-score.
03

Find the p-value

For such a Z-value, you need to refer to the standard Z-table to find the associated number. This value is the cumulative probability for the z value. To find the exact p value, subtract that value from 1. That gives the p-value.
04

Compare the p-value with alpha

Compare the obtained p-value with the significance level (\(\alpha = 0.02\)). If the p-value is less than \(\alpha\), you reject the null hypothesis.
05

Draw Conclusions

Based on the comparison between p-value and alpha in step 4, you will be able to conclude to either reject the null hypothesis, thus supporting the police's claim, or fail to reject the null hypothesis, thus not supporting the claim.

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Most popular questions from this chapter

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all long- distance calls made by residential customers is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02\). Does your conclusion change if \(\alpha=.05 ?\)

According to the American Diabetes Association (www.diabetes.org), \(23.1 \%\) of Americans aged 60 years or older had diabetes in 2007. A recent random sample of 200 Americans aged 60 years or older showed that 52 of them have diabetes. Using a \(5 \%\) significance level, perform a test of hypothesis to determine if the current percentage of Americans aged 60 years or older who have diabetes is higher than that in 2007 . Use both the \(p\) -value and the critical-value approaches.

Customers often complain about long waiting times at restaurants before the food is served. A restaurant claims that it serves food to its customers, on average, within 15 minutes after the order is placed. A local newspaper journalist wanted to check if the restaurant's claim is true. A sample of 36 customers showed that the mean time taken to serve food to them was \(15.75\) minutes with a standard deviation of \(2.4\) minutes. Using the sample mean, the journalist says that the restaurant's claim is false. Do you think the journalist's conclusion is fair to the restaurant? Use the \(1 \%\) significance level to answer this question.

According to a 2008 survey by the Royal Society of Chemistry, \(30 \%\) of adults in Great Britain stated that they typically run the water for a period of 6 to 10 minutes while taking the shower (http://www.rsc.org/ AboutUs/News/PressReleases/2008/EuropeanShowerHabits.asp). Suppose that in a recent survey of 400 adults in Great Britain, 104 stated that they typically run the water for a period of 6 to 10 minutes when they take a shower. At the \(5 \%\) significance level, can you conclude that the proportion of all adults in Great Britain who typically run the water for a period of 6 to 10 minutes when they take a shower is less than 30 ? Use both the \(p\) -value and the critical value approaches.

You read an article that states " 50 hypothesis tests of \(H_{0}: \mu=35\) versus \(H_{1}: \mu \neq 35\) were performed using \(\alpha=.05\) on 50 different samples taken from the same population with a mean of \(35 .\) Of these, 47 tests failed to reject the null hypothesis." Explain why this type of result is not surprising.
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