Math  /  Data & Statistics

QuestionSpeeding Tickets A motorist claims that the South Boro Police issue an average of 60 speeding tickets per day. The following data show the number of speeding tickets issued each day for a period of one month. Assume σ\sigma is 13.42 . Is there enough evidence to reject the motorist's claim at α=0.10\alpha=0.10 ? Use the PP-value method. Assume the variable is normally distributed. \begin{tabular}{llllllll} 57 & 60 & 83 & 26 & 72 & 58 & 87 & 48 \\ 59 & 60 & 56 & 64 & 68 & 42 & 57 & 58 \\ 63 & 49 & 73 & 75 & 42 & 63 & 57 & 60 \\ 72 & 45 & & & & & & \end{tabular} Send data to Excel
Part: 0/50 / 5
Part 1 of 5 (a) State the hypotheses and identify the claim. H0: (Choose one) H1: (Choose one) \begin{array}{l} H_{0}: \square \text { (Choose one) } \nabla \\ H_{1}: \square \text { (Choose one) } \nabla \end{array}
This hypothesis test is a (Choose one) \boldsymbol{\nabla} test. \square

Studdy Solution

STEP 1

What is this asking? A driver says the South Boro Police give out 60 speeding tickets a day, and we need to check if that's true based on a month of data, using a particular statistical method. Watch out! Don't mix up the *null* and *alternative* hypotheses, and make sure to use the *p-value method* correctly!
Also, remember the significance level α=0.10\alpha = 0.10.

STEP 2

1. State the Hypotheses
2. Calculate the Sample Mean
3. Calculate the Test Statistic
4. Find the P-value
5. Make a Decision

STEP 3

The *null hypothesis* H0H_0 is what we're trying to disprove.
The driver claims the average is 60 tickets per day, so: H0:μ=60H_0: \mu = 60

STEP 4

The *alternative hypothesis* H1H_1 is what we suspect might be true if the null hypothesis is false.
We're just checking if the average is *different* from 60, not specifically higher or lower, so we use a two-tailed test: H1:μ60H_1: \mu \ne 60

STEP 5

The driver's claim is that the average is 60, which matches our *null hypothesis*.

STEP 6

Since we're looking for a difference in either direction (more than or less than 60 tickets), this is a *two-tailed test*.

STEP 7

Add up all the ticket counts: 57+60+83++45=168057 + 60 + 83 + \dots + 45 = 1680.

STEP 8

There are 28 data points, so the sample mean xˉ\bar{x} is: xˉ=168028=60\bar{x} = \frac{1680}{28} = 60 The **sample mean** is xˉ=60\bar{x} = 60.

STEP 9

The test statistic for a z-test is: z=xˉμσnz = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} where xˉ\bar{x} is the **sample mean**, μ\mu is the **population mean** under the null hypothesis, σ\sigma is the **population standard deviation**, and nn is the **sample size**.

STEP 10

We have xˉ=60\bar{x} = 60, μ=60\mu = 60, σ=13.42\sigma = 13.42, and n=28n = 28.
Substituting these values into the formula: z=606013.4228z = \frac{60 - 60}{\frac{13.42}{\sqrt{28}}}

STEP 11

z=013.425.29=02.54=0z = \frac{0}{\frac{13.42}{5.29}} = \frac{0}{2.54} = 0 The **test statistic** is z=0z = 0.

STEP 12

The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.
Since our test statistic is 0, and this is a two-tailed test, our p-value will be the largest possible value of 1.

STEP 13

Since our z-score is 0, the p-value is 1.

STEP 14

Our p-value (1) is greater than our significance level α=0.10\alpha = 0.10.

STEP 15

Since the p-value is greater than α\alpha, we *fail to reject* the null hypothesis.

STEP 16

There is *not* enough evidence to reject the motorist's claim that the South Boro Police issue an average of 60 speeding tickets per day at the α=0.10\alpha = 0.10 significance level.
The hypotheses are H0:μ=60H_0: \mu = 60 (the claim) and H1:μ60H_1: \mu \ne 60.
This was a two-tailed test.

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