Math / Data & Statistics

QuestionAre job applicants with easy to pronounce last names less likely to get called for an interview than applicants with difficult to pronounce last names. 673 job applications were sent out with last names that are easy to pronounce and 761 identical job applications were sent out with names that were difficult to pronounce. 424 of the "applicants" with easy to pronounce names were called for an interview while 502 of the "applicants" with difficult to pronounce names were called for an interview. What can be concluded at the 0.10 level of significance? If the calculator asks, be sure to use the "Not Pooled" data option.
For this study, we should use $\square$ $z$ -test for the difference between two population proportions a. The null and alternative hypotheses would be: $\begin{array}{l} H_{0} \text { : } \\ = \\ \text { p2 } \\ \text { - }{ }^{\checkmark} \text { (please enter a decimal) } \\ H_{1} \text { : } \\ \text { p1 } \\ \text { < } \\ \text { p2 } \end{array}$ (Please enter a decimal) b. The test statistic $\square$ $\square$ (please show your answer to 3 decimal places.) c. The $p$ -value $=$ 0.1209 $\square$ (Please show your answer to 4 decimal places.)

Studdy Solution

STEP 1

What is this asking? We want to find out if having a hard-to-pronounce last name actually helps you get a job interview! Watch out! Don't mix up the groups – we're comparing easy names vs. hard names, and we need to keep track of which numbers go with which!

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Pooled Proportion
3. Calculate the Standard Error
4. Calculate the Test Statistic
5. Find the P-value

STEP 3

Our null hypothesis ( $H_0$ ) says there's no difference between the interview rates for easy and hard-to-pronounce names.
In math terms: $p_1 = p_2$ , where $p_1$ is the proportion of interviews for easy names and $p_2$ is the proportion for hard names.
So, the difference is zero: $p_1 - p_2 = 0$ .

STEP 4

Our alternative hypothesis ( $H_1$ ) says having a hard-to-pronounce name increases your chances of an interview.
Mathematically, this means $p_1 < p_2$ .

STEP 5

For easy names, we have $\frac{424}{673} \approx 0.630$ .
So, about 63% of people with easy-to-pronounce names got interviews.

STEP 6

For difficult names, it's $\frac{502}{761} \approx 0.660$ .
That's about 66% – slightly higher!

STEP 7

The pooled proportion combines both groups.
It's the total number of interviews divided by the total number of applications: $\frac{424 + 502}{673 + 761} = \frac{926}{1434} \approx 0.646$ .
So overall, about 64.6% of all applicants got interviews.

STEP 8

The standard error formula is a bit scary, but we'll break it down: $SE = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}$ Where $\hat{p}$ is our pooled proportion, $n_1$ is the number of easy name applications, and $n_2$ is the number of difficult name applications.

STEP 9

Let's plug in our numbers: $SE = \sqrt{0.646(1-0.646)(\frac{1}{673} + \frac{1}{761})}$

STEP 10

$SE = \sqrt{0.646 \cdot 0.354 \cdot (\frac{1}{673} + \frac{1}{761})} \approx \sqrt{0.2287 \cdot 0.0028} \approx \sqrt{0.00064} \approx 0.0253$ Our standard error is approximately 0.0253.

STEP 11

The test statistic ( $z$ ) tells us how far apart our proportions are in terms of standard errors: $z = \frac{\hat{p_1} - \hat{p_2}}{SE}$

STEP 12

$z = \frac{0.630 - 0.660}{0.0253}$

STEP 13

$z \approx \frac{-0.03}{0.0253} \approx -1.186$ Our test statistic is approximately -1.186.

STEP 14

The p-value (0.1209) tells us the probability of seeing a difference as big as ours (or bigger) if the null hypothesis were true.
Since our p-value is greater than our significance level (0.10), we fail to reject the null hypothesis.

STEP 15

We fail to reject the null hypothesis.
There's not enough evidence to say that having a hard-to-pronounce last name makes a difference in getting an interview.
Our p-value is 0.1209, and our test statistic is -1.186.

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Studdy Solution

STEP 1

What is this asking? We want to find out if having a hard-to-pronounce last name actually helps you get a job interview! Watch out! Don't mix up the groups – we're comparing easy names vs. hard names, and we need to keep track of which numbers go with which!

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Pooled Proportion
3. Calculate the Standard Error
4. Calculate the Test Statistic
5. Find the P-value

STEP 3

Our null hypothesis ( $H_0$ ) says there's no difference between the interview rates for easy and hard-to-pronounce names.
In math terms: $p_1 = p_2$ , where $p_1$ is the proportion of interviews for easy names and $p_2$ is the proportion for hard names.
So, the difference is zero: $p_1 - p_2 = 0$ .

STEP 4

Our alternative hypothesis ( $H_1$ ) says having a hard-to-pronounce name increases your chances of an interview.
Mathematically, this means $p_1 < p_2$ .

STEP 5

For easy names, we have $\frac{424}{673} \approx 0.630$ .
So, about 63% of people with easy-to-pronounce names got interviews.

STEP 6

For difficult names, it's $\frac{502}{761} \approx 0.660$ .
That's about 66% – slightly higher!

STEP 7

The pooled proportion combines both groups.
It's the total number of interviews divided by the total number of applications: $\frac{424 + 502}{673 + 761} = \frac{926}{1434} \approx 0.646$ .
So overall, about 64.6% of all applicants got interviews.

STEP 8

The standard error formula is a bit scary, but we'll break it down: $SE = \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}$ Where $\hat{p}$ is our pooled proportion, $n_1$ is the number of easy name applications, and $n_2$ is the number of difficult name applications.

STEP 9

Let's plug in our numbers: $SE = \sqrt{0.646(1-0.646)(\frac{1}{673} + \frac{1}{761})}$

STEP 10

$SE = \sqrt{0.646 \cdot 0.354 \cdot (\frac{1}{673} + \frac{1}{761})} \approx \sqrt{0.2287 \cdot 0.0028} \approx \sqrt{0.00064} \approx 0.0253$ Our standard error is approximately 0.0253.

STEP 11

The test statistic ( $z$ ) tells us how far apart our proportions are in terms of standard errors: $z = \frac{\hat{p_1} - \hat{p_2}}{SE}$

STEP 12

$z = \frac{0.630 - 0.660}{0.0253}$

STEP 13

$z \approx \frac{-0.03}{0.0253} \approx -1.186$ Our test statistic is approximately -1.186.

STEP 14

The p-value (0.1209) tells us the probability of seeing a difference as big as ours (or bigger) if the null hypothesis were true.
Since our p-value is greater than our significance level (0.10), we fail to reject the null hypothesis.

STEP 15

We fail to reject the null hypothesis.
There's not enough evidence to say that having a hard-to-pronounce last name makes a difference in getting an interview.
Our p-value is 0.1209, and our test statistic is -1.186.

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Studdy Solution

STEP 1

What is this asking? We want to find out if having a hard-to-pronounce last name actually *helps* you get a job interview! Watch out! Don't mix up the groups – we're comparing easy names vs. hard names, and we need to keep track of which numbers go with which!

STEP 2

1. Set up the Hypothesis Test2. Calculate the Pooled Proportion3. Calculate the Standard Error4. Calculate the Test Statistic5. Find the P-value

STEP 3

STEP 4

Our **alternative hypothesis** (H1H_1H1​) says having a hard-to-pronounce name *increases* your chances of an interview.Mathematically, this means p1<p2p_1 < p_2p1​<p2​.

STEP 5

For easy names, we have 424673≈0.630\frac{424}{673} \approx 0.630673424​≈0.630.So, about **63%** of people with easy-to-pronounce names got interviews.

STEP 6

For difficult names, it's 502761≈0.660\frac{502}{761} \approx 0.660761502​≈0.660.That's about **66%** – slightly higher!

STEP 7

STEP 8

STEP 9

Let's plug in our numbers: SE=0.646(1−0.646)(1673+1761)SE = \sqrt{0.646(1-0.646)(\frac{1}{673} + \frac{1}{761})}SE=0.646(1−0.646)(6731​+7611​)​

STEP 10

STEP 11

The **test statistic** (zzz) tells us how far apart our proportions are in terms of standard errors: z=p1^−p2^SEz = \frac{\hat{p_1} - \hat{p_2}}{SE}z=SEp1​^​−p2​^​​

STEP 12

z=0.630−0.6600.0253z = \frac{0.630 - 0.660}{0.0253}z=0.02530.630−0.660​

STEP 13

z≈−0.030.0253≈−1.186z \approx \frac{-0.03}{0.0253} \approx -1.186z≈0.0253−0.03​≈−1.186 Our **test statistic** is approximately **-1.186**.

STEP 14

The **p-value** (0.1209) tells us the probability of seeing a difference as big as ours (or bigger) if the null hypothesis were true.Since our p-value is *greater* than our significance level (0.10), we *fail to reject* the null hypothesis.

STEP 15

We *fail to reject* the null hypothesis.There's not enough evidence to say that having a hard-to-pronounce last name makes a difference in getting an interview.Our p-value is **0.1209**, and our test statistic is **-1.186**.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We want to find out if having a hard-to-pronounce last name actually helps you get a job interview! Watch out! Don't mix up the groups – we're comparing easy names vs. hard names, and we need to keep track of which numbers go with which!

1. Set up the Hypothesis Test
2. Calculate the Pooled Proportion
3. Calculate the Standard Error
4. Calculate the Test Statistic
5. Find the P-value

Our alternative hypothesis ( $H_1$ ) says having a hard-to-pronounce name increases your chances of an interview.
Mathematically, this means $p_1 < p_2$ .

For easy names, we have $\frac{424}{673} \approx 0.630$ .
So, about 63% of people with easy-to-pronounce names got interviews.

For difficult names, it's $\frac{502}{761} \approx 0.660$ .
That's about 66% – slightly higher!

Let's plug in our numbers: $SE = \sqrt{0.646(1-0.646)(\frac{1}{673} + \frac{1}{761})}$

The test statistic ( $z$ ) tells us how far apart our proportions are in terms of standard errors: $z = \frac{\hat{p_1} - \hat{p_2}}{SE}$

$z = \frac{0.630 - 0.660}{0.0253}$

$z \approx \frac{-0.03}{0.0253} \approx -1.186$ Our test statistic is approximately -1.186.

The p-value (0.1209) tells us the probability of seeing a difference as big as ours (or bigger) if the null hypothesis were true.
Since our p-value is greater than our significance level (0.10), we fail to reject the null hypothesis.

We fail to reject the null hypothesis.
There's not enough evidence to say that having a hard-to-pronounce last name makes a difference in getting an interview.
Our p-value is 0.1209, and our test statistic is -1.186.