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 z-test for the difference between two population proportions \square a. The null and alternative hypotheses would be: b. The test statistic \square zθ2=z \theta^{2}= (please show your answer to 3 decimal places.) c. The p -value == \square (Please show your answer to 4 decimal places.) d. The pp-value is \square α\alpha e. Based on this, we should \square fail to reject \checkmark \checkmark the null hypothesis.

Studdy Solution

STEP 1

What is this asking? We want to find out if having a hard-to-pronounce last name affects your chances of getting a job interview, and we're using some fancy math to do it! Watch out! Don't mix up the groups (easy vs. hard names) or the numbers of applicants and interviewees.
Also, remember that a small p-value means there *is* a significant difference!

STEP 2

1. Set up the hypotheses
2. Calculate the pooled proportion
3. Calculate the standard error
4. Calculate the test statistic (z-score)
5. Calculate the p-value
6. Interpret the p-value

STEP 3

Our **null hypothesis** (H0H_0) is that there's *no difference* in interview rates between easy and hard-to-pronounce last names.
In math terms: p1=p2p_1 = p_2, where p1p_1 is the proportion of interviewees with easy names, and p2p_2 is the proportion for hard names.

STEP 4

Our **alternative hypothesis** (H1H_1) is that there *is* a difference.
Mathematically: p1p2p_1 \ne p_2.

STEP 5

We're assuming, for the moment, that the null hypothesis is true (no difference).
So, we combine the data to get a single overall proportion of interviewees.

STEP 6

p^=x1+x2n1+n2=424+502673+761=92614340.6456\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{\textbf{424} + \textbf{502}}{\textbf{673} + \textbf{761}} = \frac{926}{1434} \approx \textbf{0.6456} Here, x1x_1 and x2x_2 are the numbers of interviewees in each group, and n1n_1 and n2n_2 are the total applicants in each group.

STEP 7

It measures how much our sample proportions are likely to vary from the true population proportions.

STEP 8

The **standard error** formula is: SE=p^(1p^)(1n1+1n2)SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}

STEP 9

SE=0.6456(10.6456)(1673+1761)0.22870.00280.000640.0253SE = \sqrt{\textbf{0.6456}(1 - \textbf{0.6456})(\frac{1}{\textbf{673}} + \frac{1}{\textbf{761}})} \approx \sqrt{0.2287 \cdot 0.0028} \approx \sqrt{0.00064} \approx \textbf{0.0253}

STEP 10

It tells us how far apart our sample proportions are, in terms of standard errors.

STEP 11

First, we calculate the individual proportions: p^1=4246730.6300\hat{p}_1 = \frac{424}{673} \approx 0.6300 and p^2=5027610.6597\hat{p}_2 = \frac{502}{761} \approx 0.6597.

STEP 12

z=p^1p^2SEz = \frac{\hat{p}_1 - \hat{p}_2}{SE}

STEP 13

z=0.63000.65970.02530.02970.0253-1.174z = \frac{\textbf{0.6300} - \textbf{0.6597}}{\textbf{0.0253}} \approx \frac{-0.0297}{0.0253} \approx \textbf{-1.174}

STEP 14

It's the probability of getting our results (or more extreme results) if the null hypothesis were actually true.

STEP 15

Using a z-table or calculator (for a two-tailed test), we find the p-value corresponding to our **z-score** of **-1.174**.
This gives us a p-value of approximately 0.2401\textbf{0.2401}.

STEP 16

Our p-value (0.2401\textbf{0.2401}) is *greater* than our significance level (α=0.10\alpha = 0.10).

STEP 17

Since the p-value is high, we **fail to reject** the null hypothesis.
This means we don't have enough evidence to say that having a hard-to-pronounce last name affects your chances of getting an interview.

STEP 18

Test Statistic z=1.174z = -1.174 p-value =0.2401= 0.2401 We fail to reject the null hypothesis.

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