Math  /  Data & Statistics

QuestionIs there a doctor in the house? A market research firm reported the mean annual earnings of all family practitioners in the United States was $178,258\$ 178,258. A random sample of 52 family practitioners in Los Angeles had mean earnings of xˉ=$193,130\bar{x}=\$ 193,130 with a standard deviation of $42,047\$ 42,047. Do the data provide sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles is greater than the national average? Use the α=0.10\alpha=0.10 level of significance and the PP-value method with the TI-84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0:μ=178,258H1:μ>178,258\begin{array}{l} H_{0}: \mu=178,258 \\ H_{1}: \mu>178,258 \end{array}
This hypothesis test is a \square right-tailed test.
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. t=2.55t=2.55
Part: 2/52 / 5
Part 3 of 5 (c) Compute the PP-value. Round the answer to at least four decimal places. P-value =P \text {-value }= \square Start over

Studdy Solution

STEP 1

What is this asking? Are Los Angeles family doctors making more money than the national average? Watch out! Don't mix up the Los Angeles sample data with the national average!

STEP 2

1. Set up the Hypothesis Test
2. Calculate the Test Statistic
3. Compute the P-value
4. Make a Decision

STEP 3

We want to see if LA doctors earn *more* than the national average.
This means we're dealing with a **right-tailed test**, since we're looking at the *upper* end of the income distribution.
Our **null hypothesis** (H0H_0) says that the LA mean income is the same as the national average: μ=$178,258\mu = \$178,258.
The **alternative hypothesis** (H1H_1) says LA doctors make *more*: μ>$178,258\mu > \$178,258.

STEP 4

We've got our **sample mean** xˉ=$193,130\bar{x} = \$193,130, **sample standard deviation** s=$42,047s = \$42,047, and **sample size** n=52n = 52.
The **national average** is our population mean μ=$178,258\mu = \$178,258.

STEP 5

Since we're using a sample standard deviation, we use the *t*-distribution and calculate the **t-statistic** with the formula: t=xˉμsnt = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} This formula tells us how far our **sample mean** is from the **population mean**, in terms of **standard errors**.

STEP 6

Let's plug in the numbers: t=$193,130$178,258$42,04752t = \frac{\$193,130 - \$178,258}{\frac{\$42,047}{\sqrt{52}}} t=$14,872$42,0477.211t = \frac{\$14,872}{\frac{\$42,047}{7.211}}t=$14,872$5,830.73t = \frac{\$14,872}{\$5,830.73}t2.55t \approx 2.55So, our **test statistic** is **2.55**.
This tells us our sample mean is roughly **2.55 standard errors** above the national average.

STEP 7

The **P-value** tells us the probability of observing a sample mean as extreme as ours (or even *more* extreme) if the **null hypothesis** were actually true (i.e., if LA doctors *really* made the same as the national average).
Since it's a **right-tailed test**, we're looking for the area to the *right* of our **test statistic** (**2.55**) under the *t*-distribution curve.

STEP 8

Using a *t*-distribution table or calculator (with degrees of freedom df=n1=521=51df = n - 1 = 52 - 1 = 51), we find the **P-value**.
For t=2.55t = 2.55 and df=51df = 51, the **P-value** is approximately **0.0070**.

STEP 9

Our **P-value** (**0.0070**) is *less than* our **significance level** (α=0.10\alpha = 0.10).
This means it's *very unlikely* to observe such a high sample mean if the null hypothesis were true.
Therefore, we **reject the null hypothesis**!

STEP 10

There *is* sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles is greater than the national average.
The P-value is approximately **0.0070**.

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