Math / Data & Statistics

QuestionYou wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.02$ . $\begin{array}{c} H_{o}: \mu=82.2 \\ H_{a}: \mu<82.2 \end{array}$
You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size $n=18$ with mean $M=78.3$ and a standard deviation of $S D=18.2$ .
What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = $\square$ What is the p-value for this sample? (Report answer accurate to four decimal places.) p -value $=$ $\square$

Studdy Solution

STEP 1

What is this asking? We want to see if there's enough evidence to say that the true average of a population is less than $\$82.2$ , even though we only have data from a small sample. Watch out! We don't know how spread out the entire population is, so we'll have to use the spread of our sample as an estimate, which makes things a little trickier!

STEP 2

1. Calculate the test statistic.
2. Calculate the p-value.

STEP 3

Let's dive in!
We're dealing with a t-test here because we don't know the population's standard deviation.
The formula for the t-statistic is:
$t = \frac{M - \mu}{\frac{SD}{\sqrt{n}}}$ Where $M$ is the sample mean, $\mu$ is the population mean we're testing against, $SD$ is the sample standard deviation, and $n$ is the sample size.

STEP 4

Let's plug in our values!
We have $M = 78.3$ , $\mu = 82.2$ , $SD = 18.2$ , and $n = 18$ .
$t = \frac{78.3 - 82.2}{\frac{18.2}{\sqrt{18}}}$

STEP 5

Time to crunch the numbers!
First, let's simplify the numerator:
$78.3 - 82.2 = -3.9$ So,
$t = \frac{-3.9}{\frac{18.2}{\sqrt{18}}}$

STEP 6

Now, let's tackle the denominator:
$\frac{18.2}{\sqrt{18}} \approx \frac{18.2}{4.243} \approx 4.289$ So,
$t = \frac{-3.9}{4.289}$

STEP 7

Finally, divide to get our t-statistic:
$t \approx -0.909$ Rounding to three decimal places, we get $t \approx -0.909$ .

STEP 8

Alright, now for the p-value!
This tells us how likely it is to get a sample as extreme as ours (or more extreme) if the true population mean really is $\$82.2$ .
Since our alternative hypothesis ( $H_a$ ) is $\mu < 82.2$ , we're looking for a one-tailed p-value.

STEP 9

We need the degrees of freedom, which is $n - 1 = 18 - 1 = 17$ .
With our $t$ -statistic of $-0.909$ and $17$ degrees of freedom, we can look up the p-value in a t-table (or use a calculator or software).

STEP 10

Looking it up, we find a p-value of approximately $0.1880$ .

STEP 11

Test statistic: $-0.909$ p-value: $0.1880$

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Studdy Solution

STEP 1

STEP 2

1. Calculate the test statistic.
2. Calculate the p-value.

STEP 3

STEP 4

Let's plug in our values!
We have $M = 78.3$ , $\mu = 82.2$ , $SD = 18.2$ , and $n = 18$ .
$t = \frac{78.3 - 82.2}{\frac{18.2}{\sqrt{18}}}$

STEP 5

Time to crunch the numbers!
First, let's simplify the numerator:
$78.3 - 82.2 = -3.9$ So,
$t = \frac{-3.9}{\frac{18.2}{\sqrt{18}}}$

STEP 6

Now, let's tackle the denominator:
$\frac{18.2}{\sqrt{18}} \approx \frac{18.2}{4.243} \approx 4.289$ So,
$t = \frac{-3.9}{4.289}$

STEP 7

Finally, divide to get our t-statistic:
$t \approx -0.909$ Rounding to three decimal places, we get $t \approx -0.909$ .

STEP 8

Alright, now for the p-value!
This tells us how likely it is to get a sample as extreme as ours (or more extreme) if the true population mean really is $\$82.2$ .
Since our alternative hypothesis ( $H_a$ ) is $\mu < 82.2$ , we're looking for a one-tailed p-value.

STEP 9

We need the degrees of freedom, which is $n - 1 = 18 - 1 = 17$ .
With our $t$ -statistic of $-0.909$ and $17$ degrees of freedom, we can look up the p-value in a t-table (or use a calculator or software).

STEP 10

Looking it up, we find a p-value of approximately $0.1880$ .

STEP 11

Test statistic: $-0.909$ p-value: $0.1880$

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Studdy Solution

STEP 1

STEP 2

1. Calculate the test statistic.2. Calculate the p-value.

STEP 3

STEP 4

Let's **plug in** our values!We have M=78.3M = 78.3M=78.3, μ=82.2\mu = 82.2μ=82.2, SD=18.2SD = 18.2SD=18.2, and n=18n = 18n=18.t=78.3−82.218.218t = \frac{78.3 - 82.2}{\frac{18.2}{\sqrt{18}}}t=18​18.2​78.3−82.2​

STEP 5

Time to **crunch the numbers**!First, let's simplify the numerator:78.3−82.2=−3.978.3 - 82.2 = -3.978.3−82.2=−3.9So,t=−3.918.218t = \frac{-3.9}{\frac{18.2}{\sqrt{18}}}t=18​18.2​−3.9​

STEP 6

Now, let's **tackle the denominator**:18.218≈18.24.243≈4.289\frac{18.2}{\sqrt{18}} \approx \frac{18.2}{4.243} \approx 4.28918​18.2​≈4.24318.2​≈4.289So,t=−3.94.289t = \frac{-3.9}{4.289}t=4.289−3.9​

STEP 7

Finally, **divide** to get our *t*-statistic:t≈−0.909t \approx -0.909t≈−0.909Rounding to **three decimal places**, we get t≈−0.909t \approx -0.909t≈−0.909.

STEP 8

Alright, now for the *p*-value!This tells us how likely it is to get a sample *as extreme as ours* (or *more extreme*) if the *true* population mean really *is* $82.2\$82.2$82.2.Since our alternative hypothesis (HaH_aHa​) is μ<82.2\mu < 82.2μ<82.2, we're looking for a *one-tailed* *p*-value.

STEP 9

We need the **degrees of freedom**, which is n−1=18−1=17n - 1 = 18 - 1 = 17n−1=18−1=17.With our ttt-statistic of −0.909-0.909−0.909 and 171717 degrees of freedom, we can look up the *p*-value in a *t*-table (or use a calculator or software).

STEP 10

Looking it up, we find a *p*-value of approximately 0.18800.18800.1880.

STEP 11

Test statistic: −0.909-0.909−0.909 p-value: 0.18800.18800.1880

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1. Calculate the test statistic.
2. Calculate the p-value.

Let's plug in our values!
We have $M = 78.3$ , $\mu = 82.2$ , $SD = 18.2$ , and $n = 18$ .
$t = \frac{78.3 - 82.2}{\frac{18.2}{\sqrt{18}}}$

Time to crunch the numbers!
First, let's simplify the numerator:
$78.3 - 82.2 = -3.9$ So,
$t = \frac{-3.9}{\frac{18.2}{\sqrt{18}}}$

Now, let's tackle the denominator:
$\frac{18.2}{\sqrt{18}} \approx \frac{18.2}{4.243} \approx 4.289$ So,
$t = \frac{-3.9}{4.289}$

Finally, divide to get our t-statistic:
$t \approx -0.909$ Rounding to three decimal places, we get $t \approx -0.909$ .

Alright, now for the p-value!
This tells us how likely it is to get a sample as extreme as ours (or more extreme) if the true population mean really is $\$82.2$ .
Since our alternative hypothesis ( $H_a$ ) is $\mu < 82.2$ , we're looking for a one-tailed p-value.

We need the degrees of freedom, which is $n - 1 = 18 - 1 = 17$ .
With our $t$ -statistic of $-0.909$ and $17$ degrees of freedom, we can look up the p-value in a t-table (or use a calculator or software).

Looking it up, we find a p-value of approximately $0.1880$ .

Test statistic: $-0.909$ p-value: $0.1880$