Math / Data & Statistics

Questionc. Is there statistically significant evidence of a difference in the mean amount purchased by all visitors to page $A$ and page $B$ ? $\qquad$ Why or why not? A large test stats
2. You also can find the $t$ -statistic on StatCrunch. If you are given summary statistics, go to Stat $>$ T Stats $>$ Two Sample $>$ With Summary. p-value
To test if two populations have different means, a random sample of 74 people from the first population had a mean of 17.3 standard deviation of 4.1. A random sample of 83 people from the second population had a mean of 19.2 and standard deviation of 3.7 . Use these summary statistics to compute $t$ and the p-value on StatCrunch. Round to four decimal places. $t=$ $\qquad$ p -value $=$ $\qquad$ 70

Studdy Solution

STEP 1

What is this asking? We're checking if there's a real difference in the average purchases between two web pages, A and B, using data from samples of visitors. Watch out! Don't mix up the two groups' data!
Also, remember a small p-value means a big difference!

STEP 2

1. Set up the hypothesis test
2. Calculate the t-statistic
3. Find the p-value
4. Interpret the results

STEP 3

We're testing if the means are different, not if one is bigger or smaller.
This means it's a two-tailed test!
Our null hypothesis ( $H_0$ ) is that the means are the same, and our alternative hypothesis ( $H_a$ ) is that they're different.
Let's write that down!

STEP 4

$\begin{array}{l} H_0: \mu_A = \mu_B \\ H_a: \mu_A \ne \mu_B \end{array}$ Where $\mu_A$ is the true mean of page A and $\mu_B$ is the true mean of page B.

STEP 5

Here's the formula for the two-sample t-statistic: It looks scary, but we'll break it down! $t = \frac{(\bar{x}_A - \bar{x}_B)}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}$

STEP 6

Let's plug in our values.
For page A, we have $\bar{x}_A = 17.3$ , $s_A = 4.1$ , and $n_A = 74$ .
For page B, we have $\bar{x}_B = 19.2$ , $s_B = 3.7$ , and $n_B = 83$ .

STEP 7

$t = \frac{(17.3 - 19.2)}{\sqrt{\frac{4.1^2}{74} + \frac{3.7^2}{83}}}$

STEP 8

$t = \frac{-1.9}{\sqrt{\frac{16.81}{74} + \frac{13.69}{83}}}$

STEP 9

$t = \frac{-1.9}{\sqrt{0.22716 + 0.16542}}$

STEP 10

$t = \frac{-1.9}{\sqrt{0.39258}}$

STEP 11

$t = \frac{-1.9}{0.62656}$

STEP 12

$t \approx -3.0322$ Our t-statistic is approximately $-3.0322$ .

STEP 13

We'll use a t-distribution with degrees of freedom.
We can approximate the degrees of freedom using a complicated formula, but since we're using StatCrunch, it will calculate it for us.
The p-value is the probability of getting a t-statistic as extreme as ours (or more extreme) if the null hypothesis is true.
Since it's a two-tailed test, we're looking at both tails of the distribution.

STEP 14

Using StatCrunch with the given summary statistics, we find a p-value of approximately $0.0029$ .

STEP 15

Our p-value ( $0.0029$ ) is much smaller than the typical significance level of $0.05$ .
This means it's super unlikely to see such a big difference in means if there wasn't a real difference between the pages.

STEP 16

$t \approx -3.0322$ p-value $\approx 0.0029$ Yes, there is statistically significant evidence of a difference in mean amounts purchased.
The p-value is less than 0.05, so we reject the null hypothesis.

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Studdy Solution

STEP 1

What is this asking? We're checking if there's a real difference in the average purchases between two web pages, A and B, using data from samples of visitors. Watch out! Don't mix up the two groups' data!
Also, remember a small p-value means a big difference!

STEP 2

1. Set up the hypothesis test
2. Calculate the t-statistic
3. Find the p-value
4. Interpret the results

STEP 3

We're testing if the means are different, not if one is bigger or smaller.
This means it's a two-tailed test!
Our null hypothesis ( $H_0$ ) is that the means are the same, and our alternative hypothesis ( $H_a$ ) is that they're different.
Let's write that down!

STEP 4

$\begin{array}{l} H_0: \mu_A = \mu_B \\ H_a: \mu_A \ne \mu_B \end{array}$ Where $\mu_A$ is the true mean of page A and $\mu_B$ is the true mean of page B.

STEP 5

Here's the formula for the two-sample t-statistic: It looks scary, but we'll break it down! $t = \frac{(\bar{x}_A - \bar{x}_B)}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}$

STEP 6

Let's plug in our values.
For page A, we have $\bar{x}_A = 17.3$ , $s_A = 4.1$ , and $n_A = 74$ .
For page B, we have $\bar{x}_B = 19.2$ , $s_B = 3.7$ , and $n_B = 83$ .

STEP 7

$t = \frac{(17.3 - 19.2)}{\sqrt{\frac{4.1^2}{74} + \frac{3.7^2}{83}}}$

STEP 8

$t = \frac{-1.9}{\sqrt{\frac{16.81}{74} + \frac{13.69}{83}}}$

STEP 9

$t = \frac{-1.9}{\sqrt{0.22716 + 0.16542}}$

STEP 10

$t = \frac{-1.9}{\sqrt{0.39258}}$

STEP 11

$t = \frac{-1.9}{0.62656}$

STEP 12

$t \approx -3.0322$ Our t-statistic is approximately $-3.0322$ .

STEP 13

STEP 14

Using StatCrunch with the given summary statistics, we find a p-value of approximately $0.0029$ .

STEP 15

Our p-value ( $0.0029$ ) is much smaller than the typical significance level of $0.05$ .
This means it's super unlikely to see such a big difference in means if there wasn't a real difference between the pages.

STEP 16

$t \approx -3.0322$ p-value $\approx 0.0029$ Yes, there is statistically significant evidence of a difference in mean amounts purchased.
The p-value is less than 0.05, so we reject the null hypothesis.

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Studdy Solution

STEP 1

What is this asking? We're checking if there's a *real* difference in the average purchases between two web pages, A and B, using data from samples of visitors. Watch out! Don't mix up the two groups' data!Also, remember a small p-value means a *big* difference!

STEP 2

1. Set up the hypothesis test2. Calculate the t-statistic3. Find the p-value4. Interpret the results

STEP 3

We're testing if the *means* are different, not if one is bigger or smaller.This means it's a *two-tailed test*!Our **null hypothesis** (H0H_0H0​) is that the means are the same, and our **alternative hypothesis** (HaH_aHa​) is that they're different.Let's write that down!

STEP 4

H0:μA=μBHa:μA≠μB\begin{array}{l} H_0: \mu_A = \mu_B \\ H_a: \mu_A \ne \mu_B \end{array}H0​:μA​=μB​Ha​:μA​=μB​​ Where μA\mu_AμA​ is the **true mean** of page A and μB\mu_BμB​ is the **true mean** of page B.

STEP 5

Here's the formula for the **two-sample t-statistic**: It looks scary, but we'll break it down! t=(xˉA−xˉB)sA2nA+sB2nBt = \frac{(\bar{x}_A - \bar{x}_B)}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}t=nA​sA2​​+nB​sB2​​​(xˉA​−xˉB​)​

STEP 6

Let's plug in our **values**.For page A, we have xˉA=17.3\bar{x}_A = 17.3xˉA​=17.3, sA=4.1s_A = 4.1sA​=4.1, and nA=74n_A = 74nA​=74.For page B, we have xˉB=19.2\bar{x}_B = 19.2xˉB​=19.2, sB=3.7s_B = 3.7sB​=3.7, and nB=83n_B = 83nB​=83.

STEP 7

t=(17.3−19.2)4.1274+3.7283t = \frac{(17.3 - 19.2)}{\sqrt{\frac{4.1^2}{74} + \frac{3.7^2}{83}}}t=744.12​+833.72​​(17.3−19.2)​

STEP 8

t=−1.916.8174+13.6983t = \frac{-1.9}{\sqrt{\frac{16.81}{74} + \frac{13.69}{83}}}t=7416.81​+8313.69​​−1.9​

STEP 9

t=−1.90.22716+0.16542t = \frac{-1.9}{\sqrt{0.22716 + 0.16542}}t=0.22716+0.16542​−1.9​

STEP 10

t=−1.90.39258t = \frac{-1.9}{\sqrt{0.39258}}t=0.39258​−1.9​

STEP 11

t=−1.90.62656t = \frac{-1.9}{0.62656}t=0.62656−1.9​

STEP 12

t≈−3.0322t \approx -3.0322t≈−3.0322 Our **t-statistic** is approximately −3.0322-3.0322−3.0322.

STEP 13

STEP 14

Using StatCrunch with the given summary statistics, we find a **p-value** of approximately 0.00290.00290.0029.

STEP 15

Our **p-value** (0.00290.00290.0029) is *much* smaller than the typical significance level of 0.050.050.05.This means it's *super* unlikely to see such a big difference in means if there wasn't a *real* difference between the pages.

STEP 16

t≈−3.0322t \approx -3.0322t≈−3.0322 p-value ≈0.0029\approx 0.0029≈0.0029 Yes, there is statistically significant evidence of a difference in mean amounts purchased.The p-value is less than 0.05, so we reject the null hypothesis.

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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're checking if there's a real difference in the average purchases between two web pages, A and B, using data from samples of visitors. Watch out! Don't mix up the two groups' data!
Also, remember a small p-value means a big difference!

1. Set up the hypothesis test
2. Calculate the t-statistic
3. Find the p-value
4. Interpret the results

We're testing if the means are different, not if one is bigger or smaller.
This means it's a two-tailed test!
Our null hypothesis ( $H_0$ ) is that the means are the same, and our alternative hypothesis ( $H_a$ ) is that they're different.
Let's write that down!

$\begin{array}{l} H_0: \mu_A = \mu_B \\ H_a: \mu_A \ne \mu_B \end{array}$ Where $\mu_A$ is the true mean of page A and $\mu_B$ is the true mean of page B.

Here's the formula for the two-sample t-statistic: It looks scary, but we'll break it down! $t = \frac{(\bar{x}_A - \bar{x}_B)}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}}$

Let's plug in our values.
For page A, we have $\bar{x}_A = 17.3$ , $s_A = 4.1$ , and $n_A = 74$ .
For page B, we have $\bar{x}_B = 19.2$ , $s_B = 3.7$ , and $n_B = 83$ .

$t = \frac{(17.3 - 19.2)}{\sqrt{\frac{4.1^2}{74} + \frac{3.7^2}{83}}}$

$t = \frac{-1.9}{\sqrt{\frac{16.81}{74} + \frac{13.69}{83}}}$

$t = \frac{-1.9}{\sqrt{0.22716 + 0.16542}}$

$t = \frac{-1.9}{\sqrt{0.39258}}$

$t = \frac{-1.9}{0.62656}$

$t \approx -3.0322$ Our t-statistic is approximately $-3.0322$ .

Using StatCrunch with the given summary statistics, we find a p-value of approximately $0.0029$ .

Our p-value ( $0.0029$ ) is much smaller than the typical significance level of $0.05$ .
This means it's super unlikely to see such a big difference in means if there wasn't a real difference between the pages.

$t \approx -3.0322$ p-value $\approx 0.0029$ Yes, there is statistically significant evidence of a difference in mean amounts purchased.
The p-value is less than 0.05, so we reject the null hypothesis.