Math / Data & Statistics

QuestionDo men score higher on average compared to women on their statistics finals? Final exam scores of thirteen randomly selected male statistics students and twelve randomly selected female statistics students are shown below.
Male: $\begin{array}{lllllllllllll}93 & 77 & 64 & 71 & 72 & 62 & 84 & 70 & 89 & 73 & 63 & 88 & 94\end{array}$ $\begin{array}{lllllllllllll}\text { Female: } 83 & 46 & 58 & 81 & 49 & 74 & 62 & 70 & 69 & 66 & 58 & 68\end{array}$ Assume both follow a Normal distribution. What can be concluded at the the $\alpha=0.01$ level of significance level of significance?
For this study, we should use Select an answer a. The null and alternative hypotheses would be: $H_{0}$ : Select an answer Select an answer Select an answer (6) (please enter a decimal) $H_{1}$ : Select an answer Select an answer Select an answer (Please enter a decimal) b. The test statistic ? $0=$ $\square$ (please show your answer to 3 decimal places.) c. The $p$ -value $=$ $\square$ (Please show your answer to 4 decimal places.) d. The $p$ -value is ? $\alpha$ e. Based on this, we should Select an answer the null hypothesis. f. Thus, the final conclusion is that ... The results are statistically insignificant at $\alpha=0.01$ , so there is statistically significant evidence to conclude that the population mean statistics final exam score for men is equal to the population mean statistics final exam score for women. The results are statistically significant at $\alpha=0.01$ , so there is sufficient evidence to conclude that the population mean statistics final exam score for men is more than the population mean statistics final exam score for women. The results are statistically significant at $\alpha=0.01$ , so there is sufficient evidence to conclude that the mean final exam score for the thirteen men that were observed is more than the mean final exam score for the twelve women that were observed. The results are statistically insignificant at $\alpha=0.01$ , so there is insufficient evidence to conclude that the population mean statistics final exam score for men is more than the population mean statistics final exam score for women.

Studdy Solution

STEP 1

What is this asking? Do men, on average, score higher than women on their statistics finals, and is the difference statistically significant? Watch out! Don't mix up the sample means with what we're really interested in: the population means!
Also, keep an eye on that significance level, $\alpha = 0.01$ .
It's stricter than the usual $\alpha = 0.05$ !

STEP 2

1. State the Hypotheses
2. Calculate the Sample Statistics
3. Perform the t-Test
4. Find the p-value
5. Make a Decision

STEP 3

We want to see if men score higher, so our alternative hypothesis is that the population mean for men ( $\mu_M$ ) is greater than the population mean for women ( $\mu_W$ ).
That's written as $H_1: \mu_M > \mu_W$ .

STEP 4

The null hypothesis is the opposite: there's no difference between the population means, or $H_0: \mu_M = \mu_W$ .
We're trying to find enough evidence to reject this null hypothesis!

STEP 5

Calculate the mean for men: Add up all the men's scores (93 + 77 + ... + 94 = 976) and divide by the number of men (13).
So, $\bar{x}_M = \frac{976}{13} \approx 75.077$ .

STEP 6

Calculate the mean for women: Add up all the women's scores (83 + 46 + ... + 68 = 768) and divide by the number of women (12).
So, $\bar{x}_W = \frac{768}{12} = 64$ .

STEP 7

Calculate the sample standard deviation for men: First, find the variance.
For each male score, subtract the mean ( $75.077$ ), square the result, and add those squared differences together.
This sum is $1519.846$ .
Divide by $n - 1 = 12$ to get the variance, $s^2_M = \frac{1519.846}{12} \approx 126.654$ .
The standard deviation is the square root of the variance: $s_M = \sqrt{126.654} \approx 11.254$ .

STEP 8

Calculate the sample standard deviation for women: Follow the same process as with the men's scores.
The sum of squared differences is $1832$ .
Divide by $n - 1 = 11$ to get the variance, $s^2_W = \frac{1832}{11} \approx 166.545$ .
The standard deviation is $s_W = \sqrt{166.545} \approx 12.905$ .

STEP 9

We're using a two-sample t-test because we're comparing the means of two independent groups.
The formula is: $t = \frac{\bar{x}_M - \bar{x}_W}{\sqrt{\frac{s^2_M}{n_M} + \frac{s^2_W}{n_W}}}$

STEP 10

Plug in our values: $t = \frac{75.077 - 64}{\sqrt{\frac{126.654}{13} + \frac{166.545}{12}}} \approx \frac{11.077}{\sqrt{9.743 + 13.879}} \approx \frac{11.077}{\sqrt{23.622}} \approx \frac{11.077}{4.86} \approx 2.279$

STEP 11

Our t-statistic is $2.279$ .
We need to find the probability of getting a t-statistic this large or larger if the null hypothesis is true.
This is our p-value.

STEP 12

Using a t-table or calculator (with degrees of freedom calculated as $\approx 22.9$ and rounded down to 22), we find that the p-value is approximately $0.016$ .

STEP 13

Our p-value ( $0.016$ ) is greater than our significance level ( $\alpha = 0.01$ ).

STEP 14

Since the p-value is greater than alpha, we fail to reject the null hypothesis.

STEP 15

a. $H_0: \mu_M = \mu_W$ and $H_1: \mu_M > \mu_W$ b. The test statistic is $t \approx 2.279$ . c. The p-value is $\approx 0.016$ . d. The p-value is $>$ $\alpha$ . e. Based on this, we should fail to reject the null hypothesis. f. Thus, there is insufficient evidence at the $\alpha = 0.01$ level to conclude that the population mean statistics final exam score for men is greater than the population mean score for women.

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Studdy Solution

STEP 1

STEP 2

1. State the Hypotheses
2. Calculate the Sample Statistics
3. Perform the t-Test
4. Find the p-value
5. Make a Decision

STEP 3

We want to see if men score higher, so our alternative hypothesis is that the population mean for men ( $\mu_M$ ) is greater than the population mean for women ( $\mu_W$ ).
That's written as $H_1: \mu_M > \mu_W$ .

STEP 4

The null hypothesis is the opposite: there's no difference between the population means, or $H_0: \mu_M = \mu_W$ .
We're trying to find enough evidence to reject this null hypothesis!

STEP 5

Calculate the mean for men: Add up all the men's scores (93 + 77 + ... + 94 = 976) and divide by the number of men (13).
So, $\bar{x}_M = \frac{976}{13} \approx 75.077$ .

STEP 6

Calculate the mean for women: Add up all the women's scores (83 + 46 + ... + 68 = 768) and divide by the number of women (12).
So, $\bar{x}_W = \frac{768}{12} = 64$ .

STEP 7

STEP 8

Calculate the sample standard deviation for women: Follow the same process as with the men's scores.
The sum of squared differences is $1832$ .
Divide by $n - 1 = 11$ to get the variance, $s^2_W = \frac{1832}{11} \approx 166.545$ .
The standard deviation is $s_W = \sqrt{166.545} \approx 12.905$ .

STEP 9

We're using a two-sample t-test because we're comparing the means of two independent groups.
The formula is: $t = \frac{\bar{x}_M - \bar{x}_W}{\sqrt{\frac{s^2_M}{n_M} + \frac{s^2_W}{n_W}}}$

STEP 10

Plug in our values: $t = \frac{75.077 - 64}{\sqrt{\frac{126.654}{13} + \frac{166.545}{12}}} \approx \frac{11.077}{\sqrt{9.743 + 13.879}} \approx \frac{11.077}{\sqrt{23.622}} \approx \frac{11.077}{4.86} \approx 2.279$

STEP 11

Our t-statistic is $2.279$ .
We need to find the probability of getting a t-statistic this large or larger if the null hypothesis is true.
This is our p-value.

STEP 12

Using a t-table or calculator (with degrees of freedom calculated as $\approx 22.9$ and rounded down to 22), we find that the p-value is approximately $0.016$ .

STEP 13

Our p-value ( $0.016$ ) is greater than our significance level ( $\alpha = 0.01$ ).

STEP 14

Since the p-value is greater than alpha, we fail to reject the null hypothesis.

STEP 15

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Studdy Solution

STEP 1

STEP 2

1. State the Hypotheses2. Calculate the Sample Statistics3. Perform the t-Test4. Find the p-value5. Make a Decision

STEP 3

We want to see if men score higher, so our **alternative hypothesis** is that the population mean for men (μM\mu_MμM​) is greater than the population mean for women (μW\mu_WμW​).That's written as H1:μM>μWH_1: \mu_M > \mu_WH1​:μM​>μW​.

STEP 4

The **null hypothesis** is the opposite: there's no difference between the population means, or H0:μM=μWH_0: \mu_M = \mu_WH0​:μM​=μW​.We're trying to find enough evidence to *reject* this null hypothesis!

STEP 5

**Calculate the mean for men**: Add up all the men's scores (93 + 77 + ... + 94 = 976) and divide by the number of men (13).So, xˉM=97613≈75.077\bar{x}_M = \frac{976}{13} \approx 75.077xˉM​=13976​≈75.077.

STEP 6

**Calculate the mean for women**: Add up all the women's scores (83 + 46 + ... + 68 = 768) and divide by the number of women (12).So, xˉW=76812=64\bar{x}_W = \frac{768}{12} = 64xˉW​=12768​=64.

STEP 7

STEP 8

STEP 9

We're using a **two-sample t-test** because we're comparing the means of two independent groups.The formula is: t=xˉM−xˉWsM2nM+sW2nWt = \frac{\bar{x}_M - \bar{x}_W}{\sqrt{\frac{s^2_M}{n_M} + \frac{s^2_W}{n_W}}}t=nM​sM2​​+nW​sW2​​​xˉM​−xˉW​​

STEP 10

STEP 11

Our **t-statistic** is 2.2792.2792.279.We need to find the probability of getting a t-statistic this large or larger *if the null hypothesis is true*.This is our **p-value**.

STEP 12

Using a t-table or calculator (with degrees of freedom calculated as ≈22.9\approx 22.9≈22.9 and rounded down to 22), we find that the p-value is approximately 0.0160.0160.016.

STEP 13

Our **p-value** (0.0160.0160.016) is greater than our **significance level** (α=0.01\alpha = 0.01α=0.01).

STEP 14

Since the p-value is greater than alpha, we *fail to reject* the null hypothesis.

STEP 15

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1. State the Hypotheses
2. Calculate the Sample Statistics
3. Perform the t-Test
4. Find the p-value
5. Make a Decision

We want to see if men score higher, so our alternative hypothesis is that the population mean for men ( $\mu_M$ ) is greater than the population mean for women ( $\mu_W$ ).
That's written as $H_1: \mu_M > \mu_W$ .

The null hypothesis is the opposite: there's no difference between the population means, or $H_0: \mu_M = \mu_W$ .
We're trying to find enough evidence to reject this null hypothesis!

Calculate the mean for men: Add up all the men's scores (93 + 77 + ... + 94 = 976) and divide by the number of men (13).
So, $\bar{x}_M = \frac{976}{13} \approx 75.077$ .

Calculate the mean for women: Add up all the women's scores (83 + 46 + ... + 68 = 768) and divide by the number of women (12).
So, $\bar{x}_W = \frac{768}{12} = 64$ .

We're using a two-sample t-test because we're comparing the means of two independent groups.
The formula is: $t = \frac{\bar{x}_M - \bar{x}_W}{\sqrt{\frac{s^2_M}{n_M} + \frac{s^2_W}{n_W}}}$

Our t-statistic is $2.279$ .
We need to find the probability of getting a t-statistic this large or larger if the null hypothesis is true.
This is our p-value.

Using a t-table or calculator (with degrees of freedom calculated as $\approx 22.9$ and rounded down to 22), we find that the p-value is approximately $0.016$ .

Our p-value ( $0.016$ ) is greater than our significance level ( $\alpha = 0.01$ ).

Since the p-value is greater than alpha, we fail to reject the null hypothesis.