Math / Data & Statistics

Question8. The mean age of students at a certain college was 26.6 years old in 2005. An administrator thinks the mean age for online students differs from this value. She randomly surveys 56 online students and finds that the sample mean is 29.4 years with a standard deviation of 2.1 years. At the $1 \%$ level of significance $(\alpha=0.01)$ , what can be concluded about the administrator's claim? Include the following steps in your test. - State the null and alternative hypotheses - Calculate the test statistic - Find the p-value - Make a decision whether to reject $\mathrm{H}_{0}$ - Write an interpretation of the results

Studdy Solution

STEP 1

1. The sample of 56 online students is randomly selected and represents the population of online students.
2. The sample mean is 29.4 years, with a standard deviation of 2.1 years.
3. The population mean age in 2005 was 26.6 years.
4. The level of significance is $\alpha = 0.01$ .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Find the p-value.
4. Make a decision whether to reject $H_0$ .
5. Write an interpretation of the results.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean age of online students is 26.6 years. $H_0: \mu = 26.6$
- Alternative hypothesis ( $H_a$ ): The mean age of online students differs from 26.6 years. $H_a: \mu \neq 26.6$

STEP 4

Calculate the test statistic using the formula for the t-statistic:
$t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$
where: - $\bar{x} = 29.4$ (sample mean) - $\mu = 26.6$ (population mean) - $s = 2.1$ (sample standard deviation) - $n = 56$ (sample size)
Substitute the values:
$t = \frac{29.4 - 26.6}{\frac{2.1}{\sqrt{56}}}$
Calculate the denominator:
$\frac{2.1}{\sqrt{56}} \approx 0.2805$
Calculate the t-statistic:
$t \approx \frac{2.8}{0.2805} \approx 9.98$

STEP 5

Find the p-value for the calculated t-statistic. Since this is a two-tailed test with $n - 1 = 55$ degrees of freedom, use a t-distribution table or calculator to find the p-value for $t \approx 9.98$ .
The p-value is very small (much less than 0.01).

STEP 6

Make a decision whether to reject $H_0$ .
Since the p-value is less than $\alpha = 0.01$ , we reject the null hypothesis $H_0$ .

STEP 7

Write an interpretation of the results.
At the 1% level of significance, there is sufficient evidence to conclude that the mean age of online students differs from 26.6 years. The administrator's claim is supported.

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Studdy Solution

STEP 1

1. The sample of 56 online students is randomly selected and represents the population of online students.
2. The sample mean is 29.4 years, with a standard deviation of 2.1 years.
3. The population mean age in 2005 was 26.6 years.
4. The level of significance is $\alpha = 0.01$ .

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Find the p-value.
4. Make a decision whether to reject $H_0$ .
5. Write an interpretation of the results.

STEP 3

State the null and alternative hypotheses:
- Null hypothesis ( $H_0$ ): The mean age of online students is 26.6 years. $H_0: \mu = 26.6$
- Alternative hypothesis ( $H_a$ ): The mean age of online students differs from 26.6 years. $H_a: \mu \neq 26.6$

STEP 4

STEP 5

Find the p-value for the calculated t-statistic. Since this is a two-tailed test with $n - 1 = 55$ degrees of freedom, use a t-distribution table or calculator to find the p-value for $t \approx 9.98$ .
The p-value is very small (much less than 0.01).

STEP 6

Make a decision whether to reject $H_0$ .
Since the p-value is less than $\alpha = 0.01$ , we reject the null hypothesis $H_0$ .

STEP 7

Write an interpretation of the results.
At the 1% level of significance, there is sufficient evidence to conclude that the mean age of online students differs from 26.6 years. The administrator's claim is supported.

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Studdy Solution

STEP 1

1. The sample of 56 online students is randomly selected and represents the population of online students.2. The sample mean is 29.4 years, with a standard deviation of 2.1 years.3. The population mean age in 2005 was 26.6 years.4. The level of significance is α=0.01 \alpha = 0.01 α=0.01.

STEP 2

1. State the null and alternative hypotheses.2. Calculate the test statistic.3. Find the p-value.4. Make a decision whether to reject H0 H_0 H0​.5. Write an interpretation of the results.

STEP 3

STEP 4

STEP 5

Find the p-value for the calculated t-statistic. Since this is a two-tailed test with n−1=55 n - 1 = 55 n−1=55 degrees of freedom, use a t-distribution table or calculator to find the p-value for t≈9.98 t \approx 9.98 t≈9.98.The p-value is very small (much less than 0.01).

STEP 6

Make a decision whether to reject H0 H_0 H0​.Since the p-value is less than α=0.01 \alpha = 0.01 α=0.01, we reject the null hypothesis H0 H_0 H0​.

STEP 7

Write an interpretation of the results.At the 1% level of significance, there is sufficient evidence to conclude that the mean age of online students differs from 26.6 years. The administrator's claim is supported.

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1. The sample of 56 online students is randomly selected and represents the population of online students.
2. The sample mean is 29.4 years, with a standard deviation of 2.1 years.
3. The population mean age in 2005 was 26.6 years.
4. The level of significance is $\alpha = 0.01$ .

1. State the null and alternative hypotheses.
2. Calculate the test statistic.
3. Find the p-value.
4. Make a decision whether to reject $H_0$ .
5. Write an interpretation of the results.

Find the p-value for the calculated t-statistic. Since this is a two-tailed test with $n - 1 = 55$ degrees of freedom, use a t-distribution table or calculator to find the p-value for $t \approx 9.98$ .
The p-value is very small (much less than 0.01).

Make a decision whether to reject $H_0$ .
Since the p-value is less than $\alpha = 0.01$ , we reject the null hypothesis $H_0$ .

Write an interpretation of the results.
At the 1% level of significance, there is sufficient evidence to conclude that the mean age of online students differs from 26.6 years. The administrator's claim is supported.