Question

mini - practice exam 2
question 4 of 5 (28 points) | question attempt: 1 of 1
time remaining: 25:43
savana
$h_0:mu_1$ not claim
$h_1:mu_2$ claim
this hypothesis test is a two - tailed test.
part: 1 / 5
part 2 of 5
(b) compute the test value. round the answer to at least three decimal places.
$t =$

Explanation:

To compute the test value (t - value) for a two - sample t - test (assuming this is a two - sample scenario with population means \(\mu_1\) and \(\mu_2\)), we need the following information: sample means (\(\bar{x}_1\), \(\bar{x}_2\)), sample standard deviations (\(s_1\), \(s_2\)), and sample sizes (\(n_1\), \(n_2\)). The formula for the two - sample t - test (pooled variance or unpooled) is:

Case 1: Pooled variance t - test (when \(\sigma_1^2=\sigma_2^2\))

The formula for the t - statistic is:

$$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$

where \(s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\) and under the null hypothesis \(H_0:\mu_1=\mu_2\) (so \(\mu_1-\mu_2 = 0\))

Case 2: Unpooled variance t - test (when \(\sigma_1^2

eq\sigma_2^2\))
The formula for the t - statistic is:

$$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$

and under the null hypothesis \(H_0:\mu_1=\mu_2\) (so \(\mu_1-\mu_2 = 0\))

Since the problem does not provide the sample data (sample means, sample standard deviations, sample sizes), we cannot calculate the exact value of \(t\). If you provide the values of \(\bar{x}_1\), \(\bar{x}_2\), \(s_1\), \(s_2\), \(n_1\), \(n_2\), we can follow these steps:

Step 1: Identify the type of t - test (pooled or unpooled)

• If we assume equal variances, use the pooled variance formula. First, calculate the pooled standard deviation \(s_p\) using the formula \(s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\)
• If we assume unequal variances, use the unpooled variance formula.

Step 2: Calculate the numerator and denominator

• Numerator: \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)\). Under \(H_0:\mu_1=\mu_2\), this simplifies to \(\bar{x}_1-\bar{x}_2\)
• Denominator:
• For pooled: \(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)
• For unpooled: \(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\)

Step 3: Compute the t - value

Divide the numerator by the denominator to get the t - value and round to at least three decimal places.

For example, if we have the following data:

• Sample 1: \(\bar{x}_1 = 10\), \(s_1 = 2\), \(n_1=20\)
• Sample 2: \(\bar{x}_2 = 12\), \(s_2 = 3\), \(n_2 = 25\)
• Assume unequal variances (\(\sigma_1^2

eq\sigma_2^2\)) and \(H_0:\mu_1=\mu_2\)

Step 1: Calculate the numerator

\(\bar{x}_1-\bar{x}_2=10 - 12=- 2\)

Step 2: Calculate the denominator

\(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{\frac{2^2}{20}+\frac{3^2}{25}}=\sqrt{\frac{4}{20}+\frac{9}{25}}=\sqrt{0.2 + 0.36}=\sqrt{0.56}\approx0.7483\)

Step 3: Calculate the t - value

\(t=\frac{-2}{0.7483}\approx - 2.673\)

But since we don't have the actual data from your problem, please provide the sample statistics (sample means, sample standard deviations, sample sizes) so that we can compute the t - value.

Answer:

To compute the test value (t - value) for a two - sample t - test (assuming this is a two - sample scenario with population means \(\mu_1\) and \(\mu_2\)), we need the following information: sample means (\(\bar{x}_1\), \(\bar{x}_2\)), sample standard deviations (\(s_1\), \(s_2\)), and sample sizes (\(n_1\), \(n_2\)). The formula for the two - sample t - test (pooled variance or unpooled) is:

Case 1: Pooled variance t - test (when \(\sigma_1^2=\sigma_2^2\))

The formula for the t - statistic is:

$$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$

where \(s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\) and under the null hypothesis \(H_0:\mu_1=\mu_2\) (so \(\mu_1-\mu_2 = 0\))

Case 2: Unpooled variance t - test (when \(\sigma_1^2

eq\sigma_2^2\))
The formula for the t - statistic is:

$$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$

and under the null hypothesis \(H_0:\mu_1=\mu_2\) (so \(\mu_1-\mu_2 = 0\))

Since the problem does not provide the sample data (sample means, sample standard deviations, sample sizes), we cannot calculate the exact value of \(t\). If you provide the values of \(\bar{x}_1\), \(\bar{x}_2\), \(s_1\), \(s_2\), \(n_1\), \(n_2\), we can follow these steps:

Step 1: Identify the type of t - test (pooled or unpooled)

• If we assume equal variances, use the pooled variance formula. First, calculate the pooled standard deviation \(s_p\) using the formula \(s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\)
• If we assume unequal variances, use the unpooled variance formula.

Step 2: Calculate the numerator and denominator

• Numerator: \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)\). Under \(H_0:\mu_1=\mu_2\), this simplifies to \(\bar{x}_1-\bar{x}_2\)
• Denominator:
• For pooled: \(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)
• For unpooled: \(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\)

Step 3: Compute the t - value

Divide the numerator by the denominator to get the t - value and round to at least three decimal places.

For example, if we have the following data:

• Sample 1: \(\bar{x}_1 = 10\), \(s_1 = 2\), \(n_1=20\)
• Sample 2: \(\bar{x}_2 = 12\), \(s_2 = 3\), \(n_2 = 25\)
• Assume unequal variances (\(\sigma_1^2

eq\sigma_2^2\)) and \(H_0:\mu_1=\mu_2\)

Step 1: Calculate the numerator

\(\bar{x}_1-\bar{x}_2=10 - 12=- 2\)

Step 2: Calculate the denominator

\(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{\frac{2^2}{20}+\frac{3^2}{25}}=\sqrt{\frac{4}{20}+\frac{9}{25}}=\sqrt{0.2 + 0.36}=\sqrt{0.56}\approx0.7483\)

Step 3: Calculate the t - value

\(t=\frac{-2}{0.7483}\approx - 2.673\)

But since we don't have the actual data from your problem, please provide the sample statistics (sample means, sample standard deviations, sample sizes) so that we can compute the t - value.