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Answer :
We wish to test
[tex]$$
H_0: \mu_1=\mu_2 \quad \text{vs.} \quad H_a: \mu_1<\mu_2
$$[/tex]
at a significance level of [tex]$\alpha=0.01$[/tex]. Given the two samples (Sample #1 and Sample #2), we use a twoâ€sample [tex]$t$[/tex]–test that does not assume equal variances (often called Welch’s [tex]$t$[/tex]–test).
Step 1. Compute the sample statistics.
For Sample #1 we have:
- Sample size: [tex]$n_1=56$[/tex]
- Sample mean: [tex]$\bar{x}_1\approx 77.1768$[/tex]
- Sample variance: [tex]$s_1^2\approx 144.9905$[/tex] (with sample standard deviation [tex]$s_1\approx 12.0412$[/tex])
For Sample #2 we have:
- Sample size: [tex]$n_2=57$[/tex]
- Sample mean: [tex]$\bar{x}_2\approx 77.7333$[/tex]
- Sample variance: [tex]$s_2^2\approx 189.4001$[/tex] (with sample standard deviation [tex]$s_2\approx 13.7623$[/tex])
Step 2. Compute the standard error of the difference in means.
The standard error ([tex]$SE$[/tex]) is given by
[tex]$$
SE = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}.
$$[/tex]
Substituting the values we obtain
[tex]$$
SE\approx 2.4314.
$$[/tex]
Step 3. Calculate the test statistic.
The test statistic for the difference in means is
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{SE}.
$$[/tex]
Substituting the values:
[tex]$$
t = \frac{77.1768 - 77.7333}{2.4314} \approx -0.2289.
$$[/tex]
Step 4. Determine the degrees of freedom.
For Welch’s [tex]$t$[/tex]–test the degrees of freedom are approximated by
[tex]$$
df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]
Evaluating this gives approximately
[tex]$$
df\approx 109.5487.
$$[/tex]
Step 5. Compute the p–value.
Since the alternative hypothesis is [tex]$H_a:\mu_1<\mu_2$[/tex], we need the one–tailed p–value:
[tex]$$
p\text{-value}=P(T < t) \quad \text{where} \quad T\sim t_{df}.
$$[/tex]
With [tex]$t\approx -0.2289$[/tex] and [tex]$df\approx109.5487$[/tex], the resulting p–value is
[tex]$$
p\text{-value}\approx 0.4097.
$$[/tex]
Thus, the final answers (rounded appropriately) are:
Test statistic [tex]$= -0.229$[/tex]
[tex]$$
p\text{-value}= 0.4097.
$$[/tex]
Since the p–value exceeds the significance level of [tex]$0.01$[/tex], we would not reject the null hypothesis that the two population means are equal.
[tex]$$
H_0: \mu_1=\mu_2 \quad \text{vs.} \quad H_a: \mu_1<\mu_2
$$[/tex]
at a significance level of [tex]$\alpha=0.01$[/tex]. Given the two samples (Sample #1 and Sample #2), we use a twoâ€sample [tex]$t$[/tex]–test that does not assume equal variances (often called Welch’s [tex]$t$[/tex]–test).
Step 1. Compute the sample statistics.
For Sample #1 we have:
- Sample size: [tex]$n_1=56$[/tex]
- Sample mean: [tex]$\bar{x}_1\approx 77.1768$[/tex]
- Sample variance: [tex]$s_1^2\approx 144.9905$[/tex] (with sample standard deviation [tex]$s_1\approx 12.0412$[/tex])
For Sample #2 we have:
- Sample size: [tex]$n_2=57$[/tex]
- Sample mean: [tex]$\bar{x}_2\approx 77.7333$[/tex]
- Sample variance: [tex]$s_2^2\approx 189.4001$[/tex] (with sample standard deviation [tex]$s_2\approx 13.7623$[/tex])
Step 2. Compute the standard error of the difference in means.
The standard error ([tex]$SE$[/tex]) is given by
[tex]$$
SE = \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}.
$$[/tex]
Substituting the values we obtain
[tex]$$
SE\approx 2.4314.
$$[/tex]
Step 3. Calculate the test statistic.
The test statistic for the difference in means is
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{SE}.
$$[/tex]
Substituting the values:
[tex]$$
t = \frac{77.1768 - 77.7333}{2.4314} \approx -0.2289.
$$[/tex]
Step 4. Determine the degrees of freedom.
For Welch’s [tex]$t$[/tex]–test the degrees of freedom are approximated by
[tex]$$
df = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]
Evaluating this gives approximately
[tex]$$
df\approx 109.5487.
$$[/tex]
Step 5. Compute the p–value.
Since the alternative hypothesis is [tex]$H_a:\mu_1<\mu_2$[/tex], we need the one–tailed p–value:
[tex]$$
p\text{-value}=P(T < t) \quad \text{where} \quad T\sim t_{df}.
$$[/tex]
With [tex]$t\approx -0.2289$[/tex] and [tex]$df\approx109.5487$[/tex], the resulting p–value is
[tex]$$
p\text{-value}\approx 0.4097.
$$[/tex]
Thus, the final answers (rounded appropriately) are:
Test statistic [tex]$= -0.229$[/tex]
[tex]$$
p\text{-value}= 0.4097.
$$[/tex]
Since the p–value exceeds the significance level of [tex]$0.01$[/tex], we would not reject the null hypothesis that the two population means are equal.
Thank you for reading the article You wish to test the following claim tex H a tex at a significance level of tex alpha 0 01 tex begin array l H. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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