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Answer :
We wish to test
[tex]$$
H_0: \mu_1 = \mu_2 \quad \text{versus} \quad H_a: \mu_1 < \mu_2,
$$[/tex]
using a significance level of [tex]$\alpha=0.01$[/tex]. Two independent samples were collected with the following summary statistics:
- For Sample 1 (with [tex]$n_1=56$[/tex] observations):
- Sample mean: [tex]$\bar{x}_1 \approx 77.177$[/tex]
- Sample standard deviation: [tex]$s_1 \approx 12.041$[/tex]
- For Sample 2 (with [tex]$n_2=57$[/tex] observations):
- Sample mean: [tex]$\bar{x}_2 \approx 77.733$[/tex]
- Sample standard deviation: [tex]$s_2 \approx 13.762$[/tex]
Because the two standard deviations are not assumed to be equal, we use the Welch’s [tex]$t$[/tex]-test. The test statistic is given by
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}.
$$[/tex]
Substituting the computed values, we obtain a test statistic of
[tex]$$
t \approx -0.229.
$$[/tex]
This negative value is consistent with our alternative hypothesis [tex]$H_a: \mu_1 < \mu_2$[/tex] (meaning Sample 1 has a smaller mean than Sample 2).
The degrees of freedom for the Welch test are approximated by the Welch–Satterthwaite formula:
[tex]$$
\nu \approx \frac{\left(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}\right)^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}},
$$[/tex]
which in this case evaluates approximately to [tex]$\nu \approx 109.549$[/tex].
Using the computed [tex]$t$[/tex]-value and the degrees of freedom, the two-tailed [tex]$p$[/tex]-value is found to be approximately [tex]$0.8194$[/tex]. However, because the alternative hypothesis is one-sided (specifically testing if [tex]$\mu_1 < \mu_2$[/tex]), we use half the two-tailed [tex]$p$[/tex]-value when the test statistic is negative. Therefore, the one-sided [tex]$p$[/tex]-value is
[tex]$$
p\text{-value} \approx \frac{0.8194}{2} \approx 0.4097.
$$[/tex]
Thus, the final answers are:
[tex]$$
\text{Test statistic} = -0.229 \quad \text{(to three decimal places)},
$$[/tex]
[tex]$$
p\text{-value} = 0.4097 \quad \text{(to four decimal places)}.
$$[/tex]
Since the [tex]$p$[/tex]-value is much larger than [tex]$\alpha=0.01$[/tex], there is insufficient evidence to conclude that [tex]$\mu_1 < \mu_2$[/tex].
[tex]$$
H_0: \mu_1 = \mu_2 \quad \text{versus} \quad H_a: \mu_1 < \mu_2,
$$[/tex]
using a significance level of [tex]$\alpha=0.01$[/tex]. Two independent samples were collected with the following summary statistics:
- For Sample 1 (with [tex]$n_1=56$[/tex] observations):
- Sample mean: [tex]$\bar{x}_1 \approx 77.177$[/tex]
- Sample standard deviation: [tex]$s_1 \approx 12.041$[/tex]
- For Sample 2 (with [tex]$n_2=57$[/tex] observations):
- Sample mean: [tex]$\bar{x}_2 \approx 77.733$[/tex]
- Sample standard deviation: [tex]$s_2 \approx 13.762$[/tex]
Because the two standard deviations are not assumed to be equal, we use the Welch’s [tex]$t$[/tex]-test. The test statistic is given by
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}.
$$[/tex]
Substituting the computed values, we obtain a test statistic of
[tex]$$
t \approx -0.229.
$$[/tex]
This negative value is consistent with our alternative hypothesis [tex]$H_a: \mu_1 < \mu_2$[/tex] (meaning Sample 1 has a smaller mean than Sample 2).
The degrees of freedom for the Welch test are approximated by the Welch–Satterthwaite formula:
[tex]$$
\nu \approx \frac{\left(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}\right)^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}},
$$[/tex]
which in this case evaluates approximately to [tex]$\nu \approx 109.549$[/tex].
Using the computed [tex]$t$[/tex]-value and the degrees of freedom, the two-tailed [tex]$p$[/tex]-value is found to be approximately [tex]$0.8194$[/tex]. However, because the alternative hypothesis is one-sided (specifically testing if [tex]$\mu_1 < \mu_2$[/tex]), we use half the two-tailed [tex]$p$[/tex]-value when the test statistic is negative. Therefore, the one-sided [tex]$p$[/tex]-value is
[tex]$$
p\text{-value} \approx \frac{0.8194}{2} \approx 0.4097.
$$[/tex]
Thus, the final answers are:
[tex]$$
\text{Test statistic} = -0.229 \quad \text{(to three decimal places)},
$$[/tex]
[tex]$$
p\text{-value} = 0.4097 \quad \text{(to four decimal places)}.
$$[/tex]
Since the [tex]$p$[/tex]-value is much larger than [tex]$\alpha=0.01$[/tex], there is insufficient evidence to conclude that [tex]$\mu_1 < \mu_2$[/tex].
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 tex begin array r. 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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Rewritten by : Jeany
