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Answer :
We wish to test
[tex]$$
\begin{array}{l}
H_0:\ \mu_1=\mu_2 \\
H_a:\ \mu_1>\mu_2
\end{array}
$$[/tex]
at a significance level of [tex]$\alpha=0.005$[/tex]. Two independent samples are available. In this case the two-sample t-test (assuming unequal variances, i.e. Welch’s t-test) is used. Here is how the calculations are carried out step by step.
1. First, the data are divided into the two samples. In each row of the provided table, the first two numbers form Sample \#1 and the last two numbers form Sample \#2. In the end, both samples have 26 observations.
2. The sample means and sample standard deviations are computed. They are found to be
[tex]$$
\bar{x}_1 = 75.0385,\quad \bar{x}_2 = 77.8500,
$$[/tex]
[tex]$$
s_1 = 18.5157,\quad s_2 = 19.1579.
$$[/tex]
3. The standard error (SE) for the difference in means is given by
[tex]$$
SE = \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }.
$$[/tex]
Substituting the values,
[tex]$$
SE = \sqrt{ \frac{(18.5157)^2}{26} + \frac{(19.1579)^2}{26} } \approx 5.2251.
$$[/tex]
4. The test statistic is computed using
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{SE}.
$$[/tex]
Substitute the values:
[tex]$$
t = \frac{75.0385 - 77.8500}{5.2251} \approx -0.538.
$$[/tex]
5. The degrees of freedom for Welch’s t-test are calculated using the Welch–Satterthwaite equation (the computed value is approximately 49.942). Although the exact degrees of freedom are used in software, we note here for completeness that the degrees of freedom can be calculated as
[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]
6. Since the alternative hypothesis is [tex]$H_a:\ \mu_1>\mu_2$[/tex], the p-value is the probability of seeing a test statistic as large or larger than the observed value assuming the null hypothesis is true. However, because our test statistic turns out to be negative (i.e. [tex]$t\approx -0.538$[/tex]), it falls in the opposite direction than expected under [tex]$H_a$[/tex]. Consequently, the p-value is given by
[tex]$$
\text{p-value} = 1 - F(t),
$$[/tex]
where [tex]$F(t)$[/tex] is the cumulative distribution function (CDF) of the [tex]$t$[/tex]-distribution with the calculated degrees of freedom. For [tex]$t \approx -0.538$[/tex], the upper tail probability turns out to be approximately
[tex]$$
\text{p-value} \approx 0.7035.
$$[/tex]
Thus, the final answers are:
[tex]$$
\text{test statistic} = -0.538
$$[/tex]
[tex]$$
\text{p-value} = 0.7035.
$$[/tex]
Since the p-value is much higher than the significance level ([tex]$0.7035 > 0.005$[/tex]), there is not enough evidence to reject [tex]$H_0$[/tex].
[tex]$$
\begin{array}{l}
H_0:\ \mu_1=\mu_2 \\
H_a:\ \mu_1>\mu_2
\end{array}
$$[/tex]
at a significance level of [tex]$\alpha=0.005$[/tex]. Two independent samples are available. In this case the two-sample t-test (assuming unequal variances, i.e. Welch’s t-test) is used. Here is how the calculations are carried out step by step.
1. First, the data are divided into the two samples. In each row of the provided table, the first two numbers form Sample \#1 and the last two numbers form Sample \#2. In the end, both samples have 26 observations.
2. The sample means and sample standard deviations are computed. They are found to be
[tex]$$
\bar{x}_1 = 75.0385,\quad \bar{x}_2 = 77.8500,
$$[/tex]
[tex]$$
s_1 = 18.5157,\quad s_2 = 19.1579.
$$[/tex]
3. The standard error (SE) for the difference in means is given by
[tex]$$
SE = \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }.
$$[/tex]
Substituting the values,
[tex]$$
SE = \sqrt{ \frac{(18.5157)^2}{26} + \frac{(19.1579)^2}{26} } \approx 5.2251.
$$[/tex]
4. The test statistic is computed using
[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{SE}.
$$[/tex]
Substitute the values:
[tex]$$
t = \frac{75.0385 - 77.8500}{5.2251} \approx -0.538.
$$[/tex]
5. The degrees of freedom for Welch’s t-test are calculated using the Welch–Satterthwaite equation (the computed value is approximately 49.942). Although the exact degrees of freedom are used in software, we note here for completeness that the degrees of freedom can be calculated as
[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]
6. Since the alternative hypothesis is [tex]$H_a:\ \mu_1>\mu_2$[/tex], the p-value is the probability of seeing a test statistic as large or larger than the observed value assuming the null hypothesis is true. However, because our test statistic turns out to be negative (i.e. [tex]$t\approx -0.538$[/tex]), it falls in the opposite direction than expected under [tex]$H_a$[/tex]. Consequently, the p-value is given by
[tex]$$
\text{p-value} = 1 - F(t),
$$[/tex]
where [tex]$F(t)$[/tex] is the cumulative distribution function (CDF) of the [tex]$t$[/tex]-distribution with the calculated degrees of freedom. For [tex]$t \approx -0.538$[/tex], the upper tail probability turns out to be approximately
[tex]$$
\text{p-value} \approx 0.7035.
$$[/tex]
Thus, the final answers are:
[tex]$$
\text{test statistic} = -0.538
$$[/tex]
[tex]$$
\text{p-value} = 0.7035.
$$[/tex]
Since the p-value is much higher than the significance level ([tex]$0.7035 > 0.005$[/tex]), there is not enough evidence to reject [tex]$H_0$[/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 0 005 tex tex begin array l. 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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